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School District of Palm Beach County Summer Packet Algebra EOC Review Summer 2013
Students and Parents, This Summer Packet for Algebra 1 EOC Review is designed to provide an opportunity to review and remediate skills in preparation to retake the Algebra 1 End‐Of‐Course exam. These materials include instruction and problem solving on each worksheet. The focus of each selected worksheet is a skill designed to prepare students for the Algebra 1 End‐Of‐Course exam. The source of the worksheets is the Pearson‐Prentice Hall Algebra 1 Textbook series. All of the contents of this packet have been copied with permission. We hope you are able to utilize the resources included in this packet to make your summer both educational as well as relaxing. Thank you! Name
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Variables and Expressions
You can represent mathematical phrases
and real-world relationships using symbols
and operations. This is called an algebraic
expression.
For example, the phrase 3 plus a number n
can be expressed using symbols and
operations as 3 1 n.
Problem
What is the phrase 5 minus a number d as an algebraic expression?
5
minus
a number d
5
2
d
The phrase 5 minus a number d, rewritten as an algebraic expression, is 5 2 d.
The left side of the table below gives some common phrases used to express
mathematical relationships, and the right side of the table gives the related
symbol.
Phrase
Symbol
sum
1
difference
2
product
3
quotient
4
less than
2
more than
1
Exercises
Write an algebraic expression for each word phrase.
1. 5 plus a number d
2. the product of 5 and g
3. 11 fewer than a number f
4. 17 less than h
5. the quotient of 20 and t
6. the sum of 12 and 4
Write a word phrase for each algebraic expression.
7. h 1 6
35
10. r
8. m 2 5
9. q 3 10
11. h 1 m
12. 5n
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Variables and Expressions
Multiple operations can be combined into a single phrase.
Problem
What is the phrase 11 minus the product of 3 and d as an algebraic expression?
11
minus
the product of 3 and a number d
11
2
33d
The phrase 11 minus the product of 3 and a number d, rewritten as an algebraic
expression, is 11 – 3d.
Exercises
Write an algebraic expression for each phrase.
13. 12 less than the quotient of 12 and a number z
14. 5 greater than the product of 3 and a number q
15. the quotient of 5 1 h and n 1 3
22
16. the difference of 17 and t
Write an algebraic expression or equation to model the relationship expressed
in each situation below.
17. Jane is building a model boat. Every inch on her model is equivalent to 3.5 feet
on the real boat her model is based on. What would be the mathematical rule
to express the relationship between the length of the model, m, and the length
of the boat, b?
18. Lyn is putting away savings for his college education. Every time Lyn puts
money in his fund, his parents put in $2. What is the expression for the
amount going into Lyn’s fund if Lyn puts in L dollars?
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Order of Operations and Evaluating Expressions
Exponents are used to represent repeated multiplication of the same number. For
example, 4 3 4 3 4 3 4 3 4 5 45 . The number being multiplied by itself is
called the base; in this case, the base is 4. The number that shows how many times
the base appears in the product is called the exponent; in this case, the exponent
is 5. 45 is read four to the fifth power.
Problem
How is 6 3 6 3 6 3 6 3 6 3 6 3 6 written using an exponent?
The number 6 is multiplied by itself 7 times. This means that the base is 6 and the
exponent is 7. 6 3 6 3 6 3 6 3 6 3 6 3 6 written using an exponent is 67 .
Exercises
Write each repeated multiplication using an exponent.
1. 4 3 4 3 4 3 4 3 4
2. 2 3 2 3 2
3. 1.1 3 1.1 3 1.1 3 1.1 3 1.1
4. 3.4 3 3.4 3 3.4 3 3.4 3 3.4 3 3.4
5. (27) 3 (27) 3 (27) 3 (27)
6. 11 3 11 3 11
Write each expression as repeated multiplication.
7. 43
8. 54
24
10. Q 7 R
9. 1.52
11. x7
12. (5n)5
13. Trisha wants to determine the volume of a cube with sides of length s. Write
an expression that represents the volume of the cube.
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Order of Operations and Evaluating Expressions
The order of operations is a set of guidelines that make it possible to be sure that
two people will get the same result when evaluating an expression. Without this
standard order of operations, two people might evaluate an expression differently
and arrive at different values. For example, without the order of operations,
someone might evaluate all expressions from left to right, while another person
performs all additions and subtractions before all multiplications and divisions.
You can use the acronym P.E.M.A. (Parentheses, Exponents, Multiplication and
Division, and Addition and Subtraction) to help you remember the order of
operations.
Problem
How do you evaluate the expression 3 1 4 3 2 2 10 4 5?
3 1 8 2 10 4 5
531822
5 11 2 2
5 9
There are no parentheses or exponents, so first,
do any multiplication or division from left to right.
Do any addition or subtraction from left to right.
Exercises
Simplify each expression.
14. (5 1 3)2
15. (8 2 5)(14 2 6)
16. (15 2 3) 4 4
17. Q
18. 40 2 15 4 3
19. 20 1 12 4 2 2 5
20. (42 1 52)2
21. 4 3 5 2 32 3 2 4 6
22 1 3
5 R
Write and simplify an expression to model the relationship expressed in the
situation below.
22. Manuela has two boxes. The larger of the two boxes has dimensions of 15 cm
by 25 cm by 20 cm. The smaller of the two boxes is a cube with sides that are 10
cm long. If she were to put the smaller box inside the larger, what would be the
remaining volume of the larger box?
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Real Numbers and the Number Line
A number that is the product of some other number with itself, or a number to the
second power, such as 9 5 3 3 3 5 32 , is called a perfect square. The number
that is raised to the second power is called the square root of the product. In this
case, 3 is the square root of 9. This is written in symbols as !9 5 3. Sometimes
square roots are whole numbers, but in other cases, they can be estimated.
Problem
What is an estimate for the square root of 150?
There is no whole number that can be multiplied by itself to give the product of 150.
10 3 10 5 100
11 3 11 5 121
12 3 12 5 144
13 3 13 5 169
You cannot find the exact value of !150, but you can estimate it by comparing
150 to perfect squares that are close to 150.
150 is between 144 and 169, so !150 is between !144 and !169.
!144 , !150 , !169
12 , !150 , 13
The square root of 150 is between 12 and 13. Because 150 is closer to 144 than it is
to 169, we can estimate that the square root of 150 is slightly greater than 12.
Exercises
Find the square root of each number. If the number is not a perfect square,
estimate the square root to the nearest integer.
1. 100
2. 49
3. 9
4. 25
5. 81
6. 169
7. 15
8. 24
9. 40
10. A square mat has an area of 225 cm2 . What is the length of each side of the
mat?
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Real Numbers and the Number Line
The real numbers can be separated into smaller, more specific groups, called
subsets. Each of these subsets has certain characteristics. For example, a rational
number can be expressed as a fraction of two integers, with the denominator of
the fraction not equal to 0. Irrational numbers cannot be expressed as a fraction of
two integers.
Every real number belongs to at least one subset of the real numbers. Some real
numbers belong to multiple subsets.
Problem
To which subsets of the real numbers does 17 belong?
17 is a natural number, a whole number, and an integer.
17
But 17 is also a rational number because it can be written as 1 , a fraction of two
integers with the denominator not equal to 0.
A number cannot belong to both the subset of rational numbers and the subset of
irrational numbers, so 17 is not an irrational number.
Exercises
List the subsets of the real numbers to which each of the given numbers
belongs.
11. 5
12. 116
13. !3
14. 17.889
15. 225
16. 268
217
17. 20
18. 0
19. !16
20. !20
21. !6.25
77
22. 10
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Properties of Real Numbers
Equivalent algebraic expressions are expressions that have the same value for all
values for the variable(s). For example x 1 x and 2x are equivalent expressions
since, regardless of what number is substituted in for x, simplifying each
expression will result in the same value. Certain properties of real numbers lead to
the creation of equivalent expressions.
Commutative Properties
The commutative properties of addition and multiplication state that changing the
order of the addends does not change the sum and that changing the order of factors
does not change the product.
Addition: a 1 b 5 b 1 a
Multiplication: a ? b 5 b ? a
To help you remember the commutative properties, you can think about the root
word “commute.” To commute means to move. If you think about commuting or
moving when you think about the commutative properties, you will remember
that the addends or factors move or change order.
Problem
Do the following equations illustrate commutative properties?
a. 3 1 4 5 4 1 3
b. (5 3 3) 3 2 5 5 3 (3 3 2)
c. 1 2 3 5 3 2 1
3 1 4 and 4 1 3 both simplify to 7, so the two sides of the equation in part (a) are
equal. Since both sides have the same two addends but in a different order, this
equation illustrates the Commutative Property of Addition.
The expression on each side of the equation in part (b) simplifies to 30. Both
sides contain the same 3 factors. However, this equation does not illustrate the
Commutative Property of Multiplication because the terms are in the same order
on each side of the equation.
1 2 3 and 3 2 1 do not have the same value, so the equation in part (c) is
not true. There is not a commutative property for subtraction. Nor is there a
commutative property for divison.
Associative Properties
The associative properties of addition and multiplication state that changing the
grouping of addends does not change the sum and that changing the grouping of
factors does not change the product.
Addition: (a 1 b) 1 c 5 a 1 (b 1 c)
Multiplication: (a ? b) ? c 5 a ? (b ? c)
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Properties of Real Numbers
Problem
Do the following equations illustrate associative properties?
a. (1 1 5) 1 4 5 1 1 (5 1 4)
b. 4 3 (2 3 7) 5 4 3 (7 3 2)
(1 1 5) 1 4 and 1 1 (5 1 4) both simplify to 10, so the two sides of the equation
in part (a) are equal. Since both sides have the same addends in the same order but
grouped differently, this equation illustrates the Associative Property of Addition.
The expression on each side of the equation in part (b) simplifies to 56. Both sides
contain the same 3 factors. However, the same factors that were grouped together
on the left side have been grouped together on the right side; only the order has
changed. This equation does not illustrate the Associative Property of Multiplication.
Other properties of real numbers include:
a. Identity property of addition:
b. Identity property of multiplication:
c. Zero property of multiplication:
d. Multiplicative property of negative one:
a1050
a?15a
a?050
21 ? a 5 2a
12 1 0 5 12
32 ? 1 5 32
6?050
21 ? 7 5 27
Exercises
What property is illustrated by each statement?
1. (m 1 7.3) 1 4.1 5 m 1 (7.3 1 4.1)
2. 5p ? 1 5 5p
3. 12x 1 4y 1 0 5 12x 1 4y
4. (3r)(2s) 5 (2s)(3r)
5. 17 1 (22) 5 (22) 1 17
6. 2(23) 5 3
Simplify each expression. Justify each step.
7. (12 1 8x) 1 13
8. (5 ? m) ? 7
9. (7 2 7) 1 12
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Adding and Subtracting Real Numbers
You can add real numbers using a number line or using the following rules.
Rule 1: To add two numbers with the same sign, add their absolute values. The
sum has the same sign as the addends.
Problem
What is the sum of 27 and 24?
Use a number line.
21121029 28 27 26 25 24 23 22 21
0
Start at zero.
Move 7 spaces to the left to represent 27.
Move another 4 spaces to the left to
represent 24.
1
The sum is –11.
Use the rule.
27 1 (24)
The addends are both negative.
|27| 1 |24|
Add the absolute values of the addends.
7 1 4 5 11
|27| 5 7 and |24| 5 4.
27 1 (24) 5 211
The sum has the same sign as the addends.
Rule 2: To add two numbers with different signs, subtract their absolute values.
The sum has the same sign as the addend with the greater absolute value.
Problem
What is the sum of 26 and 9?
Use the rule.
9 1 (26)
The addends have different signs.
|9| 2 |26|
Subtract the absolute values of the addends.
92653
|9| 5 9 and |26| 5 6.
9 1 (26) 5 3
The positive addend has the greater absolute value.
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1-5
Adding and Subtracting Real Numbers
Exercises
Find each sum.
1. 24 1 212
2. 23 1 15
3. 29 1 1
4. 13 1 (27)
5. 8 1 (214)
6. 211 1 (25)
7. 4.5 1 (21.1)
8. 25.1 1 8.3
9. 6.4 1 9.8
Addition and subtraction are inverse operations. To subtract a real number,
add its opposite.
Problem
What is the difference 25 2 (28)?
25 2 (28) 5 25 1 8
The opposite of 28 is 8.
53
Use Rule 2.
The difference 25 2 (28) is 3.
Exercises
Find each difference.
10. 8 2 20
11. 6 2 (212)
12. 24 2 9
13. 28 2 (214)
14. 211 2 (24)
15. 17 2 25
16. 3.6 2 (22.4)
17. 21.5 2 (21.5)
18. 21.7 2 5.4
19. The temperature was 58C. Five hours later, the temperature had dropped
108C. What is the new temperature?
20. Reasoning Which is greater, 52 1 (277) or 52 2 (277)? Explain.
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Multiplying and Dividing Real Numbers
You need to remember two simple rules when multiplying or dividing real
numbers.
1. The product or quotient of two numbers with the same sign is positive.
2. The product or quotient of two numbers with different signs is negative.
Problem
What is the product –6(–30)?
26(230) 5 180
26 and 230 have the same sign so the product is positive.
Problem
What is the quotient 72 4 (26)?
72 4 (26) 5 212
72 and 26 have different signs so the quotient is negative.
Exercises
Find each product or quotient.
1. 25(26)
2. 7(220)
3. 23 3 22
4. 44 4 2
5. 81 4 (29)
6. 255 4 (211)
7. 262 4 2
8. 25 ? (24)
9. (26)2
10. 29.9 4 3
11. 27.7 4 (211)
12. 21.4(22)
1
1
13. 22 3 3
2 3
14. 23 Q 25 R
3
1
15. 4 ? Q 23 R
16. The temperature dropped 2°F each hour for 6 hours. What was the total
change in temperature?
17. Reasoning Since 52 5 25 and (25)2 5 25, what are the two values for the
square root of 25?
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Multiplying and Dividing Real Numbers
The product of 7 and 17 is 1. Two numbers whose product is 1 are called
reciprocals. To divide a number by a fraction, multiply by its reciprocal.
Problem
5
What is the quotient 23 4 Q 27 R ?
5
7
2
2
3 4 Q 27 R 5 3 3 Q 25 R
5 214
15
To divide by a fraction, multiply by its reciprocal.
The signs are different so the answer is negative.
Exercises
Find each quotient.
1
1
18. 2 4 3
2
19. 26 4 3
2
2
20. 25 4 Q 23 R
1
1
21. 2 4 Q 24 R
5
1
22. Q 27 R 4 Q 22 R
2
1
23. 23 4 4
1
2
a
24. Writing Another way of writing is a 4 b. Explain how you could evaluate 1 .
b
6
What is the value of this expression?
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The Distributive Property
The Distributive Property states that the product of a sum and another factor can be
rewritten as the sum of two products, each term in the sum multiplied by the other
factor. For example, the Distributive Property can be used to rewrite the product
3(x 1 y) as the sum 3x 1 3y. Each term in the sum x 1 y is multiplied by 3; then
the new products are added.
Problem
What is the simplified form of each expression?
a. 4(x 1 5)
b. (2x 2 3)(23)
5 4(x) 1 4(5) Distributive Property
5 4x 1 20
Simplify.
5 2x(23) 2 3(23) Distributive Property
5 26x 1 9
Simplify.
The Distributive Property can be used whether the factor being multiplied by a sum
or difference is on the left or right.
The Distributive Property is sometimes referred to as the Distributive Property of
Multiplication over Addition. It may be helpful to think of this longer name for the
property, as it may remind you of the way in which the operations of multiplication
and addition are related by the property.
Exercises
Use the Distributive Property to simplify each expression.
1. 6(z 1 4)
2. 2(22 2 k)
3. (5x 1 1)4
4. (7 2 11n)10
5. (3 2 8w)4.5
6. (4p 1 5)2.6
7. 4(y 1 4)
8. 6(q 2 2)
Write each fraction as a sum or difference.
9.
2m 2 5
9
10.
8 1 7z
11
11.
24f 1 15
9
12.
12d 2 16
6
Simplify each expression.
13. 2(6 1 j)
14. 2(29h 2 4)
15. 2(2n 1 11)
16. 2(6 2 8f)
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The Distributive Property
The previous problem showed how to write a product as a sum using the Distributive
Property. The property can also be used to go in the other order, to convert a sum
into a product.
Problem
How can the sum of like terms 15x 1 6x be simplified using the Distributive Property?
Each term of 15x 1 6x has a factor of x. Rewrite 15x 1 6x as 15(x) 1 6(x).
Now use the Distributive Property in reverse to write 15(x) 1 6(x) as (15 1 6)x,
which simplifies to 21x.
Exercises
Simplify each expression by combining like terms.
17. 16x 1 12x
18. 25n 2 17n
19. 24p 1 6p
20. 215a 2 9a
21. 29k2 2 5k2
22. 12t2 2 20t2
By thinking of or rewriting numbers as sums or differences of other numbers that are easier
to use in multiplication, the Distributive Property can be used to make calculations easier.
Problem
How can you multiply 78 by 101 using the Distributive Property and mental math?
78 3 101
78 3 (100 1 1)
78(100) 1 78(1)
7800 1 78
7878
Write the product.
Rewrite 101 as sum of two numbers that are easy to use in multiplication.
Use the Distributive Property to write the product as a sum.
Multiply.
Simplify.
Exercises
Use mental math to fintd each product.
23. 5.1 3 7
24. 24.95 3 4
25. 999 3 11
26. 12 3 95
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An Introduction to Equations
An equation is a mathematical sentence with an equal sign. An equation can be true, false,
or open. An equation is true if the expressions on both sides of the equal sign are equal, for
example 2 1 5 5 4 1 3. An equation is false if the expressions on both sides of the equal
sign are not equal, for example 2 1 5 5 4 1 2.
An equation is considered open if it contains one or more variables, for example
x 1 2 5 8. When a value is substituted for the variable, you can then decide whether the
equation is true or false for that particular value. If an open sentence is true for a value of
the variable, that value is called a solution of the equation. For x 1 2 5 8, 6 is a solution
because when 6 is substituted in the equation for x, the equation is true: 6 1 2 5 8.
Problem
Is the equation true, false, or open? Explain.
a. 15 1 21 5 30 1 6
b. 24 4 8 5 2 ? 2
c. 2n 1 4 5 12
The equation is true, because both expressions equal 36.
The equation is false, because 24 4 8 5 3 and 2 ? 2 5 4; 3 2 4.
The equation is open, because there is a variable in the expression
on the left side.
Tell whether each equation is true, false, or open. Explain.
1. 2(12) 2 3(6) 5 12
2. 3x 1 12 5 219
3. 14 2 19 5 25
4. 2(28) 1 4 5 12
5. 7 2 9 1 3 5 x
6. (22 1 12) 4 22 5 220 1 3
7. 14 2 (28) 2 14 5 8
8. (13 2 16) 4 3 5 1
9. 42 4 27 1 3 5 (3)(24) 1 9
Problem
Is x 5 23 a solution of the equation 4x 1 5 5 27?
4x 1 5 5 27
4(23) 1 5 5 27
27 5 27
Substitute 23 for x.
Simplify.
Since 27 5 27, 23 is a solution of the equation 4x 1 5 5 27.
Tell whether the given number is a solution of each equation.
10. 4x 2 1 5 227; 27
11. 18 2 2n 5 14; 2
12. 21 5 3p 2 5; 9
13. k 5 (26)(28) 2 14; 262
14. 20v 1 36 5 2156; 26
15. 8y 1 13 5 21; 1
16. 224 2 17t 5 258; 2
1
17. 226 5 3 m 1 5; 27
3
1
18. 4 g 2 8 5 2 ; 38
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An Introduction to Equations
A table can be used to find or estimate a solution of an open equation. You will have to
choose a value to begin your table. If you choose the value that makes the equation true,
you have found the solution and are done. If your choice is not the solution, make another
choice based on the values of both sides of the equation for your first choice. If you choose
one value that makes one side of the equation too high and then another value that makes
that same side too low, you know that the solution must lie between the two values you
chose. It may not be possible to determine an exact solution for each equation; estimating
the solution to be between two integers may be all that is possible in some cases.
Problem
What is the solution of 6n 1 8 5 28?
If n 5 2, then the left side of the equation is 6(2) 1 8 or 20, which is too low.
If n 5 5, then the left side of the equation is 6(5) 1 8 or 38, which is too high.
The solution must lie between 2 and 5, so keep trying values between them.
If n 5 3, then the left side of the equation is 6(3) 1 8 or 26, which is too low.
If n 5 4, then the left side of the equation is 6(4) 1 8 or 32, which is too high.
The solution must lie between 3 and 4, but there are no other integers between 3 and 4.
You can give an estimate for the solution of 6n 1 8 5 28 as being between the
integers 3 and 4.
Write an equation for each sentence.
19. 13 times the sum of a number and 5 is 91.
20. Negative 8 times a number minus 15 is equal to 30.
21. Jared receives $23 for each lawn he mows. What is an equation that relates the
number of lawns w that Jared mows and his pay p?
22. Shariff has been working for a company 2 years longer than Patsy. What is an equation
that relates the years of employment of Shariff S and the years of employment of Patsy P?
Use mental math to find the solution of each equation.
23. h 1 6 5 13
24. 211 5 n 1 2
25. 6 2 k 5 14
26. 5 5 28 1 t
z
27. 5 5 22
j
28. 26 5 12
29. 8c 5 248
30. 215a 5 245
Use a table to find the solution of each equation.
31. 23b 2 12 5 15
32. 15y 1 6 5 21
33. 28 5 5y 1 22
34. 6t 2 1 5 249
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Patterns, Equations, and Graphs
Tables, equations, and graphs are some of the ways that a relationship between
two quantities can be represented. You can use the information provided by one
representation to produce one of the other representations; for example, you can use
data from a table to produce a graph. You can also use any of the representations to draw
conclusions about the relationship.
Problem
Are (2, 11) and (5, 3) solutions of the equation y 5 3x 1 5?
For each ordered pair, you can substitute the x- and y- coordinates into the equation for
x and y and then simplify to see if the values satisfy the equation.
For (5, 3):
For (2, 11):
11 5 3(2) 1 5
11 5 11
3 5 3(5) 1 5
3 2 20
Substitute for x and y.
Multiply and then add.
Since both sides of the equation have the same value, the ordered pair (2, 11)
is a solution of the equation y 5 3x 1 5. Since the two sides of the equation
have different values, the ordered pair (5, 3) is not a solution of the equation
y 5 3x 1 5.
Problem
The table shows the relationship between the number of hours Kaya
works at her job and the amount of pay she receives. Extend the
pattern. How much money would Kaya earn if she worked 40 hours?
Hours
Worked
Money
Earned ($)
3
37.50
Method 1: Write an equation.
6
75
y 5 12.50x
5 12.50(40)
9
5 500
Kaya earns $12.50 per hour.
Substitute 40 for x.
112.50
150
12
Simplify.
She would earn $500 in 40 hours.
She would earn $500 in
40 hours.
Money Earned ($)
Method 2: Draw a graph.
500 y
450
400
350
300
250
200
150
100
50
O
x
4
8
12
16
20
Hours worked
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28
32
36
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Reteaching (continued)
Patterns, Equations, and Graphs
Exercises
Tell whether the equation has the given ordered pair as a solution.
1. y 5 x 2 7; (2, 25)
2. y 5 x 1 6; (25, 11)
3. y 5 2x 1 1; (21, 0)
4. y 5 25x; (23, 215)
5. y 5 x 2 8; (7, 21)
3
1
6. y 5 x 1 4; (21, 24 )
Use a table, an equation, and a graph to represent each relationship.
7. Tickets to the fair cost $17.
8. Brian is 5 years older than Sam.
Use the table to draw a graph and answer the question.
Cakes
Earnings
($)
5
120
10
240
15
360
9. The table shows Jake’s earnings for the number of cakes he
baked. What are his earnings for baking 75 cakes?
Use the table to write an equation and answer the question.
10. The table shows the number
11. The table shows the amount of
of miles that Kate runs on a
weekly basis while training for a
race. How many total miles will
she have run after 15 weeks?
money Kevin receives for
items that he sells. How much
will he earn if he sells 30 items?
Items
Sold
Earnings
($)
15
1125
40
20
1500
2
80
25
1875
3
120
Training
Weeks
Miles
Run
1
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Solving One-Step Equations
You can use the properties of equality to solve equations. Subtraction is the
inverse of addition.
Problem
What is the solution of x 1 5 5 33?
In the equation, x 1 5 5 33, 5 is added to the variable. To solve the equation, you
need to isolate the variable, or get it alone on one side of the equal sign. Undo
adding 5 by subtracting 5 from each side of the equation.
Drawing a diagram can help you write an
equation to solve the problem.
Whole
Part
33
X
Part
5
x 1 5 5 33
Solve
x 1 5 2 5 5 33 2 5
Undo adding 5 by subtracting 5.
x 5 28
Simplify. This isolates x.
x 1 5 5 33
Check
Check your solution in the original equation.
28 1 5 0 33
Substitute 28 for x.
33 5 33 3
The solution to x 1 5 5 33 is 28.
Division is the inverse of multiplication.
Problem
x
What is the solution of 5 5 12?
x
In the equation, 5 5 12, the variable is divided by 5. Undo
12
12
X
12
dividing by 5 by multiplying by 5 on each side of the equation.
x
5 5 12
Solve
x
5 ? 5 5 12 ? 5
Undo dividing by 5 by multiplying by 5.
x 5 60
Simplify. This isolates x.
x
The solution to 5 5 12 is 60.
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Solving One-Step Equations
Exercises
Solve each equation using addition or subtraction. Check your answer.
1. 23 5 n 1 9
2. f 1 6 5 26
3. m 1 12 5 22
4. r 1 2 5 7
5. b 1 1.1 5 211
6. t 1 9 5 4
Define a variable and write an equation for each situation. Then solve.
7. A student is taking a test. He has 37 questions left. If the test has 78 questions,
how many questions has he finished?
8. A friend bought a bouquet of flowers. The bouquet had nine daisies and some
roses. There were a total of 15 flowers in the bouquet. How many roses were in
the bouquet?
Solve each equation using multiplication or division. Check your answer.
z
9. 8 5 2
a
12. 23 5 18
c
10. 226 5 13
q
11. 11 5 26
g
13. 225 5 5
s
14. 20.4 5 2.5
15. A student has been typing for 22 minutes and has typed a total of 1496 words.
Write and solve an equation to determine the average number of words she
can type per minute.
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Solving Two-Step Equations
Properties of equality and inverse operations can be used to solve equations that
involve more than one step to solve. To solve a two-step equation, identify the
operations and undo them using inverse operations. Undo the operations in the
reverse order of the order of operations.
Problem
What is the solution of 5x 2 8 5 32?
5x 2 8 1 8 5 32 1 8
To get the variable term alone on the left side, add 8 to each side.
5x 5 40
Simplify.
5x
40
5 5 5
Divide each side by 5 since x is being multiplied by 5 on the left
side. This isolates x.
x58
Check
Simplify.
5x 2 8 5 32
Check your solution in the original equation.
5(8) 2 8 5 32
Substitute 8 for x.
32 5 32 3
Simplify.
x
To solve 216 5 3 1 5, you can use subtraction first to undo the addition, and
then use multiplication to undo the division.
Problem
x
What is the solution of 216 5 3 1 5?
x
216 2 5 5 3 1 5 2 5
x
221 5 3
Simplify.
x
3(221) 5 3a 3 b
263 5 x
To get the variable term alone on the right, subtract 5
from each side.
Since x is being divided by 3, multiply each side by 3 to
undo the division. This isolates x.
Simplify.
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Solving Two-Step Equations
Solve each equation. Check your answer.
1. 4f 2 8 5 20
2. 25 2 6b 5 55
3. 2z 1 7 5 28
w
4. 29 1 7 5 10
n
5. 25 5 8 1 2
6.
y28
3 5 27
Solve each equation. Justify each step.
7. 6d 2 5 5 31
8.
p27
22 5 5
Define a variable and write an equation for each situation. Then solve.
9. Ray’s birthday is 8 more than four times the number of days away from today
than Jane’s birthday. If Ray’s birthday is 24 days from today, how many days
until Jane’s birthday?
10. Jerud weighs 15 pounds less than twice Kate’s weight. How much does Kate
weigh if Jerud weighs 205 pounds?
11. A phone company charges a flat fee of $17 per month, which includes free
local calling plus $0.08 per minute for long distance calls. The Taylor’s phone
bill for the month is $31.80. How many minutes of long distance calling did
they use during the month?
12. A delivery company charges a flat rate of $3 for a large envelope plus an
additional $0.25 per ounce for every ounce over a pound the package weighs.
The postage for the package is $5.50. How much does the package weigh?
(Hint: remember the first pound is included in the $3.)
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Solving Multi-Step Equations
To solve multi-step equations, use properties of equality, inverse operations, the
Distributive Property, and properties of real numbers to isolate the variable. Like
terms on either side of the equation should be combined first.
Problem
a) What is the solution of 23y 1 8 1 13y 5 252?
23y 1 13y 1 8 5 252
Group the terms with y together so that the like terms are
grouped together.
10y 1 8 5 252
Add the coefficients to combine like terms.
10y 1 8 2 8 5 252 2 8
10y 5 260
To get the variable term by itself on the left side, subtract 8
from each side.
Simplify.
10y
260
10 5 10
Divide each side by 10 since y is being multiplied by 10 on
the left side. This isolates y.
y 5 26
Simplify.
b) What is the solution of 22(3n 2 4) 5 210?
26n 1 8 5 210
Distribute the 22 into the parentheses by multiplying each
term inside by 22.
26n 1 8 2 8 5 210 2 8
To get the variable term by itself on the left side, subtract 8
from each side.
26n 5 218
Simplify.
26n
218
26 5 26
Divide each side by 26 since n is being multiplied by 26 on
the left side. This isolates n.
n53
Simplify.
Solve each equation. Check your answer.
1. 4 2 6h 2 8h 5 60
2. 232 5 27n 2 12 1 3n
3. 14 1 12 5 215x 1 2x
4. 8(23d 1 2) 5 88
5. 222 5 2(x 2 4)
6. 35 5 25(2k 1 5)
7. 3m 1 6 2 2m 5 222
8. 4(3r 1 2) 2 3r 5 210
9. 218 5 15 2 3(6t 1 5)
10. 25 1 2(10b 2 2) 5 31
11. 7 5 5x 1 3(x 2 2) 1 5
12. 218 5 3(2z 1 6) 1 2z
13. Reasoning Solve the equation 14 5 7(2x 2 4) using two different methods.
Show your work. Which method do you prefer? Explain.
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Solving Multi-Step Equations
Equations with fractions can be solved by using a common denominator or by
eliminating the fractions altogether.
Problem
x
7
What is the solution of 4 2 23 5 12 ?
Method 1
Method 2
Get a common denominator first.
Multiply by the common denominator first.
3 x
7
4 2
3 Q 4 R 2 4 Q 3 R 5 12
x
7
12 Q 4 2 23 R 5 12 Q 12 R
3x
8
7
12 2 12 5 12
3
x
7
12 Q R 24 12 Q 2 R 5 12 Q R
4
3x
15
12 5 12
3
12
3x 2 8 5 7
3x 12
15 12
12 ? 3 5 12 ? 3
3x 5 15
x55
x55
Decimals can be cleared from the equation by multiplying by a power of ten with
the same number of zeros as the number of digits to the right of the decimal. For
instance, if the greatest number of digits after the decimal is 3, like 4.586, you
multiply by 1000.
Problem
What is the solution of 2.8x 2 4.25 5 5.55?
100(2.8x 2 4.25 5 5.55)
280x 2 425 5 555
280x 5 980
x 5 3.5
Multiply by 100 because the most number of digits after the decimal
is two.
Simplify by moving the decimal point to the right 2 places in each
term.
Add 425 to each side to get the term with the variable by itself on the
left side.
Divide each side by 280 to isolate the variable.
Solve each equation. Check your answer.
x
3
1
14. 16 2 2 5 8
2a
8
15. 3 1 9 5 4
3n
1
16. 7 2 1 5 8
17. 21.68j 1 1.24 5 13
18. 4.6 5 3.5w 2 6.6
19. 5.23y 1 3.02 5 22.21
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Solving Equations With Variables on Both Sides
To solve equations with variables on both sides, you can use the properties of
equality and inverse operations to write a series of simpler equivalent equations.
Problem
What is the solution of 2m 2 4 1 5m 5 13 2 6m 2 4?
7m 2 4 5 26m 1 9
Add the terms with variables together on the left side and the
constants on the right side to combine like terms.
7m 2 4 1 6m 5 26m 1 9 1 6m
13m 2 4 5 9
To move the variables to the left side, add 6m to each side.
Simplify.
13m 2 4 1 4 5 9 1 4
To get the variable term alone on the left, add 4 to each side.
13m 5 13
Simplify.
13m
13
13 5 13
Divide each side by 13 since x is being multiplied by 13 on the
left side. This isolates x.
m51
Simplify.
Problem
What is the solution of 3(5x 2 2) 5 23(x 1 6)?
15x 2 6 5 23x 2 18
15x 2 6 1 6 5 23x 2 18 1 6
15x 5 23x 2 12
15x 1 3x 5 23x 2 12 1 3x
Distribute 3 on the left side and 23 on the right side into the
parentheses by multiplying them by each term inside.
To move all of the terms without a variable to the right side, add
6 to each side.
Simplify.
To get the variable terms to the left side, add 3x to each side.
18x 5 212
Simplify.
18x
12
18 5 218
Divide each side by 18 since x is being multiplied by 18 on the
left side. This isolates x.
x 5 223
Simplify and reduce the fraction.
Solve each equation. Check your answer.
1. 25x 1 9 5 23x 1 1
2. 14 1 7n 5 14n 1 28
3. 22(g 2 1) 5 2g 1 8
4. 2d 1 12 2 3d 5 5d 2 6
5. 4(m 2 2) 5 22(3m 1 3)
6. 2(4y 2 8) 5 2(y 1 4)
7. 5a 2 2(4a 1 5) 5 7a
8. 11w 1 2(3w 2 1) 5 15w
9. 4(3 2 5p) 5 25(3p 1 3)
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Solving Equations With Variables on Both Sides
An equation that is true for every value of the variable for which the equation is
defined is an identity. For example, x 2 5 5 x 2 5 is an identity because the
equation is true for any value of x. An equation has no solution if there is no value
of the variable that makes the equation true. The equation x 1 6 5 x 1 3 has no
solution.
Problem
What is the solution of each equation?
a) 3(4x 2 2) 5 22(26x 1 3)
12x 2 6 5 12x 2 6
12x 2 6 2 12x 5 12x 2 6 2 12x
26 5 26
Distribute 3 on the left side and 22 on the right side into the
parentheses by multiplying them by each term inside.
To get the variable terms to the left side, subtract 12x from each
side.
Simplify.
Because 26 5 26 is always true, there are infinitely many solutions of the
original equation. The equation is an identity.
b) 2n 1 4(n 2 2) 5 8 1 6n
2n 1 4n 2 8 5 8 1 6n
6n 2 8 5 8 1 6n
6n 2 8 2 6n 5 8 1 6n 2 6n
28 5 8
Distribute 4 into the parentheses by multiplying it by each term
inside.
Add the variable terms on the left side to combine like terms.
To get the variable terms to the left side, subtract 6n from each
side.
Simplify.
Since 28 2 8 , the equation has no solution.
Determine whether each equation is an identity or whether it has no solution.
10. 23(2x 1 1) 5 2(23x 2 1) 11. 4(23x 1 4) 5 22(6x 2 8) 12. 3n 1 3(2n 1 3) 5 3
Solve each equation. If the equation is an identity, write identity. If it has no
solution, write no solution.
13. 2(4n 1 2) 5 22(2n 2 1) 14. 2(2d 1 4) 5 2d 1 8
15. 2k 2 18 5 25 2 k 2 13
16. Open-Ended Write three equations with variables on both sides of the equal
sign with one having no solution, one having exactly one solution, and one
being an identity.
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Literal Equations and Formulas
A literal equation is an equation that involves two or more variables. When you
work with literal equations, you can use the methods you have learned in this
chapter to isolate any particular variable. To solve for specific values of a variable,
simply substitute the values into your equation and simplify.
Problem
What is the solution of 4x 2 5y 5 3 for y? What is the value of y when x 5 10?
4x 2 5y 2 4x 5 3 2 4x
To get the y-term by itself on the left side, subtract 4x from
each side.
25y 5 24x 1 3
Simplify.
25y
24x 1 3
25 5
25
Divide each side by 25 since y is being multiplied by 25 on
the left side. This isolates y.
3
y 5 45x 2 5
Simplify by dividing each term by 25. Notice, this changes
the sign of each term.
3
y 5 45(10) 2 5
To find the value of y when x 5 10, substitute 10 in for x.
y 5 725
Simplify by multiplying first, then subtracting.
When you rewrite literal equations, you may have to divide by a variable or
variable expression. When you do so in this lesson, assume that the variable or
variable expression is not equal to zero because division by zero is not defined.
Problem
Solve the equation ab 2 bc 5 cd for b.
b(a 2 c) 5 cd
b(a 2 c)
cd
a2c 5a2c
Since b is a factor of each term on the left side, it can be
factored out using the Distributive Property.
To get b by itself, divide each side by a 2 c since b is being
multiplied by a 2 c. Remember a 2 c 2 0.
cd
b5a2c
Simplify.
Solve each equation for y. Then find the value of y for each value of x.
1. y 1 5x 5 2; 21, 0, 1
2. 6x 5 2y 2 4; 1, 2, 4
3. 6x 2 3y 5 29; 22, 0, 2
4. 4y 5 5x 2 8; 22, 21, 0
5. 3y 1 2x 5 25; 0, 2, 3
6. 5x 5 8y 2 6; 21, 0, 1
7. 3(y 2 2) 1 x 5 1; 21, 0, 1
x12
8. y 2 3 5 1; 21, 0, 1
y14
9. x 2 5 5 23; 22, 2, 4
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Literal Equations and Formulas
A formula is an equation that states a relationship among quantities. Formulas
are special types of literal equations. Some common formulas are shown below.
Notice that some of the formulas use the same variables, but the definitions of the
variables are different. For instance, r is the radius in the area and circumference
of a circle and the rate in the distance formula.
Formula Name
Formula
Perimeter of a rectangle
P 5 2l 1 2w
Circumference of a circle
C 5 2πr
Area of a rectangle
A 5 lw
Area of a triangle
A 5 2bh
Area of a circle
A 5 πr2
Distance traveled
d 5 rt
1
Each of the formulas can be solved for any of the other unknowns in the equation
C
to produce a new formula. For example, r 5 2π is a formula for the radius of a
circle in terms of its circumference.
Problem
What is the length of a rectangle with width 24 cm and area 624 cm2 ?
A 5 lw
Formula for the area of a rectangle.
lw
A
w5 w
Since you are trying to get l by itself, divide each side by w.
A
l5w
624
l 5 24
l 5 26 cm
Simplify.
Substitute 624 for A and 24 for w.
Simplify.
Solve each problem. Round to the nearest tenth, if necessary. Use 3.14 for π.
10. A triangle has base 6 cm and area 42 cm2 . What is the height of the triangle?
11. What is the radius of a circle with circumference 56 in.?
12. A rectangle has perimeter 80 m and length 27 m. What is the width?
13. What is the length of a rectangle with area 402 ft2 and width 12 ft?
14. What is the radius of a circle with circumference 27 in.?
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Ratios, Rates, and Conversions
12 in.
A unit rate is a rate with denominator 1. For example, 1 ft is a unit rate. Unit rates
can be used to compare quantities and convert units.
Problem
Which is greater, 74 inches or 6 feet?
It is helpful to convert to the same units. Conversion factors, a ratio of two
equivalent measures in different units, are used to do conversions.
Multiply the original quantity by the conversion factor(s) so that units cancel out,
leaving you with the desired units.
12
6 ft 3 1 ftin 5 72 in.
Since 72 in. is less than 74 in., 74 in. is greater than 6 ft.
Rates, which involve two different units, can also be converted. Since rates involve
two different units, you must multiply by two conversion factors to change both of
the units.
Problem
Jared’s car gets 26 mi per gal. What is his fuel efficiency in kilometers per liter?
You need to convert miles to kilometers and gallons to liters. This will involve
multiplying by two conversion factors.
1.6 km
1 mi
There are 1.6 km in 1 mi. The conversion factor is either 1 mi or 1.6 km .
1.6 km
Since miles is in the numerator of the original quantity, use 1 mi as the
conversion factor so that miles will cancel.
1.6 km
mi
26gal
3 1 mi
3.8 L
1 gal
There are 3.8 L in 1 gal. The conversion factor is either 1 gal or 3.8 L .
1 gal
Since gallons is in the denominator of the original quantity, use 3.8 L as the
conversion factor so that gallons will cancel.
1.6 km
1 gal
mi
26gal
3 1 mi 3 3.8 L < 10.9 km
L
Jared’s vehicle gets 10.9 kilometers per liter.
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Ratios, Rates, and Conversions
Exercises
Convert the given amount to the given unit.
1. 12 hours; minutes
2. 1000 cm; km
3. 45 ft; yd
4. 32 cups; gallons
5. 30 m; cm
6. 15 lbs; kilograms
7. 42 in.; cm
8. 10 miles; km
9. 25 ft; in.
10. Serra rode 15 mi in 1.5 hr. Phaelon rode 38 mi in 3.5 h. Justice rode 22 mi in
2.25 hr. Who had the fastest average speed?
11. Mr. Hintz purchased 12 gallons of drinking water for his family for $14.28. He
knows that this should last for 2 weeks. What is the average cost per day for
drinking water for the family?
12. The unit price for a particular herb is 49 cents for 6 ounces. What is the price of
the herb in dollars per pound?
Copy and complete each statement.
13. 45 mi/h 5 ____ft/s
14. 7 g/s 5 ____kg/min
15. 50 cents/min 5 ____$/h
16. 22 m/h 5 ____cm/s
17. 15 km/min 5 ____mi/h
18. 6 gal/min 5 ____qt/h
19. Writing Describe the conversion factor you would use to convert feet to
miles. How do you determine which units to place in the numerator and the
denominator?
20. Writing Describe a unit rate. How do you determine the unit rate if the rate is
not given as a unit rate. Illustrate using an example.
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Solving Proportions
A proportion is an equation that states that two ratios are equal. If a quantity in a
proportion is unknown, you can solve a proportion to find the unknown quantity
as shown below.
Problem
x
What is the solution of 34 5 14 ?
There are two methods for solving proportions—using the Multiplication Property
of Equality and the Cross Products Property.
1) The multiplication Property of Equality says that you can multiply both sides
of an equation by the same number without changing the value.
3
x
4 5 14
3
x
14 Q 4 R 5 Q 14 R 14
42
4 5x
To isolate x, multiply each side by 14.
Simplify.
10.5 5 x
Divide 42 by 4.
2) The Cross Products Property says that you can multiply diagonally across the
proportion and these products are equal.
3
x
4 5 14
(4)(x) 5 (3)(14)
Multiply diagonally across the proportion.
4x 5 42
Multiply.
4x
42
4 5 4
To isolate x, divide each side by 4.
x 5 10.5
Simplify.
Real world situations can be modeled using proportions.
Problem
A bakery can make 6 dozen donuts every 21 minutes. How many donuts can the
bakery make in 2 hours?
A proportion can be used to answer this question. It is key for you to set up the
proportion with matching units in both numerators and both denominators.
For this problem, you know that 2 hours is 120 minutes and 6 dozen is 72 donuts.
Correct:
Incorrect:
72 donuts
x donuts
21 min 5 120 min
72 donuts
120 min
21 min 5 x donuts
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Solving Proportions
This proportion can be solved using the Multiplication Property of Equality or the
Cross Products Property.
Problem
Solve this proportion using the cross products.
72 donuts
x donuts
21 min 5 120 min
21x 5 (72)(120)
Cross Products Property
21x 5 8640
Multiply.
21x
8640
21 5 21
Divide each side by 21.
x 5 411.43
Simplify.
Since you cannot make 0.43 donuts, the correct answer is 411 donuts.
Exercises
Solve each proportion using the Multiplication Property of Equality.
3
n
1. 4 5 7
t
1
2. 3 5 10
n
8
3. 5 5 20
9
z
4. 6 5 8
15
a
5. 5 5 11
7
d
6. 2 5 8
Solve each proportion using the Cross Products Property.
3
b
7. 5 5 8
7
14
10. v 5 3
9
z
9. 2 5 6
8
12
8. m 5 3
f
24
11. 29 5 212
12.
13
2
5 26
h
13. A cookie recipe calls for a half cup of chocolate chips per 3 dozen cookies.
How many cups of chocolate chips should be used for 10 dozen cookies?
Solve each proportion using any method.
14.
x23
4
22 5 5
y16
12
15. 10 5 13
5
2
16. x 2 3 5 26
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Proportions and Similar Figures
In similar figures, the measures of
corresponding angles are equal, and the
ratios of corresponding side lengths are
equal. It is important to be able to identify the
corresponding parts in similar figures.
E
B
A
Since /A > /D, /B > /E, and /C > /F ,
C
D
F
BC AB
AC
AB
DE 5 EF , DE 5 DF . This fact can help you
to find missing lengths.
Problem
What is the missing length in the similar figures?
First, determine which sides correspond. The side
with length 14 corresponds to the side with length
16. The side with length x corresponds to the side
with length 12. These can be set into a proportion.
x
14
16 5 12
14
16
x
12
Write a proportion using corresponding lengths.
(16)(x) 5 (14)(12)
Cross Products Property
16x 5 168
Multiply.
x 5 10.5
Divide each side by 16 and simplify.
Exercises
The figures in each pair are similar. Identify the corresponding sides and angles.
F
1.
B
G
Q
2.
C
N
40 m
A
D
E
M
H
60 m
41 m
48 m
O
P
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2-8
Proportions and Similar Figures
Exercises
The figures in each pair are similar. Find the missing length.
3.
4.
5m
4m
14 cm
11 cm
8m
x
9 cm
5.
10 ft
4 ft
x
6.
5 ft
23 in.
27 in.
x
x
18 in.
Problem
A map shows the distance between two towns is 3.5 inches where the scale on the
map is 0.25 in. : 5 mi. What is the actual distance between the towns?
map distance
Map scale: actual distance
If you let x be the actual distance between the towns, you can set up and solve a
the proportion to answer the question.
0.25 in.
3.5 in.
5 mi 5 x mi
0.25x 5 17.5
x 5 70
The towns are 70 miles apart.
Exercises
The scale of a map is 1.5 in. : 50 mi. Find the actual distance corresponding to
each map distance.
7. 10 in.
8. 4.25 in.
9. 6.75 in.
10. The blueprints of an octagonal shaped hot tub are drawn with a 1 in. : 5 ft
scale. In the drawing the sides are 3.5 inches long. What is the perimeter of the
hot tub?
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Percents
Percents compare whole quantities, represented by 100%, and parts of the whole.
Problem
What percent of 90 is 27?
There are two ways presented for finding percents.
p
a
1) You can use the percent proportion 5 100 . The percent is represented by
b
p
100 . The base, b, is the whole quantity and must be the denominator of the
other fraction in the proportion. The part of the quantity is represented by a.
p
27
90 5 100
Substitute given values into the percent proportion. Since you are
looking for percent, p is the unknown.
27(100) 5 (90)(p)
Cross Products Property
2700 5 90p
Multiply.
30 5 p
Divide each side by 90 and simplify.
27 is 30% of 90.
2) The other way to find percents is to use the percent equation. The percent
equation is a 5 p% 3 b, where p is the percent, a is the part, and b is the base.
27 5 p% 3 90
Substitute 27 for a and 90 for b.
0.3 5 p%
Divide each side by 90.
30% 5 p%
Write the decimal as a percent.
27 is 30% of 90.
Exercises
Find each percent.
1. What percent of 125 is 50?
2. What percent of 14 is 35?
3. What percent of 24 is 18?
4. What percent of 50 is 75?
Problem
75% of 96 is what number?
In this problem you are given the percent p and the whole quantity (base) b.
a 5 p% 3 b
Write the percent equation.
a 5 75% 3 96 5 72
Substitute 75 for p and 96 for b. Multiply.
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Percents
Problem
28% of what number is 42?
You are given the percent p and the partial quantity a. You are looking for the base b.
a 5 p% 3 b
Write the percent equation.
42 5 28% 3 b
Substitute 28 for p and 42 for a.
42 5 0.28 3 b
Write 28% as a decimal, 0.28.
150 5 b
Divide each side by 0.28.
Exercises
Find each part.
5. What is 32% of 250?
6. What is 78% of 130?
Find each base.
7. 45% of what number is 90?
8. 70% of what number is 35?
Problems involving simple interest can be solved using the formula I 5 Prt ,
where I is the interest, P is the principal, r is the annual interest rate written as a
decimal, and t is the time in years.
Problem
You deposited $2200 in a savings account that earns a simple interest rate of
2.8% per year. You want to keep the money in the account for 3 years. How much
interest will you earn?
I 5 Prt
Simple Interest Formula
I 5 (2200)(2.8%)(3)
Substitute 2200 for P, 2.8% for r, and 3 for t.
I 5 184.8
Multiply.
You will earn $184.80 in interest.
Exercises
9. If you deposit $11,000 in a savings account that earns simple interest at a rate
of 3.5% per year, how much interest will you have earned after 5 years?
10. If you deposit $500 in a savings account that earns simple interest at a rate of
4.25% per year, how much interest will you have earned after 10 years?
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Change Expressed as a Percent
A percent change occurs when the original amount changes and the change
is expressed as a percent of the original amount. There are two possibilities for
percent change: percent increase or perent decrease. The following formula can
be used to find percents of increase/decrease.
percent change 5
amount of increase or decrease
original amount
Problem
In its first year, membership of the community involvement club was 32 members.
The second and third years there were 28 members and 35 members respectively.
Determine the percent change in membership each year.
From the first to the second year, the membership went down from 32 to 28
members, representing a percent decrease. The amount of decrease can be found
by subtracting the new amount from the original amount.
percent change 5
5
original amount 2 new amount
original amount
32 2 28
32
Percent Change Formula for percent
decrease.
Substitute 32 for the original number
and 28 for the new number.
4
5 32
5 0.125
Subtract. Then divide.
Membership decreased by 12.5% from the first year to the second year.
From the second to the third year, the membership increased from 28 to 35
members, representing a percent increase. The amount of increase can be found
by subtracting the original amount from the new amount.
percent change 5
5
original amount 2 new amount
original amount
35 2 28
28
Percent Change Formula for percent
increase.
Substitute 28 for the original number
and 35 for the new number.
7
5 32 < 0.22
Subtract. Then divide.
Membership increased by about 22% from the second year to the third year.
Exercises
Tell whether each percent change is an increase or decrease. Then find the
percent change. Round to the nearest percent.
1. Original amount: 25
New amount: 45
2. Original amount: 17
New amount: 10
3. Original amount: 22
New amount: 21
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Change Expressed as a Percent
Errors can occur when making measurements or estimations. Percents can be
used to compare estimated or measured values to exact values. This is called
relative error. Relative error can be determined with the following formula
comparing the estimated value and the actual value.
Percent error 5
u measured or estimated value 2 actual value u
actual value
Problem
Mrs. Desoto estimated that her class would earn an average of $126 per person
for the fundraiser. When the money was counted after the fundraiser ended, each
student had raised an average of $138 per person. What is the percent error?
There are two values given in this situation. The estimated value is $126 per
person. The actual value that each person raised was $138.
u measured or estimated value 2 actual value u
actual value
u 126 2 138 u
5
138
Percent error 5
Percent Error Formula
Substitute 126 for the estimated
value and 138 for the actual value.
u 212 u
5 138
12
5 138
Subtract.
u 212 u 5 12
< 0.09
Divide.
There was a 9% error in her estimation.
Exercises
Find the percent error in each estimation. Round to the nearest percent.
4. You estimate that your baby sister weighs 22 lbs. She is actually 26 lbs.
5. You estimate that the bridge is 60 ft long. The bridge is actually 53 ft long.
6. You estimate the rope length to be 80 ft. The rope measures 72 ft long.
7. A carpenter estimates the roof to be 375 ft2 . The rectangular roof measures 18
feet wide by 22 feet long. What is the percent error?
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3-1
Inequalities and Their Graphs
You use the following symbols for inequalities.
. is greater than
$ is greater than or equal to
, is less than
# is less than or equal to
Problem
What inequality represents “5 plus a number y is less than 210”?
5 plus a number y
is less than
210
,
210
51y
The inequality 5 1 y , 210 represents the phrase.
Exercises
Write an inequality that represents each verbal expression.
1. p is greater than or equal to 5
2. a is less than or equal to 24
3. 2 times d is less than 10
4. r divided by 5 is greater than 0
Problem
Is 22 a solution of 3t 1 10 $ 5?
3t 1 10 $ 5
Original inequality
?
3(22) 1 10 $ 5
Substitute 22 for t.
?
26 1 10 $ 5
Simplify.
445
22 is not a solution.
Exercises
Determine whether each number is a solution of the given inequality.
5. 5b 2 7 . 13
a. 24
b. 4
c. 8
6. 2(m 1 1) , 26
a. 26
b. 24
c. 22
a. 6
b. 8
c. 10
7.
81h
2 #8
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Inequalities and Their Graphs
When graphing an inequality on a number line, an open circle means the number
is not included in the inequality. A closed circle means the number is included in
the inequality.
Problem
What is the graph of w $ 21?
Since w is greater than or equal to 21, place a closed circle at 21.
Draw a dark line with an arrow to the right of the closed circle to show the
numbers greater than 21.
25
0
5
Exercises
Graph each inequality.
8. y # 0
9. p . 24
10. a $ 22
Problem
What inequality represents the graph?
25
0
5
The circle is open so 4 is not included in the inequality.
The dark line and arrow are to the left indicating less than.
The graph represents “x is less than 4” or x , 4.
Exercises
Write an inequality for each graph.
11.
13.
25
0
5
25
0
5
12.
14.
25
0
5
25
0
5
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Solving Inequalities Using Addition or Subtraction
You can add the same number to each side of an equation. You can also add the
same number to each side of an inequality.
Problem
What are the solutions of b 2 4 . 22? Graph and check the solutions.
b 2 4 . 22
Original inequality.
b 2 4 1 4 . 22 1 4
b.2
Add 4 to each side.
Simplify.
To graph b . 2, place an open circle at 2 and shade to the right.
25
0
5
To check the endpoint of b . 2, make sure
that 2 is the solution of the related equation
b 2 4 5 22.
b 2 4 5 22
Then check to see if a number greater
than 2 is a solution of the inequality.
5 is greater than 2.
b 2 4 . 22
?
2 2 4 0 22
5 2 4 . 22
1 . 22 3
22 5 22 3
Exercises
Solve each inequality. Graph and check your solutions.
1. m 2 14 $ 210
2. t 2 2 , 4
3. y 2 3 # 4
4. d 2 9 $ 212
5. w 2 17 . 13
6. a 2 22 , 27
7. Writing Explain how you would solve t 2 15 # 5.
8. Anita is baking dinner rolls and pumpkin bread. She needs 4 cups of flour for
the rolls. She needs at least 7 cups of flour left for the pumpkin bread. Write
and solve an inequality to determine how much flour Anita needs before she
starts baking.
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Solving Inequalities Using Addition or Subtraction
You can subtract the same number from each side of an equation. You can also
subtract the same number from each side of an inequality.
Problem
What are the solutions of h 1 7 # 4? Graph and check the solutions.
h17#4
Original inequality.
h1727#427
h # 23
Subtract 7 from each side.
Simplify.
To graph h # 23, place a closed circle at 23 and shade to the left.
25
0
5
To check the endpoint of h # 23, make
sure that 23 is the solution of the related
equation h 1 7 5 4.
Then check to see if a number less than
23 is a solution of the inequality. 24 is
less than 23.
h1754
h17#4
23 1 7 0 4
24 1 7 # 4
?
3#43
4 5 43
Exercises
Solve each inequality. Graph and check your solutions.
9. s 1 7 $ 12
10. p 1 3 , 21
11. b 1 5 # 24
12. n 1 1 $ 8
13. v 1 18 . 212
14. k 1 26 , 6
15. A boat can hold up to 1000 pounds. Two friends get in the boat. Together they
weigh 285 pounds. Write and solve an inequality to determine how much
more weight can be added to the boat.
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Solving Inequalities Using Multiplication or Division
You can solve inequalities using multiplication or division using these two
important rules.
• You can multiply or divide each side of an inequality by a positive number.
• You can multiply or divide each side of an inequality by a negative number
only if you reverse the inequality sign.
Problem
c
What are the solutions of 5 # 22? Graph the solutions.
c
Original inequality
5 # 22
c
5 Q 5 R # 5(22)
Multiply each side by 5. Keep the inequality symbol the same.
c # 210
Simplify.
To graph c # 210, place a closed circle at 210 and shade to the left.
16 12 8 4
0
4
Problem
What are the solutions of 2 23 t . 9? Graph the solutions.
223 t . 4
Original inequality
3
3
22 Q223 t R , 22(4)
3
Multiply each side by 22 . Reverse the inequality symbol.
t , 26
Simplify.
To graph t , 26, place an open circle at 26 and shade to the left.
8 6 4 2
0
2
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Solving Inequalities Using Multiplication or Division
Problem
What are the solutions of 26h # 239? Graph the solutions.
26h # 239
Original inequality
26h
239
26 $ 26
h $ 612
Divide each side by 26. Reverse the inequality symbol.
Simplify.
To graph h $ 612 , place closed circle at 612 and shade to the right.
2
0
2
4
6
8
Exercises
Solve each inequality. Graph and check your solutions.
x
1. 7 . 22
2. 8p # 32
2
3. 5 r $ 6
k
4. 22 , 25
5. 23f # 12
3
6. 5t . 29
7. 22w . 28
z
8. 25 $ 4
3
3
9. 24d , 28
10. 24n $ 14
11. A bus company charges $2 for each trip. It also sells monthly passes for $50.
Write and solve an inequality to find how many trips you could make before
the monthly pass is cheaper.
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Solving Multi-Step Inequalities
Solving inequalities is similar to solving equations. However, if you multiply
or divide each side of an inequality by a negative number, the direction of the
inequality is reversed.
Problem
What are the solutions of 6 2 3k . 45?
6 2 3k . 45
6 2 3k 2 6 . 45 2 6
Original inequality
Subtract 6 from each side.
23k . 39
Simplify.
23k
39
23 , 23
Divide each side by 23 and reverse the sign.
k , 213
Simplify.
Problem
What are the solutions of 6(n 2 3) 1 4n # 42?
6(n 2 3) 1 4n # 42
Original inequality
6n 2 18 1 4n # 42
Distributive Property
10n 2 18 # 42
Combine like terms.
10n 2 18 1 18 # 42 1 18
Add 18 to each side.
10n # 60
Simplify.
10n
60
10 # 10
Divide each side by 10.
n#6
Simplify.
Problem
What are the solutions of 7p 1 12 . 6p 2 15?
7p 1 12 . 6p 2 15
Original inequality
7p 1 12 2 6p . 6p 2 15 2 6p
p 1 12 . 215
p 1 12 2 12 . 215 2 12
p . 227
Subtract 6p from each side.
Simplify.
Subtract 12 from each side.
Simplify.
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Solving Multi-Step Inequalities
Exercises
Solve each inequality.
1. 8w 1 9 , 231
2. 5h 2 6 $ 24
3. 17 2 2a # 29
4. 5 2 3t . 27
d
5. 7 1 4 . 22
2x
6. 4 2 3 # 28
7. 5(y 2 2) 2 2y $ 5
8. 8(2f 1 3) 1 4f # 216
9. 3(p 2 2) 2 7p , 6
10. 2(3b 1 5) 2 10b . 30
11. 7z 2 4 # 6z 1 18
12. 8m 1 7 $ 6m 2 9
13. 12c 1 6 . 9c 2 15
14. 7d 1 2 , 17 2 3d
15. A student had $45 when she went to the mall. She spent $9 on a pair of
earrings. Then she wants to by some CDs that cost $12 each. Write and solve
an inequality to determine how many CDs she can buy.
16. A friend needs at least $125 to go on the class trip. He has saved $45. He makes
$20 for each lawn he mows. Write and solve an inequality to determine how
many lawns he needs to mow to go on the trip.
17. You have earned 85, 92, 95, and 88 on tests this grading period. You have one
last test and want an average of at least 90. Write and solve an inequality to
determine what scores you can earn to achieve your goal.
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Working With Sets
There are two ways to write a set.
• Roster form lists the elements of a set within braces, { }.
• Set-builder notation describes the properties an element must have to be
included in a set.
Problem
How do you write “R is the set of even whole numbers less than 10” in roster form
and in set-builder notation?
Roster Form
List the numbers 0, 2, 4, 6, and 8 in braces.
R 5 {0, 2, 4, 6, 8}
Set-Builder Notation
Describe the properties.
R 5 5x u x is an even whole number, x , 10}
This is read as “the set of all even whole numbers x such that x is less than 10.”
Exercises
Write each set in roster form and in set-builder notation.
1. D is the set of integers greater than 25 and less than 5.
2. N is the set of odd natural numbers less than 14.
3. P is the set of natural numbers less than or equal to 7.
4. T is the set of real numbers that are factors of 18.
5. A is the set of integers between 23 and 5.
Write the solutions of each inequality in set-builder notation.
6. 4b 1 8 . 212
7. 7n 2 14 $ 28
8. 5s 2 15 # 18 2 2s
9. 2(3p 2 5) 2 7p , 22
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Working With Sets
Set A is a subset of set B if each element of A is also an element of B.
Problem
What are all the subsets of the set {r, s, t}?
;
Start with the empty set because it is a subset of every set.
5r6, 5s6, 5t6
Each element of the given set is a subset of the given set, so list the subsets with one
element.
5r, s6, 5r, t6, 5s, t6
List the subsets with two elements from the given set.
5r, s, t6
List the original set. It is always a subset of itself.
The subsets of 5r, s, t6 are ; , 5r6, 5s6, 5t6 , 5r, s6, 5r, t6, 5s, t6 and 5r, s, t6 .
Exercises
List all the subsets of each set.
10. {5, 10}
11. {mom, dad, child}
12. 522, 21, 06
13. {a, b, c, d}
The universal set U represents the largest set in a problem. The complement of
a set S is the set of all the elements in the universal set but not in the set S. The
complement of S is written Sr.
Problem
If U 5 {days of the week} and D 5 {Sunday, Friday}, what is Dr?
The days of the week that are not in D are Monday, Tuesday, Wednesday,
Thursday, and Saturday.
Dr 5 {Monday, Tuesday, Wednesday, Thursday, Saturday}
Exercises
14. If U 5 {natural numbers less than 20} and N 5 {factors of 18}, what is Nr?
15. If U 5 {months of the year} and M 5 {months starting with J}, what is Mr?
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Compound Inequalities
A compound inequality with the word or means one or both inequalities must be
true. The graph of the compound inequality a , 24 or a $ 3 is shown below.
25
0
5
A compound inequality with the word and means both inequalities must be true.
The graph of the compound inequality b # 4 and b . 21 is shown below.
4 2
0
2
4
6
To solve a compound inequality, solve the simple inequalities from which
it is made.
Problem
What are the solutions of 17 # 2x 1 7 # 29? Graph the solutions.
17 # 2x 1 7 # 29 is the same as 17 # 2x 1 7 and 2x 1 7 # 29. You can solve it
as two inequalities.
17 # 2x 1 7
and
2x 1 7 # 29
17 2 7 # 2x 1 7 2 7
and
2x 1 7 2 7 # 29 2 7
10 # 2x
and
2x # 22
10
2x
2 # 2
and
2x
22
2 # 2
5#x
and
x # 11
To graph the compound inequality, place closed circles at 5 and 11. Shade
between the two circles.
0
2
4
6
8 10 12
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Compound Inequalities
Problem
What are the solutions of 3t 2 5 , 28 or 2t 1 5 . 17? Graph the solutions.
Solve each inequality.
3t 2 5 , 28
or
2t 1 5 . 17
3t 2 5 1 5 , 28 1 5
or
2t 1 5 2 5 . 17 2 5
3t , 23
or
2t . 12
3t
23
3 , 3
or
2t
12
2 . 2
t , 21
or
t.6
To graph the compound inequality, place open circles at 21 and at 6. Shade to the
left of 21 and to the right of 6.
23
0
3
7
Exercises
Solve each compound inequality. Graph the solutions.
1. h 2 7 $ 25 and h 1 4 , 10
2. r 1 2 # 21 or r 2 3 . 2
3. 27 , w 2 4 , 2
y
4. 22 # 2 # 1
5. 5p 1 3 # 22 or 3p 2 6 $ 3
6. 22n 2 5 $ 1 or 5n 1 7 . 2
3
2
7. 4a 2 6 , 0 and 3a 1 4 . 2
8. 24 # 4d 1 24 # 4
w
10. 2 1 1 $ 2 and w 2 5 # 1
9. 5m 2 2 , 8 or 6m 2 2 . 6 1 5m
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Unions and Intersections of Sets
The union of two or more sets contains all elements of the sets. The symbol for
union is < .
Problem
If A 5 {1, 2, 3, 4, 5, 6} and B 5 {2, 4, 6, 8, 10}, what is A < B?
The union of A and B is the set of elements that are in either A or B or both. List
all elements that are in either or both of the sets, but do not repeat any elements.
Even if an element is in both sets, it only appears once in the union.
A < B 5 {1, 2, 3, 4, 5, 6, 8, 10}
The intersection of two or more sets contains elements that are in every set. The
symbol for intersection is >.
Problem
If M 5 {m k m is a factor of 24} and N 5 {n k n is an even number}, what is M > N ?
List the numbers in each set.
M 5 {1, 2, 3, 4, 6, 8, 12, 24}
N 5 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, …}
The intersection of M and N is the set of elements that are in both M and N.
Because the largest number in M is 24, it is not necessary to list numbers greater
than 24 for set N. The numbers that are in both sets are 2, 4, 6, 8, 12, and 24.
M > N 5 {2, 4, 6, 8, 12, 24}
Exercises
Find each union or intersection. Let P 5 {1, 2, 3, 4, 5}, R 5 {r z r is an odd
natural number, r R 10}, S 5 {5, 10, 15}, and T 5 {t z t is a factor of 8}.
1. P < R
2. P > R
3. R > S
4. S < R
5. P < T
6. P > T
7. S < T
8. S > T
9. S > P
10. S < P
11. R > T
12. R < T
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Unions and Intersections of Sets
The solutions of compound inequalities are unions or intersections of sets.
Problem
What are the solutions of x 1 5 . 28 and x 2 7 , 12? Write the solutions as
either the union or the intersection of two sets.
First solve the two inequalities separately.
x 1 5 . 28
and
x 2 7 , 12
x 1 5 2 5 . 28 2 5
and
x 2 7 2 7 , 12 2 7
x . 213
x,5
and
The word and indicates the intersection of two sets. The solution is numbers that
satisfy both of the final inequalities, numbers between 213 and 5.
The solution can be written as {x k x . 213} > {x k x , 5}.
Problem
What are the solutions of u 2yu $ 7? Write the solutions as either the union or the
intersection of two sets.
First write the absolute value inequality as two inequalities and solve the two
inequalities separately.
2y $ 7
or
2y # 27
2y
7
2 $2
or
2y
27
2 # 2
y $ 3.5
or
y # 23.5
The word or indicates the union of two sets. The solution is numbers that satisfy
either of the final inequalities or both of them, numbers greater than or equal to
3.5 or numbers less than or equal to 23.5.
The solution can be written as {y k y $ 3.5} < {y k y # 23.5}.
Exercises
Solve. Write the solutions as either the union or intersection of two sets.
13. p 1 4 $ 3 or p 2 2 , 24
14. 5t $ 220 and 4t # 24
15. k k 2 4 k $ 15
n
16. ` 6 ` , 5
17. k 3t 2 6 k , 15
18. k 5d 1 10 k $ 5
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Using Graphs to Relate Two Quantities
An important life skill is to be able to a read graph. When looking at a graph, you
should check the title, the labels on the axes, and the general shape of the graph.
Problem
What information can you determine from the graph?
• The title tells you that the graph describes Trina’s trip.
• In general, the more time that has elapsed, the closer
Trina gets to her destination. In the middle of the trip,
the distance does not change, showing she stops for a
while.
Trina’s Trip
Distance
to Destination
• The axes tell you that the graph relates the variable of
time to the variable of distance to the destination.
Time
Exercises
What are the variables in each graph? Describe how the variables are related at
various points on the graph.
2.
Dion’s Growth
Chart
3.
Height
Tiles Installed
Tiling Job
Time
Kicked
Football
Height
1.
Age
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Using Graphs to Relate Two Quantities
A graph can show the relationship described in a table.
Problem
Which graph shown below represents the
information in the table at the right?
CDs
Purchased
Total
Cost
1
$15
2
$30
3
$45
4
$60
5
$75
Notice that for each additional CD purchased, the
11
total cost increases by $15. The points on the graph
should be in a straight line that goes up from left to 11
right. The graph that shows this trend is Graph B.
11
11
115
115
115
Total Cost
C.
Total Cost
B.
Total Cost
A.
115
CDs Purchased
CDs Purchased
CDs Purchased
Exercises
Match each graph with its related table. Explain your answers.
Tickets Sold
6.
Tickets Sold
5.
Tickets Sold
4.
Day
A.
Day
Day
Day
Tickets Sold
1
B.
Day
Tickets Sold
60
1
2
45
3
4
C.
Day
Tickets Sold
70
1
35
2
65
2
45
40
3
50
3
55
75
4
45
4
65
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Patterns and Linear Functions
A relationship can be represented in a table, as ordered pairs, in a graph, in words,
or in an equation.
Problem
Consider the relationship between the number of squares in the pattern and the
perimeter of the figure. How can you represent this relationship in a table, as
ordered pairs, in a graph, in words, and in an equation?
5
5
5 5 5
5 5
5
5
5
5
5
5 5 5 5
5
5
5 5 5
5 5
5 5 5 5 5
5
5 5 5 5
5
5
5 5 5 5 5
Table
For each number of squares determine the perimeter of the figure. Write the
values in the table. Remember to focus on the perimeter of the figure, not the
squares.
Number of squares
1
2
3
4
5
Perimeter
20
30
40
50
60
Ordered Pairs
Let x represent the number of squares and y represent the perimeter. Use the
numbers in the table to write the ordered pairs.
(1, 20), (2, 30), (3, 40), (4, 50), (5, 60)
Graph
Use the ordered pairs to draw the graph.
Perimeter
60
y
40
20
O
x
2
4
6
Number of Squares
Words
The pattern shows the perimeter is the number of squares times 10 plus 10.
Equation
Write an equation for the words.
y 5 10x 1 10
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Patterns and Linear Functions
Exercises
Consider each pattern.
1.
2.
3
3
3
3
6
3
3
6
6
3
3 3
6
3
3
6
3
6
1 1
1
1
a. Make a table to show the
1
1
1 1 1
1
1 1
1
1
1 1 1 1
1
1 1 1
a. Make a table to show
relationship between the
number of trapezoids and the
perimeter.
the relationship between
the number of cubes and
the surface area.
Number of Trapezoids
Number of Cubes
Perimeter
Surface Area
b. Write the ordered pairs for
b. Write the ordered pairs for the
the relationship.
relationship.
c. Make a graph for the
c. Make a graph for the
relationship.
relationship.
d. Use words to describe the
d. Use words to describe
relationship.
the relationship.
e. Write an equation for the
e. Write an equation for
relationship.
the relationship.
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Patterns and Nonlinear Functions
If the points of the graph of a function are in a straight line, the function is a
linear function. If the points of the graph of a function are not in a straight line,
the function is a nonlinear function.
Problem
Is the function given by the table at the right
linear or nonlinear?
x
y
1
6
Graph the function.
2
3
3
2
6
1
6
y
4
2
x
O
2
4
6
The points are not in a straight line, so the function is nonlinear.
Do you like to solve puzzles? When you are given a list of function values and you
are asked to find the rule for the function, you are solving a puzzle. You are looking
for a rule that works for all pairs of numbers.
Problem
What is a rule that represents the function given by the table below?
x
y
6
3
8
5
9
6
12
9
Try a rule. Is there an operation or sequence of operations that
relates the values in the first column of the table to the values in
the second column?
Try division: 6 4 2 5 3, but 8 4 2 2 5.
Try another rule. 6 2 3 5 3 and 8 2 3 5 5.
Check to make sure this works for all pairs of numbers.
9 2 3 5 6 and 12 2 3 5 9.
The function can be represented by the rule y 5 x 2 3.
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Patterns and Nonlinear Functions
Graph the function shown by each table. Tell whether the function is linear or
nonlinear.
1.
x
y
0
2.
x
y
1
2
2
3
3
3.
x
y
1
1
4
3
3
2
1
4
4
5
3
0
6
7
5
7
4
1
x
y
x
y
x
y
2
6
Ź4
4
0
1
3
4
Ź3
3
2
2
4
3
0
0
4
3
6
2
2
2
6
4
5.
4.
6.
Each set of ordered pairs represents a function. Write a rule that represents the
function.
7. (2, 10), (4, 20), (5, 25), (7, 35), (9, 45)
8. (2, 5), (4, 9), (5, 11), (7, 15), (10, 21)
9. (0, 0), (1, 1), (2, 8), (3, 27), (4, 64)
10. (2, 5), (3, 10), (4, 17), (5, 26), (6, 37)
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Graphing a Function Rule
By finding values that satisfy a function rule, you can graph points and discover
the shape of its graph.
Problem
What is the graph of the function rule y 5 3x 1 5?
First, choose any values for x and find the
corresponding values of y. Make a table of
your values.
x
y â 3x à5
(x, y)
Ľ2
y â 3(Ľ2) à5 âĽ1
(Ľ2, Ľ1)
Ľ1
y â 3(Ľ1) à5 â2
(Ľ1, 2)
0
y â 3(0) à5 â5
(0, 5)
1
y â 3(1) à5 â8
(1, 8)
2
y â 3(2) à5 â11
(2, 11)
Then, graph the points from your table. In
this case, the points are in a line. Draw the
line.
y
8
4
x
O
Ľ4
4
8
Problem
What is the graph of the function rule y 5 ux 2 2 u ?
First, choose any values for x and find the
corresponding values of y. Make a table of
your values.
x
y â x Ľ2
(x, y)
0
y â 0 Ľ2 â2
(0, 2)
1
y â 1 Ľ2 â1
(1, 1)
2
y â 2 Ľ2 â0
(2, 0)
3
y â 3 Ľ2 â1
(3, 1)
4
y â 4 Ľ2 â2
(4, 2)
Then, graph the points from your table. In
this case, the points make a V shape. Draw
the V.
y
4
2
x
O
2
4
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Graphing a Function Rule
Exercises
Graph each function rule.
x
1. y 5 2 1 3
x
2. y 5 2x 2 3
yâ
x
à3
2
x
(x, y)
3. y 5 x2 2 3
x
y â Ľx Ľ3
(x, y)
4. y 5 u x u 1 1
y â x2 Ľ3
x
(x, y)
y â x à1
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Writing a Function Rule
When writing function rules for verbal descriptions, you should look for key words.
Words that Suggest
Addition
plus
sum
more than
increased by
total
in all
Words that Suggest
Subtraction
minus
difference
less than
decreased by
fewer than
subtracted by
Words that Suggest
Multiplication
times
product
of
each
factors
twice
Words that Suggest
Division
divided by
quotient
rate
ratio
half
a third of
Problem
Twice a number n increased by 4 equals m. What is a function rule that represents
the sentence?
twice a number n
increased by 4
equals
m
2n
à4
â
m
The function rule is 2n 1 4 5 m.
Exercises
Write a function rule that represents each sentence.
1. t is 4 more than the product of 7 and s.
2. The ratio of a to 5 equals b.
3. 8 fewer than p times 3 equals x.
4. y is half of x plus 10.
5. k equals the sum of h and 23.
6. 15 minus twice a equals b.
7. m equals 5 times n increased by 6.
8. 17 decreased by three times d equals c.
9. 5 more than the product of 6 and n is 17.
10. d is 8 less than the quotient of b and 4.
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4-5
Writing a Function Rule
You can write functions to represent situations and then evaluate the function to
determine a particular value.
Problem
A sales associate earns $500 per week plus 4% of his sales. Write a function
rule for the amount he makes in a week if he sells s dollars of merchandise.
How much will he make if he sells $4000 worth of merchandise?
First write the function rule.
earnings
equals
500
plus
4% of sales
e
â
500
à
0.04s
Use this function rule to calculate how much he will make.
e 5 500 1 0.04s
5 500 1 0.04(4000)
5 700
He will make $700.
Exercises
11. Twelve cans of peaches are placed into each box. Write a function rule
for the number of boxes needed for c cans. How many boxes are needed for
1440 cans?
12. Tara plans to rent a car for the weekend. The cost to rent the car is $45 plus
$0.15 for each mile she drives. Write a function rule for the total cost of the
rental. How much is the rental if she travels 500 miles?
13. A plumber charges $60 for a service call plus $55 for each hour she works.
Write a function rule for the total bill for a plumbing job. What is the total bill
for a job that takes the plumber 3 hours of work?
14. Tickets to a concert cost $45 per ticket plus a $10 processing fee for each order.
Write a function rule for the total cost of ordering tickets. What is the total cost
to order 6 tickets?
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Formalizing Relations and Functions
When a relation is represented as a set of ordered pairs, the domain of the relation
is the set of x-values. The range is the set of y-values.
A relation where each value in the domain is paired with just one value in the
range is called a function.
Problem
Identify the domain and range of the relation 5(22, 3), (0, 2), (1, 3), (3, 4)6 .
Represent the relation with a mapping diagram. Is the relation a function?
The domain (or x-values) is {–2, 0, 1, 3}.
The range (or y-values) is {2, 3, 4}.
Domain
Range
22
0
1
3
2
3
4
Notice that each number in the domain is mapped to only one number in the
range. This relation is a function.
Exercises
Identify the domain and range of each relation. Use a mapping diagram to
determine whether the relation is a function.
1. {(2, 3), (4, 6), (1, 5), (2, 5), (0, 5)}
2. {(3, 4), (5, 4), (7, 4), (8, 4), (10, 4)}
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Formalizing Relations and Functions
You can determine whether or not a relation is a function by looking at the graph
of the relation. If a vertical line is drawn anywhere on the graph and passes
through two points of the relation, the relation is not a function. This is called the
vertical line test.
Problem
Is the relation shown below a function? Use a vertical line test.
Notice that two of the dashed vertical lines pass through
just one point on the graph.
y
x
O
However, one of the dashed vertical lines passes through
three points.
The relation is not a function.
Exercises
Use the vertical line test to determine whether the relation is a function.
3.
4.
y
5.
y
y
x
x
x
O
O
O
6.
7.
y
x
O
8.
y
y
x
O
x
O
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Sequences and Functions
An orderly list of numbers is called a sequence. Each number in a sequence is
called a term. Many sequences follow a pattern. To find the pattern, you will be
solving a puzzle.
Problem
Describe the pattern of the sequence 8, 4, 0,24, 28, … . What are the next two
terms of the sequence?
You can divide 8 by 2 to get 4, but 4 divided by 2 is not 0. The pattern cannot be
“divide by 2.” Look at the pattern again. You can subtract 4 from each number to
get the next number.
8,
4,
24
0,
24,
24
24
28,...
24
The pattern is “subtract 4 from the previous term.” The next two terms are 28 2 4
or 212 and 212 2 4 or 216.
Exercises
Describe the pattern in each sequence. Then find the next two terms of the
sequence.
1. 1, 5, 25, 125, …
2. 3, 9, 15, 21, …
3. 64, 32, 16, 8, …
4. 25, 23, 21, 1, …
5. 1, 23, 9, 227, …
6. 10, 3,24,211, …
7. 1000, 2100, 10,21, …
8. 23, 31, 39, 47, …
9. 24, 212, 236, 2108, …
10. 25, 29, 213, 217, …
11. 3.6, 4.1, 4.6, 5.1, …
12. 281, 227, 29, 23, …
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Sequences and Functions
An arithmetic sequence is a sequence in which the difference between
consecutive terms is constant. This difference is called the common difference.
You can find the nth term of an arithmetic sequence by using the following
formula.
A(n) 5 A(1) 1 (n 2 1)d
In this formula,
• A(n) represents the nth term
• A(1) represents the first term
• n represents the term number
• d represents the common difference
Problem
Write a formula for the arithmetic sequence 15, 10, 5, 0, 25, … .
What is the tenth term of the sequence?
The pattern is to “add –5 to the previous term.”
15,
1(25)
10,
5,
1(25)
0,
1(25)
25,...
1(25)
A(n) 5 A(1) 1 (n 2 1)d
A(n) 5 15 1 (n 2 1)(25)
A(1) 5 15 and d 5 25
Use the formula to find the tenth term.
A(n) 5 15 1 (n 2 1)(25)
A(10) 5 15 1 (10 2 1)(25)
n 5 10
A(10) 5 230
Simplify.
Exercises
Write a formula for each arithmetic sequence. Then, find the tenth term.
13. 3, 10, 17, 24, …
14. 24, 1, 6, 11, …
15. 44, 40, 36, 32, …
16. 8, 2,24,210, …
17. 22, 32, 42, 52, …
18. 55, 44, 33, 22, …
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Rate of Change and Slope
The rate of the vertical change to the horizontal change between two points on a
line is called the slope of the line.
vertical change
rise
slope 5 horizontal change 5 run
There are two special cases for slopes.
• A horizontal line has a slope of 0.
• A vertical line has an undefined slope.
Problem
What is the slope of the line?
vertical change
4
rise
slope 5 horizontal change 5 run
y
(4, 2)
2 (1, 1)
rise 5 1
x
run 5 3
5 13
O
Ź2
The slope of the line is 13 .
2
4
6
Ź2
Ź4
In general, a line that slants upward from left to right has a positive slope.
Problem
What is the slope of the line?
4
vertical change
rise
slope 5 horizontal change 5 run
5 22
1
(0, 3)
rise 5 22
Ź4
5 22
y
run 5 1
Ź2 O
(1, 1)
2
x
4
Ź2
The slope of the line is 22.
Ź4
In general, a line that slants downward from left to right has a negative slope.
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Rate of Change and Slope
Exercises
Find the slope of each line.
1.
4
2
Ź2
2.
y
(1, 0)
O
Ź2
6
x
2
4
(5, 21)
6
y
3.
(6, 5)
6
4
4
2
(1, 2)
2
O
2
y
(1, 4)
x
x
Ź2
4
O
Ź2
6
2
Ź2
Ź2
Ź4
Suppose one point on a line has the coordinates (x1, y1) and another point on the
same line has the coordinates (x2, y2). You can use the following formula to find
the slope of the line.
rise
y 2y
2
1
slope 5 run 5 x 2 x , where x2 2 x1 2 0
2
1
Problem
What is the slope of the line through R(2, 5) and S(21, 7)?
y2 2 y1
slope 5 x 2 x
2
1
Let y2 5 7 and y1 5 5.
725
5 21 2 2
2
(5, 4)
Let x2 5 21 and x1 5 2 .
2
5 23 5 23
Exercises
Find the slope of the line that passes through each pair of points.
4. (0, 0), (4, 5)
5. (2, 4), (7, 8)
6. (22, 0), (23, 2)
7. (22, 23), (1, 1)
8. (1, 4), (2,23)
9. (3, 2), (25, 3)
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Direct Variation
A direct variation is a relationship that can be represented by a function in the
form y 5 kx where k 2 0. The constant of variation for a direct variation k is the
y
coefficient of x. The equation y 5 kx can also be written as x 5 k.
Problem
Does the equation 6x 1 3y 5 9 represent a direct variation? If so, find the
constant of variation.
If the equation represents a direct variation, the equation can be rewritten in the
form y 5 kx. So, solve the equation for y to determine whether the equation can
be written in this form.
6x 1 3y 5 9
3y 5 9 2 6x
Subtract 6x from each side.
y 5 3 2 2x
Divide each side by 3.
You cannot write the equation in the form y 5 kx. So 6x 1 3y 5 9 does not
represent a direct variation.
Problem
Does the equation 5y 5 3x represent a direct variation? If so, find the constant
of variation.
Again, if the equation represents a direct variation, the equation can be rewritten
in the form y 5 kx. So, solve the equation for y to determine whether the equation
can be written in this form.
5y 5 3x
3
y 5 5x
Divide each side by 5.
The equation has the form y 5 kx, so the equation represents a direct variation.
3
3
The coefficient of x is 5 , so the constant of variation is 5 .
Exercises
Determine whether each equation represents a direct variation. If it does, find
the constant of variation.
1. 2y 5 x
2. 3x 1 2y 5 1
3. 24y 5 8x
4. 2x 5 y 2 5
5. 4x 2 3y 5 0
6. 5x 5 2y
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Direct Variation
To write an equation for direct variation, find the constant of variation k using an
ordered pair. Then use the value of k to write an equation.
Problem
Suppose y varies directly with x, and y 5 24 when x 5 8. What direct variation
equation relates x and y? What is the value of y when x 5 10?
You are given that x and y vary directly. This means that the relationship between x
and y can be written in the form y 5 kx, where k is a constant.
y 5 kx
Start with the direct variation equation.
24 5 k(8)
Substitute the given values: 8 for x and 24 for y.
35k
Divide each side by 8 to solve for k.
y 5 3x
Write the direct variation equation that relates x and y
by substituting 3 for k in y 5 kx.
The equation y 5 3x relates x and y. When x 5 10, y 5 3(10)or 30.
Exercises
Suppose y varies directly with x. Write a direct variation equation that relates x
and y. Then find the value of y when x 5 6.
7. y 5 14 when x 5 2.
8. y 5 3 when x 5 9.
9. y 5 12 whenx 5 224.
10. y 5 281 when x 5 9.
11. y 5 216 when x 5 24.
12. y 5 5 when x 5 20.
13. Consider the direct variation y 5 3x.
a. List three ordered pairs that satisfy the equation.
b. Plot your three ordered pairs from
part (a) on a coordinate grid.
c. Complete the graph of y 5 3x on the grid.
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Slope-Intercept Form
The slope-intercept form of a linear equation is y 5 mx 1 b. In this equation,
m is the slope and b is the y-intercept.
Problem
What are the slope and y-intercept of the graph of y 5 22x 2 3?
The equation is solved for y, but it is easier to determine the y-intercept if the right
side is written as a sum instead of a difference.
y 5 22x 2 3
y 5 22x 1 (23)
Write the subtraction as addition.
The slope is 22 and the y-intercept is 23.
Problem
What is an equation for the line with slope 23 and y-intercept 9?
When the slope and y-intercept are given, substitute the values into the slopeintercept form of a linear equation.
y 5 mx 1 b
y 5 23 x 1 9
Substitute 32 for m and 9 for b.
Problem
What is an equation in slope-intercept form for the line that passes through the
points (1, 23) and (3, 1)?
Substitute the two given points into the slope formula to find the slope of the line.
m5
1 2 (23)
4
321 5252
Then substitute the slope and the coordinates of one of the points into the
slope-intercept form to find b.
y 5 mx 1 b
Use slope-intercept form.
23 5 2(1) 1 b
Substitute 2 for m, 1 for x, and 23 for y.
25 5 b
Solve for b.
Substitute the slope and y-intercept into the slope-intercept form.
y 5 mx 1 b
Use slope-intercept form.
y 5 2x 1 (25)
Substitute 2 for m and 25 for b.
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Class
5-3
Date
Reteaching (continued)
Slope-Intercept Form
Exercises
Find the slope and y-intercept of the graph of each equation.
1
1. y 5 2 x 1 7
2. y 5 25x 1 1
2
3. y 5 25 x 2 3
4. y 5 x 1 5
1
5. y 5 6 x 2 2
6. y 5 4x
Write an equation for the line with the given slope m and y-intercept b.
7. m 5 23, b 5 7
2
8. m 5 3, b 5 8
9. m 5 4, b 5 23
1
10. m 5 25, b 5 21
5
11. m 5 26, b 5 0
12. m 5 7, b 5 22
Write an equation in slope-intercept form for the line that passes through the
given points.
13. (1, 3) and (2, 5)
14. (2,21) and (4, 0)
15. (1, 2) and (2,21)
16. (1,25) and (3,23)
17. (3, 3) and (6, 5)
18. (4,23) and (8,24)
19. Consider the equation y 5 22x 1 4.
a. What is the y-intercept of the graph of the equation?
b. Graph the y-intercept.
c. What is the slope of the graph of the equation?
d. Use the point you graphed in part (b) and the slope to find another point
on the graph of the equation.
e. Graph the equation.
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5-4
Point-Slope Form
The point-slope form of a nonvertical linear equation is y 2 y1 5 m(x 2 x1). In
this equation, m is the slope and (x1, y1) is a point on the graph of the equation.
Problem
A line passes through (5, 22) and has a slope 23. What is an equation for this line
in point-slope form?
y 2 y1 5 m(x 2 x1)
Use point-slope form.
y 2 (22) 5 23(x 2 5)
y 1 2 5 23(x 2 5)
Substitute (5, 22) for (x1, y1) and 23 for m.
Simplify.
Problem
A line passes through (1, 4) and (2, 9). What is an equation for this line in
point-slope form? What is an equation for this line in slope-intercept form?
First use the two given points to find the slope.
924
5
m52215155
Use the slope and one point to write an equation in
point-slope form.
y 2 y1 5 m(x 2 x1)
Use point-slope form
y 2 4 5 5(x 2 1)
y 2 4 5 5x 2 5
Substitute (1, 4) for (x1, y1) and 5 for m.
Distributive Property
y 5 5x 2 1
Add 4 to each side.
An equation in point-slope form is y 2 4 5 5(x 2 1) . An equation in slopeintercept form is y 5 5x 2 1 .
Exercises
Write an equation for the line through the given point and with the given slope m.
1
1. (21, 3); m 5 24
2. (7, 25); m 5 4
2
3. (22, 25); m 5 3
Write an equation in point-slope form of the line through the given points. Then
write the equation in slope-intercept form.
4. (1, 4) and (2, 7)
5. (2, 0) and (3, 22)
6. (4, 25) and (22, 22)
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Reteaching (continued)
Point-Slope Form
You can use the point-slope form of an equation to help graph the equation. The
point given by the point-slope form provides a place to start on the graph. Plot a
point there. Then use the slope from the point-slope form to locate another point
in either direction. Then draw a line through the points you have plotted.
Problem
What is the graph of the equation y 2 2 5 13(x 2 1)?
The equation is in point-slope form, so the line passes through (1, 2) and has a
1
slope of 3 .
Plot the point (1, 2).
4
2
1
Use the slope, 3 . From (1, 2), go up 1 unit
and then right 3 units. Draw a point.
O
Ź2
y
(1, 2)
(4, 3)
rise 5 1
run 5 3
2
4
x
6
Ź2
Draw a line through the two points.
Ź4
1
21
Because 3 5 23 , you can start at (1, 2) and go down 1 unit and to the left 3 units to
locate a third point on the line.
Exercises
Graph each equation.
7. y 2 3 5 2(x 1 1)
2
8. y 1 2 5 3(x 2 2)
1
9. y 2 4 5 22(x 1 1)
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5-5
Class
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Reteaching
Standard Form
The standard form of a linear equation is Ax 1 By 5 C, where A, B, and C are
real numbers, and A and B are not both zero. You can easily determine the x- and
y-intercepts of the graph from this form of the equation.
Each intercept occurs when one coordinate is 0. When substituting 0 for either of
x or y, one of the terms on the left side of the standard form equation disappears.
This leaves a linear equation in one variable, with a variable term on the left and a
constant on the right. Determining the other coordinate of the intercept requires
only multiplication or division.
Problem
What are the x- and y-intercepts of the graph of 6x 2 9y 5 18?
First find the x-intercept.
6x 2 9y 5 18
6x 2 9(0) 5 18
6x 5 18
x53
Substitute 0 for y.
Simplify.
Divide each side by 6.
Then find the y-intercept.
6x 2 9y 5 18
6(0) 2 9y 5 18
Substitute 0 for x.
29y 5 18
Simplify.
y 5 22
Divide each side by 29.
The x-intercept is 3 and the y-intercept is 22.
Exercises
Find the x- and y-intercepts of the graph of each equation.
1. x 2 y 5 12
2. 3x 1 2y 5 12
3. 27x 1 3y 5 42
4. 8x 2 6y 5 24
5. 5x 2 4y 5 240
6. 24x 1 y 5 28
7. 6x 1 3y 5 230
8. 7x 2 2y 5 28
9. 8x 1 2y 5 232
10. Write an equation in standard form with an x-intercept of 5 and a y-intercept
of 24.
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5-5
Standard Form
You can graph linear equations in standard form by plotting the x- and y-intercepts.
Problem
What is the graph of 2x 2 y 5 4?
Find the intercepts.
2x 2 y 5 4
2x 2 (0) 5 4
2x 5 4
x52
2x 2 y
2(0) 2 y
2y
y
5
5
5
5
4
4
4
24
y
4
2
Ź4 Ź2
The x-intercept is 2, and the y-intercept is 24. Plot the x- and
y-intercepts and draw a line through the points.
(2, 0)
O
2
Ź2
Ź4
(0, 24)
Exercises
Graph each equation using x- and y-intercepts.
11. x 1 y 5 3
12. 2x 2 3y 5 6
13. x 1 2y 5 24
14. 23x 1 4y 5 12
15. 5x 2 3y 5 15
16. 5x 1 2y 5 210
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x
4
Name
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Date
Reteaching
5-6
Parallel and Perpendicular Lines
Nonvertical lines are parallel if they have the same slope and different y-intercepts. The
graphs of y 5 2x 2 6 and y 5 2x 1 3 are parallel because they have the same slope, 2,
but different y-intercepts, 26 and 3.
Problem
What is an equation in slope-intercept form of the line that passes through
3
(8, 7) and is parallel to the graph of y 5 4 x 1 2?
3
3
The slope of y 5 4 x 1 2 is 4 . Because the desired equation is for a line parallel to
3
3
a line with slope 4 , the slope of the parallel line must also be 4 . Use the slope and
the given point in the point-slope form of a linear equation and then solve for y to
write the equation in slope-intercept form.
y 2 y1 5 m(x 2 x1) Start with the point-slope form.
3
3
y 2 7 5 4 (x 2 8)
Substitute (8, 7) for (x1, y1) and 4 for m.
3
y 2 7 5 4x 2 6
Distributive Property
3
y 5 4x 1 1
Add 7 to each side.
3
The graph of y 5 4 x 1 1 passes through (8, 7) and is parallel to the
3
graph of y 5 4 x 1 2.
Exercises
3
2
1. Writing Are the graphs of y 5 5 x 1 3 and y 5 5 x 2 4 parallel? Explain
how you know.
Write an equation in slope-intercept form of the line that passes through the
given point and is parallel to the graph of the given equation.
2. (3, 1); y 5 2x 1 4
3. (1, 3); y 5 7x 1 5
4. (1, 6); y 5 9x 2 5
1
5. (0, 0); y 5 22 y 2 4
2
6. (25, 7); y 5 25 x 2 3
1
7. (6, 6); y 5 3 x 2 1
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Reteaching (continued)
Parallel and Perpendicular Lines
Two lines that are neither horizontal nor vertical are perpendicular if the
5
4
product of their slopes is 21. The graphs of y 5 25 x 2 5 and y 5 4 x 1 4 are
4 5
perpendicular because 25 Q 4 R 5 21.
Problem
What is an equation in slope-intercept form of the line that passes through
(2, 11) and is perpendicular to the graph of y 5 14 x 2 5?
1
1
1
The slope of y 5 4 x 2 5 is 4 . Since 4 (24) 5 21, the slope of the line
perpendicular to the given line is 24.
Use this slope and the given point to write an equation in point-slope form. Then
solve for y to write the equation in slope-intercept form.
y 2 y1 5 m(x 2 x1)
Start with the point-slope form.
y 2 11 5 24(x 2 2)
y 2 11 5 24x 1 8
y 5 24x 1 19
Substitute (2, 11) for (x1, y1) and 24 for m.
Distributive Property
Add 11 to each side.
The graph of y 5 24x 1 19 passes through (2, 11) and is perpendicular to the
1
graph of y 5 4 x 2 5.
Exercises
3
2
8. Writing Are the graphs of y 5 3 x 1 6 and y 5 22 x 2 4 parallel,
perpendicular, or neither? Explain how you know.
Write an equation in slope-intercept form of the line that passes through the
given point and is perpendicular to the graph of the given equation.
9. (5, 23); y 5 5x 1 3
3
12. (6, 0); y 5 2 x 2 6
10. (4, 8); y 5 22x 2 4
11. (22, 25); y 5 x 1 3
13. (5, 3); y 5 5x 1 2
7
14. (7, 1); y 5 22 x 1 6
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Reteaching
Scatter Plots and Trend Lines
A scatter plot is a graph that relates two different sets of data by displaying them
as ordered pairs. A scatter plot can show a trend or correlation, which may be
either positive or negative. Or the scatter plot may show no trend or correlation.
It is often easier to determine whether there is a correlation by looking at a scatter
plot than it is to determine by looking at the numerical data.
If the points on a scatter plot generally slope up to the right, the two sets of data
have a positive correlation. If the points on a scatter plot generally slope down to
the right, the two sets of data have a negative correlation. If the points on a scatter
plot do not seem to generally rise or fall in the same direction, the two sets of data
have no correlation.
Problem
The table below compares the average height of girls at different ages. Make a
scatter plot of the data. What type of correlation does the scatter plot indicate?
Age in years
2
3
4
5
6
7
8
9
10
Height in Inches
34
37
40
42
45
48
50
52
54
Treat the data as ordered pairs. The average height of a
2-year old girl is 34 inches, so one ordered pair is (2, 34).
Plot this point. Then plot (3, 37), (4, 40), (5, 42),
(6, 45), (7, 48), (8, 50), (9, 52), and (10, 54).
Girl’s Growth chart
60
50
40
Notice that the height increases as the age increases.
There is a positive correlation for this data.
30
A trend line is a line on a scatter plot that is drawn near
the points. You can use a trend line to estimate other
values.
10
20
0
0
2
4
6
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8
10 12 14
Age in years
Name
Class
Date
Reteaching (continued)
5-7
Scatter Plots and Trend Lines
Problem
Draw a trend line for the scatter plot in the previous problem. What is the
equation for your trend line? What would you estimate to be the average height
of a girl who is 12 years old?
Draw a line that seems to fit the data. The line drawn for this data goes through
(4, 40) and (8, 50). Use these points to write an equation.
50 2 40
m 5 8 2 4 5 2.5
Use the point-slope form of the line.
y 2 y1 5 m(x 2 x1)
Girl’s Growth chart
60
50
40
y 2 40 5 2.5(x 2 4)
30
y 2 40 5 2.5x 2 10
20
y 5 2.5x 1 30
10
Use this equation to estimate the average height of
12-year old girls.
y 5 2.5(12) 1 30
0
0
2
4
6
8
10 12 14
Age in years
y 5 60
Exercises
Ryan practices throwing darts. From each distance listed below, he throws
10 darts and records how many times he hits the center.
Distance (in feet)
2
5
7
8
10
12
15
Number of Center Hits
10
9
8
6
5
1
2
1. Use the space at the right to make a scatter plot of the data.
2. Describe the type of correlation that is shown in the scatter plot.
3. Draw a trend line.
4. What equation represents your trend line?
5. How many hits do you estimate Ryan would make from 6 feet?
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6-1
Solving Systems by Graphing
Graphing is useful for solving a system of equations. Graph both equations and
look for a point of intersection, which is the solution of that system. If there is no
point of intersection, there is no solution.
Problem
What is the solution to the system? Solve by graphing. Check.
x1y54
2x 2 y 5 2
Solution
y 5 2x 1 4
y 5 2x 2 2
Put both equations into y-intercept form, y 5 mx 1 b.
y 5 2x 1 4
The first equation has a y-intercept of (0, 4).
0 5 2x 1 4
Find a second point by substituting in 0 for y and solve for x.
x54
You have a second point (4, 0), which is the x-intercept.
y 5 2x 2 2
The second equation has a y-intercept of (0, 22).
0 5 2(x) 2 2
Find a second point by substituting in 0 for y and solve for x.
2 5 2x, x 5 1
You have a second point for the second line, (1, 0).
8
y
4
(2, 2)
x
Ź4
Ź8
O
4
8
Ź4
Ź8
Plot both sets of points and draw both lines. The lines appear
to intersect (2, 2), so (2, 2) is the solution.
Check
If you substitute in the point (2, 2), for x and y in your original equations, you can
double-check your answer.
x1y54
2 1 2 0 4,
4 5 43
2x 2 y 5 2
2(2) 2 2 0 2,
2 5 23
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Solving Systems by Graphing
If the equations represent the same line, there is an infinite number of solutions,
the coordinates of any of the points on the line.
Problem
What is the solution to the system? Solve by graphing. Check.
2x 2 3y 5 6
4x 2 6y 5 18
Solution
y 5 23x 2 2
What do you notice about these equations? Using the y-intercepts and
solving for the x-intercepts, graph both lines using both sets of points.
y 5 23x 2 3
Graph equation 1 by finding two points: (0, 22) and (3, 0). Graph
equation 2 by finding two points (0, 23) and (4.5, 0).
8
Is there a solution? Do the lines ever intersect? Lines with the
same slope are parallel. Therefore, there is no solution to this
system of equations.
y
4
x
Ź8
Ź4
O
Ź4
Ź8
Exercises
Solve each system of equations by graphing. Check.
1. 2x 5 2 2 9y
21y 5 4 2 6x
2. 2x 5 3 2 y
y 5 4x 2 12
4. 6y 5 2x 2 14
5. 3y 5 26x 2 3
x 2 7 5 3y
y 5 2x 2 1
7. 2x 1 3y 5 11
8. 3y 5 3x 2 6
x 2 y 5 27
y5x22
3. y 5 1.5x 1 4
0.5x 1 y 5 22
6. 2x 5 3y 2 12
1
3x 5 4y 1 5
1
9. y 5 2x 1 9
2y 2 x 5 1
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8
Name
Class
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Reteaching
6-2
Solving Systems Using Substitution
You can solve a system of equations by substituting an equivalent expression for
one variable.
Problem
Solve and check the following system:
x 1 2y 5 4
2x 2 y 5 3
x 1 2y 5 4
Solution
x 5 4 2 2y
2(4 2 2y) 2 y 5 3
8 2 4y 2 y 5 3
8 2 5y 5 3
8 2 8 2 5y 5 3 2 8
25y 5 25
y51
x 1 2(1) 5 4
x12225422
x52
Check
The first equation is easiest to solve in terms of one
variable.
Get x to one side by subtracting 2y.
Substitute 4 2 2y for x in the second equation.
Distribute.
Simplify.
Subtract 8 from both sides.
Divide both sides by 25 .
You have the solution for y. Solve for x.
Substitute in 1 for y in the first equation.
Subtract 2 from both sides.
The solution is (2, 1) .
Substitute your solution into either of the given linear equations.
x 1 2y 5 4
2 1 2(1) 0 4
Substitute (2, 1) into the first equation.
4 5 43
You check the second equation.
Exercises
Solve each system using substitution. Check your answer.
2. x 2 3y 5 214
1. x 1 y 5 3
x 2 y 5 22
2x 2 y 5 0
4. 4x 1 y 5 8
3. 2x 2 2y 5 10
x 1 2y 5 5
x2y55
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Reteaching (continued)
6-2
Solving Systems Using Substitution
Problem
Solve and check the following system:
x
2 2 3y 5 10
3x 1 4y 5 26
x
2 2 3y 5 10
x
2 5 10 1 3y
Solve
First, isolate x in the first equation.
Add 3y to both sides and simplify.
x 5 20 1 6y
Multiply by 2 on both sides.
3x 1 4y 5 26
Substitute 20 1 6y for x in second equation.
3(20 1 6y) 1 4y 5 26
Simplify.
60 1 22y 5 26
Subtract 60 from both sides.
22y 5 266, y 5 23
Divide by 22 to solve for y.
x
2 2 3(23) 5 10
x
2 1 9 5 10
Substitute 23 in the first equation.
Simplify.
x52
Solve for x.
The solution is (2, 23)..
Check
3(2) 1 4(23) 0 26
26 5 26 3
Now you check the first equation.
Exercises
Solve each system using substitution. Check your answer.
5. 22x 1 y 5 8
6. 3x 2 4y 5 8
3x 1 y 5 22
2x 1 y 5 9
7. 3x 1 2y 5 25
8. 6x 2 5y 5 3
2x 1 3y 5 26
x 2 9y 5 25
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Reteaching
6-3
Solving Systems Using Elimination
Elimination is one way to solve a system of equations. Think about what the word
“eliminate” means. You can eliminate either variable, whichever is easiest.
Problem
4x 2 3y 5 24
22x 1 3y 5 34
Solution The equations are already arranged so that like terms are in columns.
Solve and check the following system of linear equations.
Notice how the coefficients of the y-variables have the opposite sign and the same
value.
4x 2 3y 5 24
2x 1 3y 5 34
6x 5 30
x55
4(5) 2 3y 5 24
20 2 3y 5 24
23y 5 224
y58
Add the equations to eliminate y.
Divide both sides by 6 to solve for x.
Substitute 5 for x in one of the original equations
and solve for y.
The solution is (5, 8).
Check
4x 2 3y 5 24
4(5) 2 3(8) 0 24
20 2 24 0 24
24 5 24 ✓
Substitute your solution into both of
the original equations to check.
You can check the other equaton.
Exercises
Solve and check each system.
2. 6x 2 3y 5 214
1. 3x 1 y 5 3
6x 2 y 5 22
23x 1 y 5 3
3. 3x 2 2y 5 10
4. 4x 1 y 5 8
x 2 2y 5 6
x1y55
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Reteaching (continued)
Solving Systems Using Elimination
If none of the variables has the same coefficient, you have to multiply before you
eliminate.
Problem
Solve the following system of linear equations.
22x 1 3y 5 21
5x 1 4y 5 6
Solution
5(22x 2 3y) 5 (21)5
2(5x 1 4y) 5 (6)2
210x 2 15y 5 25
10x 1 8y 5 12
27y 5 7
y 5 21
5x 1 4(21) 5 6
Multiply the first equation by 5 (all terms, both sides) and
the second equation by 2. You can eliminate the x variable
when you add the equations together.
Distribute, simplify and add.
Divide both sides by 7.
Substitute –1 in for y in the second equation to find the
value of x.
5x 2 4 5 6
5x 5 10
x52
Simplify.
Add 4 to both sides.
Divide by 5 to solve for x.
The solution is (2, 21).
22x 1 3y 5 21
Check
Substitute your solution into both original equations.
22(2) 2 3(21) 0 21
23 5 23 ✓
You can check the other equation.
Exercises
Solve and check each system.
5.
x 2 3y 5 23
22x 1 7y 5 10
6. 22x 2 6y 5 0
3x 1 11y 5 4
7. 3x 1 10y 5 5
8. 4x 1 y 5 8
7x 1 20y 5 11
x1y55
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Applications of Linear Systems
You can solve systems of linear equations by graphing, substitution, or
elimination. Deciding which method to use depends on the exactness needed and
the form of the equations.
Problem
You just bought a coffee shop for $153,600. The prior owner had an average
monthly revenue of $8600 and an average monthly cost of $5400. If your monthly
costs and revenues remain the same, how long will it take you to break even?
Write equations for revenue and costs, including the price you paid for the shop,
after t months. Then solve the system by graphing.
y 5 8600t
Equation for revenue
y 5 5400t 1 153,600 Equation for cost
It appears that the point of intersection is where
t is equal to 48 months. Substitute t 5 48 into either
equation to find the other coordinate (y), which is 412.8.
Therefore, your breakeven point is after you have run the
shop for 48 months, at which point your revenue and cost
are the same: $412,800.
y
512
384
256
128
t
O
Problem
16
32
48
64
A perfume is made from t ounces of 15% scented Thalia and b ounces of 40% Thalia. You
want to make 60 oz of a perfume that has a 25% blend of the Thalia. How many ounces of
each concentration of Thalia are needed to get 60 oz of perfume that is 25% strength of
Thalia?
60(0.25) 5 0.15t 1 0.4b
Write your systems of equations:
60 5 t 1 b
Solve the system by using substitution:
60(0.25) 5 0.15t 1 0.4b
Solve the second equation for t and substitute in the first equation.
15 5 0.15(60 2 b) 1 0.4b
15 5 9 2 0.15b 1 0.4b
Substitute 60 2 b for t in the first equation.
24 5 b
Solve for b.
Distributive property
Substitute 24 for b in second equation to find that t = 36. The answer is (36, 24).
The blend requires 36 oz of the 15% perfume and 24 oz of the 25% perfume.
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Applications of Linear Systems
Exercises
1. You have a coin bank that has 275 dimes and quarters that total $51.50. How
many of each type of coin do you have in the bank?
2. Open-Ended Write a break-even problem and use a system of linear
equations to solve it.
3. You earn a fixed salary working as a sales clerk making $11 per hour. You get a
weekly bonus of $100. Your expenses are $60 per week for groceries and $200
per week for rent and utilities. How many hours do you have to work in order
to break even?
4. Reasoning Find A and B so that the system below has the solution (1, 21).
Ax 1 2By 5 0
2Ax 2 4By 5 16
5. You own an ice cream shop. Your total cost for 12 double cones is $24 and you
sell them for $2.50 each. How many cones to you have to sell to break even?
6. Multi-Step A skin care cream is made with vitamin C. How many ounces of
a 30% vitamin C solution should be mixed with a 10% vitamin C solution to
make 50 ounces of a 25% vitamin C solution?
• Define the variables.
• Make a table or drawing to help organize the information.
7. Your hot-air balloon is rising at the rate of 4 feet per second. Another aircraft
nearby is at 7452 feet and is losing altitude at the rate of 30 feet per second.
In how many seconds will your hot-air balloon be at the same altitude as the
other aircraft?
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Systems of Linear Inequalities
A system of linear inequalities is a set of linear inequalities in the same plane. The
solution of the system is the region where the solution regions of the inequalities
of the system overlap.
Problem
What is the graph of the system of linear inequalities:
x 2 y . 21
?
y # 2x 1 3
Put the first inequality into slope-intercept form, y , x 1 1. Use a dashed line
since , does not include the points on the boundary line in the solution. Using
the point (0, 0), decide where to shade the first inequality. The point (0, 0) makes
the inequality true, so shade the region including (0, 0).
Then graph the boundary line of the second inequality, y # 2x 1 3. It is a solid
line because of the # sign. Use the point (0, 0) to decide where to shade the
second inequality. The point (0, 0) makes the second inequality true, so shade the
region including (0, 0).
8
y
4
x
Ź8
Ź4
O
4
Ź4
Ź8
8
The overlapping region of the 2 inequalities is the
solution to the system. It includes the points (0, 0),
(1, 1), (3, 1). You can test any point in the region in
both equations to see if it makes both equations true.
In word problems, the solutions often cannot be
negative (cars, tickets sold, etc.). Two requirements
are that x $ 0 and y $ 0. Keep this in mind when
graphing word problems.
Problem
A cash register has fewer than 200 dimes and quarters worth more than $39.95.
How many of each coin are in the register?
type of coin
quantity
value of coin
value in cents
quarters
q
$0.25
25q
dimes
d
$0.10
10d
TOTAL
200
3995
The system of inequalities that you get from the table is:
q 1 d , 200
25q 1 10d . 3995
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Systems of Linear Inequalities
Using elimination, solve for q by multiplying all terms in the first equation by 210
and eliminating d: (q 1 d , 200)(210).
210q 2 10d . 22000
Now add the 2 systems together to solve for q.
25q 1 10d . 3995
15q . 1995
q . 133
q 1 d , 200
Write first inequality.
133 1 d , 200, d , 67
Substitute in 133 for q, subtract 133 from both sides
and solve for d.
The register contains at least 133 quarters and no more than 67 dimes.
Exercises
Graph the following systems of inequalities.
1. x 2 2y , 3
y
2 . 3x 1 6
2. y $ 2x 1 5
x
4. 3y $ 4
5. 2x 2 y , 1
2y # x 1 2
2x # 22y 2 3
x 1 2y , 24
3. x 1 3y $ 24
3x 2 2y , 5
6. 5x 2 4y $ 3
2x 1 3y # 22
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Zero and Negative Exponents
For every nonzero number a, a0 5 1.
1
For every nonzero number a and integer n, a2n 5 n . In other words, when the
a
exponent is negative, raise the reciprocal of the base to the opposite of the
exponent.
Problem
What is the simplified form of each expression?
a. 3.90 5 1
Since the exponent is 0 but the base of the expression is 3.9, which is not
0, the expression has a value of 1.
1
b. 922 5 2
9
5
The exponent is negative, so raise the reciprocal of 9, or 19 , to the
exponent 2(22), or 2.
1
81
Simplify.
Problem
What is the simplified form of
7b23
7
5 2 ? b23
2
a
a
7
1
5 2? 3
a
b
7
5 2 3
a b
7b23
using only positive exponents?
a2
Rewrite the expression as a product of factors with positive exponents and
factors with negative exponents.
Rewrite the factor with the negative exponent by raising the reciprocal of
the base to a positive exponent.
Simplify by multiplying.
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Zero and Negative Exponents
Exercises
Write each expression as an integer, a simple fraction, or an expression that
contains only positive exponents. Simplify.
1. 2.30
2. 1024
3. 2a25
4. 113.70
5. 1921
6.
7. (7q)21
8. a2 b
323
p
7 22
8
9. 1.8c0
10. (29.7)0
Write each expression so that it contains only positive exponents. Simplify.
11. 2623
12. 22rs25
13. 7x28 y0
14. a
15. (28v)22 w3
16.
223
m0n21
17. (3xy)0 z
18.
2323
uv 22
5a 22
b
3b
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Scientific Notation
Scientific notation is used to write very large numbers and very small numbers in
a more compact form.
Scientific notation makes use of the fact that our number system is a base 10
system.
The number 10,000 can be expanded as 10 3 10 3 10 3 10. It can be written as
a power of 10 as 104 . The exponent 4 is the number of places the decimal point
moves to the left when the number is written in scientific notation. In scientific
notation, 10,000 is written as 1 3 104 . The number 0.0001 can be written as a
power of 10 as 1024 .
Problem
What is 0.0034 written in scientific notation?
In scientific notation, numbers are written in the form a 3 10n, where a is at
least 1, but less than 10. Move the decimal point 3 places to the right. The
exponent of 10 in scientific notation will be 23. Since 0.0034 is smaller
than 3.4, the exponent must be negative.
0.0034 5 3.4 3 1023
Problem
What is 35,100,000 written in scientific notation?
To write 35,100,000 in scientific notation first identify a. The value of a must be
greater than or equal to 1 and less than 10. So a 5 3.51. Now determine what
power of 10 you need to multiply a by to get 35,100,000.
3.51 3 10,000,000 5 35,100,000
Because 10,000,000 5 107, 35,100,000 5 3.51 3 107 .
To change a number from scientific notation to standard notation, start with the
value of a. Then move the decimal point to the left or right depending on the
exponent of 10. For example, 4.72 3 104 5 47,200. The decimal point is moved
4 places to the right, and zeros are added as placeholders.
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Scientific Notation
For each number in standard notation, first identify a and then write the
number in scientific notation.
1. 4300
2. 0.0029
3. 87,000,000
4. 0.0000402
5. 834,000
6. 0.0000090
For each number in scientific notation, rewrite in standard notation.
7. 8.3 3 105
10. 6.1 3 1022
8. 4.01 3 1028
11. 5.83 3 100
9. 5.11 3 1011
12. 3.6 3 10212
13. Error Analysis A student knew that the number 478.2 3 105 was not written
in scientific notation, but was having trouble explaining why. Using the
conditions placed on the value of a, explain why 478.2 3 105 is not written
in scientific notation. Then rewrite the expression so that it is in scientific
notation.
14. Reasoning A classmate suggests that instead of using scientific notation to
write very large and very small numbers, it would be easier to use octagonal
notation with numbers written in the form a ? 8b . Do you agree? Explain.
15. Open-Ended Describe three situations in which it is easier to use scientific
notation than standard notation. Give examples for each situation.
16. Error Analysis Two students came up with the answers 3.8 ? 105 km and
3.8 ? 108 m for the same problem. Could they both be right? Explain. What if
the answers were 2.5 ? 104 mi and 2.5 ? 107 ft?
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Multiplying Powers With the Same Base
When multiplying powers with the same base, you add the exponents. This is true
for numerical and algebraic expressions.
Problem
What is each expression written as a single power?
a. 34 ? 32 ? 33
All three powers have the same base, so this expression can be written as a single
power by adding the exponents.
34 ? 32 ? 33 5 34 1 2 1 3
All powers have the same base. Add the exponents.
5 39
Simplify the exponent.
34 represents 4 factors of 3, 32 represents 2 factors of 3, and 33 represents 3 factors
of 3. This is a total of 9 factors of 3, so the answer is reasonable.
Even when some of the exponents are negative, exponents can be added when the
bases are the same in a product of powers.
b. 1123 ? 114 ? 1125
1123 ? 114 ? 1125 5 1123 1 4 1 (25)
5 1124
All powers have the same base. Add the exponents.
Simplify the exponent.
Problem
What is the simplified form of (1.8 3 1011)(2.7 3 108)? Write the answer in
scientific notation.
Use the Associative and Commutative Properties of Multiplication to regroup and
reorder the factors so that the powers of 10 are grouped together and numbers
that are not powers of 10 are grouped separately from the powers of 10.
(1.8 3 1011)(2.7 3 108) 5 (1.8 ? 2.7)(1011 ? 108) Associative and Commutative Prop. of Mult.
5 (4.86)(1011 1 8)
Multiply the numbers in the first set of
parentheses. Add the exponents for the powers
of 10.
5 4.86 3 1019
Simplify the exponent.
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Multiplying Powers With the Same Base
Exercises
Simplify each expression.
1. a2a3
2. 3n3n5
3. 8k3 ? 3k6
4. (8p5)(6p4)
5. 21d7 ? 2d3
6. (26.1m4)(3m2)
7. h5 ? h2 ? h10
8. (29q28)(6q11)
9. (16r27)(22r)
10. (y3z13)(y2z26)
11. (23x2)(5w 8)(4x3)
12. (15fg 2)(f 3g23)(28f21g 6)
13. m26 ? m3 ? n22
14. 26j23k ? 7jk5
15. 22uvw21 ? 3u2v22w
Simplify each expression. Write each answer in scientific notation.
16. (4 3 103)(2 3 105)
17. (1 3 104)(6 3 103)
18. (7 3 102) ? 105
19. (8 3 109)(3 3 1025)
20. (2 3 105)(5 3 106)
21. (7 3 1028)(3 3 1026)
Write each answer in scientific notation.
22. The distance light travels in one year (one light-year) is about
5.87 3 1012 mi. A star called Proxima Centauri is 4.2 light-years away from
Earth. About how many miles from Earth is Proxima Centauri?
23. After the Revolutionary War, the U.S. national debt was approximately
7.5 3 107 dollars. In 2008, the debt was approximately 1.33 3 105 times the
original amount. What was the national debt in 2008?
Complete each equation.
24. 4u ? 43 5 413
25. 86 ? 85 5 8u
26. 34 ? 3u 5 310
27. k11 ? ku 5 k2
28. w u ? w 5 w4
29. x2 ? x u ? x 5 x9
30. p25 ? pu 5 p3 ? p2
31. n5 ? n217nu 5 n13
32. t5u2 ? t uuu 5 t24u3
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More Multiplication Properties of Exponents
When a power is raised to another power, like (xy)z, multiply the exponents.
Problem
What is the simplified form of (d3)4?
(d3)4 5 d3?4
5 d12
The expression is a power, d3 , raised to another power, 4. Multiply the exponents.
Simplify.
Simplifying powers may require you to use multiple properties of exponents. You
should follow the order of operations when simplifying exponential expressions.
Problem
What is the simplified form of (n23)6n4?
Using the order of operations, first simplify the power (n23)6 .
(n23)6n4 5 (n23?6)n4 5 n218n4
Next, multiply. The two powers have the same base, so simplify by adding the exponents.
n218n4 5 n21814 5 n214
Finally, write the expression using positive exponents. Rewrite the expression using the
reciprocal of the base and the opposite of the exponent.
1
n214 5 14
n
You should follow the same rules when simplifying numbers written in scientific
notation raised to a power.
Problem
What is the simplified form of (4.2 3 10 27)2 written in scientific notation?
(4.2 3 1027)2 5 (4.2)2(1027)2
This is a product raised to the exponent 2, so each factor of the
product must be raised to the exponent 2.
5 17.64 3 10214
Multiply 4.2 by itself. Multiply the exponents on the expression
with base 10.
5 1.764 3 10213
Move the decimal point one place left and adjust the exponent on
10 to write in scientific notation.
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More Multiplication Properties of Exponents
Exercises
Simplify each expression.
1. (y2)3
2. (v9)6
3. (h4)5
4. (n4)11
5. (p21)5
6. (z3)26
7. (x24)25x
8. (f 5)21f 8
9. (3a)4
13. (2y25)3(x11y210)2
10. (6c)23
11. (7k)0
12. (10s23)2
14. u29(u21v)4u25
15. (x13y6)22(y25x10)6 16. 4m0n0(6m5)2
Simplify. Write each answer in scientific notation.
17. (2 3 1028)3
18. (3 3 105)3
19. (9 3 10215)3
20. (6 3 105)2
21. (6.7 3 1011)2
22. (9.5 3 107)3
23. (4.7 3 10211)22
24. (5.14 3 106)2
25. The radius of a cylinder is 6.8 3 105 m. The height of the cylinder is 2.2 3 103
m. What is the volume of the cylinder? (Hint: V 5 3.14r 2h)
Complete each equation.
26. (y3)u 5 y6
27. (6p3qu)2 5 36p6
28. (4au)3 5 64a26
29. (k11)u 5 1
30. (t28)u 5 t16
31. 15(c21)u 5 15c10
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Division Properties of Exponents
Understanding division properties of exponents allows you to simplify
quotients involving exponents.
Problem
What is 66 divided by 64 ?
Method 1: Evaluate 66 and 64 , and then divide the results.
66 5 46,656
64 5 1296
46,656 ÷ 1296 = 36
Method 2: Expand the numerator and denominator.
66
6?6?6?6?6?6
5
6?6?6?6
64
After dividing out the common factors, you are left with 6 3 6 5 36 .
When you divide powers with the same base, subtract the exponents. In the
example above, 66 and 64 are powers with the same base and when you divided
them, the result was 36 5 62. This is the same result you get by subtracting the
66
exponents: 4 5 6624 5 62.
6
The division property of exponents also allows you to simplify quotients that
contain variables.
Problem
x5
How can you use the division property of exponents to show that x2 5 3 when x u 0?
x
Expand the numerator and denominator.
x5
x?x?x?x?x
5
3
x?x?x
x
After dividing out the common factors you are left with x ? x 5 x2 .
Division properties of exponents work whether the bases in the problem are
constants or variables. When you divide powers with the same base, subtract the
exponents. In this example, x5 and x3 are powers with the same base and when
you divided them, the result was x2 5 x523 .
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Division Properties of Exponents
Simplify each expression.
1.
75
72
2.
39
32
3.
52
5
4.
4z
44
5.
m4
m2
6.
7.
r3
r
8.
9.
a3
a5
10.
p6
p5
x5y4
x3y
10x5
15x2
11. Use properties of exponents to show that a0 5 1. (Hint: Write the quotient of
two powers that have a as their base and have the same exponent.)
12. Compare multiplying and dividing powers with the same base.
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Adding and Subtracting Polynomials
You can add and subtract polynomials by lining up like terms and then adding or
subtracting each part separately.
Problem
What is the simplified form of (3x2 2 4x 1 5) 1 (5x2 1 2x 2 8)?
Write the problem vertically, lining up the like terms.
Then add each pair of like terms.
Solve
3x2 2 4x 1 5
1 5x2 1 2x 2 8
Add the x2 terms.
Add the x terms.
Add the constant terms.
3x2 1 5x2 5 8x2
24x 1 2x 5 22x
5 1 (28) 5 23
3x2 2 4x 1 5
1 5x2 1 2x 2 8
8x2 2 2x 2 3
Check
Add the sums.
Check your solution using subtraction.
8x2 2 5x2 5 3x2
22x 2 2x 5 24x
23 2 (28) 5 5
Solution: (3x2 2 4x 1 5) 1 (5x2 1 2x 2 8) 5 8x2 2 2x 2 3
Exercises
Simplify.
1.
5b2 1 3b
1 2b2 2 5b
2.
3c2 1 3c
1 4c2 1 2c
3.
4d2 2 3d 1 6
1 2d2 1 5d 2 3
4.
23e2 2 5e 1 2
1 e2 1 2e 2 7
5.
4f 3 1 2f 2 1 5f
1 2f 3 2 4f 2 2 3f
6.
5g3 2 2g2 1 3g
1 2g3 1 5g2 2 2g
7. (3h2 1 5) 1 (25h2 2 3)
8. (2j2 1 4j 2 6) 1 (4j2 2 3j 2 3)
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Adding and Subtracting Polynomials
To subtract polynomials, follow the same steps as in addition.
Problem
What is the simplified form of (6x3 1 4x2 2 3x) 2 (2x3 1 3x2 2 5x)?
6x3 1 4x2 2 3x
2 (2x3 1 3x2 2 5x)
Write the problem vertically, lining up the like terms.
Then subtract each pair of like terms.
Solve
Subtract the x3 terms.
Subtract the x2 terms.
6x3 2 2x3 5 4x3
4x2 2 3x2 5 x2
6x3 1 4x2 2 3x
2 (2x3 1 3x2 2 5x)
4x3 1 x2 1 2x
Check
Subtract the x terms.
23x 2 (25x) 5 2x
Add the differences.
Check your solution using subtraction.
4x3 1 2x3 5 6x3
x2 1 3x2 5 4x2
2x 1 (25x) 5 23x
Solution: (6x3 1 4x2 2 3x) 2 (2x3 1 3x2 2 5x) 5 4x3 1 x2 1 2x
Exercises
Simplify.
9.
12.
4k2 1 5k
2 (3k2 1 2k)
10.
5m2 2 4m
2 (2m2 1 3m)
11.
7n2 1 4n 1 9
2 (4n2 1 3n 1 5)
5p2 1 6p 1 4
2 (7p2 1 4p 1 8)
13.
3q3 1 2q2 1 7q
2 (6q3 2 4q2 2 5q)
14.
2r3 2 2r2 1 5r
2 (4r3 1 5r2 1 3r)
15. (6s2 2 5s) 2 (22s2 1 3s)
16. (3w2 1 6w 2 5) 2 (5w2 2 4w 1 2)
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Multiplying and Factoring
You can multiply a monomial and a trinomial by solving simpler problems. You
can use the Distributive Property to make three simpler multiplication problems.
Problem
What is the simplified form of 3x(2x2 1 4x 2 1)?
Use the Distributive Property to rewrite the problem as three separate
multiplication problems.
3x(2x2 1 4x 2 1) 5 (3x ? 2x2) 1 (3x ? 4x) 1 (3x ? (21))
Remember that to multiply with exponents that you add them.
Solve
Check
3x ? 2x2 5 6x3
Multiply inside the first pair of parentheses.
3x ? 4x 5 12x2
Multiply inside the second pair of parentheses.
3x ? (21) 5 23x
Multiply inside the third pair of parentheses.
6x3 1 12x2 2 3x
Add the products.
6x3 4 2x2 5 3x
Check your solution using division.
12x2 4 4x 5 3x
23x 4 (21) 5 3x
Solution: 3x(2x2 1 4x 2 1) 5 6x3 1 12x2 2 3x
Exercises
Simplify each product.
1. 4x(2x 2 7)
2. 3y(3y 1 4)
3. 2z2(2z 2 3)
4. 3a(24a 2 6)
5. 6b(3b2 1 2b 2 4)
6. 3c2(2c2 2 4c 1 3)
7. 22d(4d2 1 3d 2 2)
8. 5e2(23e2 2 2e 2 3)
9. 4f(23f 3 1 2f 2 1 6)
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Multiplying and Factoring
To factor a polynomial, find the greatest common factor (GCF) of the coefficients and
constants and also the GCF of the variables.
Problem
What is the factored form of 8x4 1 12x2 2 16x?
Solve
Find the GCF of the coefficients. Use prime factorization.
852?2?2
12 5 2 ? 2 ? 3
16 5 2 ? 2 ? 2 ? 2
The GCF of the numbers is 4.
Each term has a variable. Remember, x 5 x1 .
The GCF is the least exponent.
The GCF of the variables is x.
Combine the GCFs.
The GCF is 4x.
Factor out the GCF of each term.
Check
4(2 1 3 2 4)
Factor the coefficients.
4x(2x3 1 3x 2 4)
Insert the variables.
4x(2x3 1 3x 2 4) 5 8x4 1 12x2 2 16x
Check by multiplying.
Solution: The factored form of 8x4 1 12x2 2 16x is 4x(2x3 1 3x 2 4).
Exercises
Find the GCF of the terms of each polynomial.
10. 12x2 2 6x
11. 4y2 1 12y 1 8
12. 6z3 1 15z2 2 9z
13. 8a 1 10
14. 12b2 2 18b
15. 9c3 1 12c2
16. 5d3 2 10d2 1 20d
17. 6e2 1 10e 2 8
18. 8g3 2 24g2 1 16g
Factor each polynomial.
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Multiplying Binomials
You can multiply binomials by using the FOIL method. FOIL stands for First,
Outer, Inner, and Last.
Problem
What is the simplified form of (4x 1 3)(2x 1 6)?
Use the FOIL method to simplify the binomial.
Solve
4x ? 2x 5 8x2
Multiply the First terms.
4x ? 6 5 24x
Multiply the Outer terms.
3 ? 2x 5 6x
Multiply the Inner terms.
3 ? 6 5 18
Multiply the Last terms.
8x2 1 24x 1 6x 1 18
Add the products.
8x2 1 30x 1 18
Add the like terms.
Check
Substitute any number for x. Try x 5 2 . If the two sides of the
equation are equal the simplification may be correct.
(4x 1 3)(2x 1 6) 0 8x2 1 30x 1 18
(4 ? 2 1 3)(2 ? 2 1 6) 0 (8 ? 22) 1 (30 ? 2) 1 18
(11)(10) 0 32 1 60 1 18
110 5 110 3
Solution: The simplified form of (4x 1 3)(2x 1 6) is 8x2 1 30x 1 18.
Exercises
Simplify each product.
1. (a 1 6)(a 2 3)
2. (b 2 4)(b 1 5)
3. (c 1 3)(c 1 7)
4. (2d 1 4)(3d 2 2)
5. (4e 2 5)(3e 1 3)
6. (3f 2 2)(2f 2 4)
7. (5g 1 3)(g 2 3)
8. (4h 1 4)(2h 1 5)
9. (3j 2 5)(4j 2 3)
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Multiplying Binomials
To multiply a trinomial by a binomial, use the same steps as you would to multiply
a 3-digit number by a 2-digit number. Find the partial products for each term of the
binomial and then add the like terms of the partial products.
Problem
What is the simplified form of (2x2 1 3x 2 4)(3x 1 2)?
Solve
Start by arranging the polynomials vertically.
Multiply each part of the trinomial by 2.
2x2 1 3x 2 4
2x2 1 3x 1 2
4x2 1 6x 2 8
2x2 ? 2 5 4x2
3x ? 2 5 6x
24 ? 2 5 28
Multiply each part of the trinomial by 3x.
6x3 1 2x2 1 3x 2 4
6x2 1 4x2 1 3x 1 2
6x3 1 4x2 1 6x 2 8
6x3 1 9x2 2 12x 2 8
2x2 ? 3x 5 6x3
3x ? 3x 5 9x2
24 ? 3x 5 212x
Add the partial products.
6x3 1 4x2 1 6x 2 8
6x3 1 9x2 2 12x 2 8
6x3 1 13x2 2 6x 2 8
Check
Substitute any number for x. Try x 5 2 . If the two sides of the
equation are equal, the simplification may be correct.
(2x2 1 3x 2 4)(3x 1 2) 0 6x3 1 13x2 2 6x 2 8
(8 1 6 2 4)(6 1 2) 0 48 1 52 2 12 2 8
80 5 80 3
Solution: The simplified form of (2x2 1 3x 2 4)(3x 1 2) is 6x3 1 13x2 2 6x 2 8.
Exercises
Simplify each product.
10. (w2 1 3w 2 4)(2w 1 3)
11. (x2 2 8x 1 6)(3x 2 4)
12. (2y2 1 4y 2 5)(4y 1 2)
13. (3z2 2 6z 1 4)(4z 1 1)
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8-4
Class
Date
Reteaching
Multiplying Special Cases
A binomial is squared when it is multiplied by itself. The square of a binomial is
the square of the first term plus the twice the product of the two terms plus the
square of the last term. This can be expressed as (a 1 b)2 5 a2 1 2ab 1 b2 .
Problem
What is the simplified form of (x 1 5)2 ?
Use the rules for squaring a binomial.
Solve
x ? x 5 x2
Square the first term.
2(5 ? x) 5 10x
Multiply the product of the two terms by 2.
5 ? 5 5 25
Square the last term.
So, (x 1 5)2 5 x2 1 10x 1 25.
Check
(x 1 5)2 5 (x 1 5)(x 1 5)
Rewrite the binomials.
x ? x 5 x2
Multiply the First addends.
x ? 5 5 5x
Multiply the Outer addends.
5 ? x 5 5x
Multiply the Inner addends.
5 ? 5 5 25
Multiply the Last addends.
x2 1 5x 1 5x 1 25
Add the products.
x2 1 10x 1 25
Combine the like terms.
Solution: The simplified form of (x 1 5)2 is x2 1 10x 1 25.
Exercises
Simplify each expression.
1. (a 1 7)2
2. (b 2 4)2
3. (2c 1 3)2
4. (3d 2 5)2
5. (4e 1 1)2
6. (2f 2 6)2
7. (g 2 10)2
8. (5h 1 8)2
9. (3j 2 3)2
10. (2k 1 4)2
11. (4m 2 2)2
12. (3n 1 6)2
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Reteaching (continued)
Multiplying Special Cases
The product of the sum and the difference of the same two terms produces a pattern
that can be expanded algebraically as (a 1 b)(a 2 b) 5 a2 2 ab 1 ab 2 b2 . The
sum of the two ab- terms is 0. Therefore, (a 1 b)(a 2 b) 5 a2 2 b2 . The product is
the square of the first term minus the square of the last term.
Problem
What is the simplified form of (2x 2 3)(2x 1 3)?
Use the rules for finding the product of the sum and the difference of the same two terms.
Solve
2x ? 2x 5 4x2
Square the first term.
3? 359
Square the last term.
Remember, the product is the difference of the two squares.
The product is 4x2 2 9.
Check
Multiply the binomials using the FOIL Method.
2x ? 2x 5 4x2
Multiply the First addends.
2x ? 3 5 6x
Multiply the Outer addends.
23 ? 2x 5 26x
Multiply the Inner addends.
23 ? 3 5 29
Multiply the Last addends.
4x2 1 6x 2 6x 2 9
Add the products.
4x2 2 9
Combine the like terms.
Solution: The simplified form of (2x 2 3)(2x 1 3) is 4x2 2 9.
Exercises
Simplify each product.
13. (p 2 4)(p 1 4)
14. (q 1 5)(q 2 5)
15. (3r 1 2)(3r 2 2)
16. (4s 2 6)(4s 1 6)
17. (2t 2 1)(2t 1 1)
18. (5u 2 3)(5u 1 3)
19. (6v 2 4)(6v 1 4)
20. (3w 2 8)(3w 1 8)
21. (7x 2 9)(7x 1 9)
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Reteaching
Factoring x2 1 bx 1 c
If a trinomial of the form x2 1 bx 1 c can be written as the product of two
binomials, then:
• The coefficient of the x-term in the trinomial is the sum of the constants in the
binomials.
• The trinomial’s constant term is the product of the constants in the binomials.
Problem
What is the factored form of x2 1 12x 1 32?
To write the factored form, you are looking for two factors of 32 that have a sum of 12.
Solve
Make a table showing the factors of 32.
Factors of 32
Sum of Factors
1 and 32
2 and 16
4 and 8
33
18
12
x2 1 12x 1 32 5 (x 1 4)(x 1 8)
Check
(x 1 4)(x 1 8)
x2 1 8x 1 4x 1 32
Use FOIL Method.
1 12x 1 32
Combine the like terms.
Solution: The factored form of x2 1 12x 1 32 is (x 1 4)(x 1 8).
x2
Exercises
Factor each expression.
1. x2 1 9x 1 20
2. y2 1 12y 1 35
3. z2 1 8z 1 15
4. a2 1 11a 1 28
5. b2 1 10b 1 16
6. c2 1 12c 1 27
7. d2 1 6d 1 5
8. e2 1 15e 1 54
9. f 2 1 11f 1 24
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Reteaching (continued)
Factoring x2 1 bx 1 c
Some factorable trinomials in the form of x2 1 bx 1 c will have negative
coefficients. The rules for factoring are the same as when the x-term and the
constant are positive.
• The coefficient of the x-term of the trinomial is the sum of the constants in the
binomials.
• The trinomial’s constant term is the product of the constants in the binomials.
However, one or both constants in the binomial factors will be negative.
Problem
What is the factored form of x2 2 3x 2 40?
To write the factored form, you are looking for two factors of 240 that have a sum
of 23. The negative constant will have a greater absolute value than the positive
constant.
Solve
Make a table showing the factors of 240.
Factors of ]40
1 and Ľ40
2 and Ľ20
4 and Ľ10
5 and Ľ8
Sum of Factors
Ľ39
Ľ18
Ľ6
Ľ3
x2 2 3x 2 40 5 (x 2 8)(x 1 5)
Check
(x 2 8)(x 1 5)
x2 1 5x 2 8x 2 40
Use FOIL Method.
x2 1 (23x) 2 40
Combine the like terms.
Solution: The factored form of x2 2 3x 2 40 is (x 2 8)(x 1 5).
Exercises
Factor each expression.
10. s2 1 2x 2 35
11. t2 2 4t 2 32
12. u2 1 6u 2 27
13. v2 2 2v 1 48
14. w2 2 8w 2 9
15. x2 1 3x 2 18
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8-6
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Reteaching
Factoring ax2 1 bx 1 c
You can use your knowledge of prime numbers to help you factor some trinomials as two
binomials. A prime number has only 1 and itself as factors. For trinomials of the form
ax2 1 bx 1 c, if a is a prime number then you already know the first term of each
binomial: ax and 1x. Then list the factors that will multiply to produce c. Use guess and check
to find the factor pair that will add to b.
Problem
What is the factored form of 7x2 1 31x 1 12?
7x2 1 31x 1 12 5 (7x
5 (7x
7x2 1 31x 1 12 5 (7x 1
)(1x
)
a is 7, which is prime, so the factors are 7 and 1.
)(x
)
You don’t need the 1 in front of the variable, so drop it.
)(x 1
) The trinomial has two plus signs, so the
binomials also have plus signs.
Because c is 12, find factor pairs that multiply to 12: (1 and 12), (2 and 6), (3 and 4).
Try each pair in the expression to see if the INNER and OUTER products add to b, or 31.
(7x 1 1)(x 1 12) 5 7x2 1 x 1 84x 5 7x2 1 85x 1 12 (NO)
(7x 1 2)(x 1 6) 5 7x2 1 2x 1 42x 5 7x2 1 44x 1 12 (NO)
(7x 1 3)(x 1 4) 5 7x2 1 3x 1 28x 5 7x2 1 31x 1 12 (YES)
The factored form of 7x2 1 31x 1 12 is (7x 1 3)(x 1 4).
Exercises
Factor each expression.
1. 3x2 1 14x 1 8
2. 5y2 1 43y 1 24
3. 2z2 1 19z 1 42
4. 11a2 1 39a 1 18
5. 13b2 1 58b 1 24
6. 23c2 1 56c 1 20
7. 7d2 1 d 2 8
8. 3e2 1 20e 2 32
9. 19f 2 1 10f 2 9
10. 5s2 2 18s 1 16
11. 17t2 2 12t 2 5
12. 29u2 1 48u 2 20
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Reteaching (continued)
Factoring ax2 1 bx 1 c
If you are given the area and one side of a rectangle, you can find the second side by
factoring the trinomial. One binomial is the width and the other binomial is the length.
Problem
The area of a rectangular swimming pool is 6x2 1 11x 1 3. The width of the pool
is 2x 1 3. What is the length of the pool?
You are given the area and length of the pool. Set up an equation with what you
are given and solve or length.
6x2 1 11x 1 3 5 (2x 1 3)( uuu )
Area = length × width.
6x2 1 11x 1 3 5 (2x 1 3)(3x uu )
6x2 5 (2x)(3x), so the first term of the second
binomial is 3x.
6x2 1 11x 1 3 5 (2x 1 3)(3x 1 u )
The trinomial has two plus signs, so the sign for
the second binomial must also be plus.
6x2 1 11x 1 3 5 (2x 1 3)(3x 1 1)
The value of c is 3. Since 3 5 3 3 1 , the second
term must be 1.
Multiply to check your answer. Use FOIL.
(2x 1 3)(3x 1 1) 5 6x2 1 2x 1 9x 1 3 5 6x2 1 11x 1 3 3
The length of the swimming pool is 3x 1 1.
Exercises
13. The area of a rectangular cookie sheet is 8x2 1 26x 1 15. The width of the
cookie sheet is 2x 1 5. What is the length of the cookie sheet?
14. The area of a rectangular lobby floor in the new office building is
15x2 1 47x 1 28. The length of one side of the lobby is 5x 1 4. What is the
width?
15. The area of a rectangular school banner is 12x2 1 13x 2 90. The width of the
banner is 3x 1 10. What is the length of the banner?
16. The distance a train has traveled is 6x2 2 23x 1 20. The train’s average speed
is 3x 2 4. How long has the train been traveling?
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Name
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8-7
Date
Reteaching
Factoring Special Cases
The area of a square is given by A 5 s2 , where s is a side length. When
the side length is a binomial, the area can be written as a perfect-square s A 5 s2
trinomial. If you are given the area of such a square, you can use
s
factoring to write an expression for a side length.
Problem
A mosaic is made of small square tiles called tesserae. Suppose the area of one
tessera is 9x2 1 12x 1 4. What is the length of one side of a tessera?
Because the tile is a square, you know the side lengths must be equal. Therefore,
the binomial factors of the trinomial must be equal.
9x2 1 12x 1 4 5 ( u u u )2
This is a perfect square trinomial and can be factored as the
square of a binomial.
9x2 5 (3x)2
9x2 and 4 are perfect squares. Write them as squares.
4 5 22
2(3x)(2) 5 12x
Check that 12x is twice the product of the first and last terms. It
is, so you are sure that you have a perfect-square trinomial.
9x2 1 12x 1 4 5 (3x 1 2)2
Rewrite the equation as the square of a binomial.
Multiply to check your answer.
(3x 1 2)(3x 1 2) 5 9x2 1 6x 1 6x 1 4 5 9x2 1 12x 1 43
The length of one side of the square is 3x 1 2.
Exercises
Factor each expression to find the side length.
1. The area of a square oil painting is 4x2 1 28x 1 49. What is the length of one
side of the painting?
2. You are installing linoleum squares in your kitchen. The area of each linoleum
square is 16x2 2 24x 1 9. What is the length of one side of a linoleum square?
3. You are building a table with a circular top. The area of the tabletop is
(25x2 2 40x 1 16)π. What is the radius of the tabletop?
4. A fabric designer is making a checked pattern. Each square in the pattern has
an area of x2 2 16x 1 64. What is the length of one side of a check?
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8-7
Class
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Reteaching (continued)
Factoring Special Cases
Some binomials are a difference of two squares. To factor these expressions, write
the factors so the x-terms cancel and you are left with two perfect squares.
Problem
What is the factored form of 4x2 2 9?
4x2 2 9 5 ( u 1 u )( u 2 u )
"4x2 5 2x
Both 4x2 and 9 are perfect squares. You know the signs of
the factors will be opposite, so the x-terms will cancel out.
Find the square root of each term.
!9 5 3
(2x 1 3)(2x 2 3)
Write each term as a binomial with opposite signs, so the
x-terms will cancel out.
Multiply to check your answer.
(2x 1 3)(2x 2 3) 5 4x2 1 6x 2 6x 2 9
5 4x2 2 93
The factored form of 4x2 2 9 is (2x 1 3)(2x 2 3).
Exercises
Factor each expression.
5. 9x2 2 4
6. 25x2 2 49
7. 144x2 2 1
8. 64x2 2 25
9. 49x2 2 16
10. 36x2 2 49
11. 81x2 2 16
12. 16x2 2 121
13. 25x2 2 144
14. 16x2 2 9
15. x2 2 81
16. 4x2 2 49
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8-8
Class
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Reteaching
Factoring by Grouping
You can factor some higher-degree polynomials by grouping terms and factoring
out the GCF to find the common binomial factor. Make sure to factor out a
common GCF from all terms first before grouping.
Problem
What is the factored form of 2b4 2 8b3 1 10b2 2 40b?
2b4 2 8b3 1 10b2 2 40b 5 2b(b3 2 4b2 1 5b 2 20)
2b is the GCF of all four terms. Factor out 2b
from each term.
5 2bfb2(b 2 4) 1 5(b 2 4)g Group terms into pairs and look for the GCF of
each pair. b2 is the GCF of the first pair, and 5
is the GCF of the second pair.
5 2b(b2 1 5)(b 2 4)
b 2 4 is the common binomial factor. Use the
Distributive Property to rewrite the expression.
Multiply to check your answer.
Multiply b2 1 5 and b 2 4.
2b(b2 1 5)(b 2 4) 5 2b(b3 1 5b 2 4b2 2 20)
5 2b4 1 10b2 2 8b3 2 40b
Multiply by 2b.
5 2b4 2 8b3 1 10b2 2 40b 3
Reorder the terms by degree.
The factored form of 2b4 2 8b3 1 10b2 2 40b is 2b(b2 1 5)(b 2 4).
Exercises
Factor completely. Show your steps.
1. 4x4 1 8x3 1 12x2 1 24x
2. 24y4 1 6y3 1 36y2 1 9y
3. 72z4 1 48z3 1 126z2 1 84z
4. 2e4 2 8e3 1 18e2 2 72e
5. 12f 3 2 36f 2 1 60f 2 180
6. 16g4 2 56g3 1 64g2 2 224g
7. 56m3 2 28m2 2 42m 1 21
8. 40n4 2 60n3 2 50n2 1 75
9. 60x3 2 90x2 2 30x 1 45
10. 12p5 1 8p4 1 18p3 1 12p2
11. 6r3 1 9r2 2 60r
12. 20s6 2 50s5 2 30s4
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Reteaching (continued)
Factoring by Grouping
Polynomials can be used to express the volume of a rectangular prism. They can
sometimes be factored into 3 expressions to represent possible dimensions of the
prism. The three factors are the length, width, and height.
Problem
The plastic storage container to the right has a volume of
12x3 1 8x2 2 15x. What linear expressions could
represent possible dimensions of the storage container?
12x3 1 8x2 2 15x 5 x(12x2 1 8x 2 15)
V = 12x3 + 8x2 -15x
Factor out x, the GCF for all three terms.
5 x(12x2 1 18x 2 10x 2 15)
ac is –180 and b is 8. Break 8x into two terms
that have a sum of 8x and a product of 2180x2 .
5 xf6x(2x 1 3) 2 5(2x 1 3)g
Group the terms into pairs and factor out the GCF
from each pair. The GCF of the first pair is 6x. The
GCF of the second pair is 25.
5 x(6x 2 5)(2x 1 3)
2x 1 3 is the common binomial term. Use the
Distributive Property to reorganize the factors.
Multiply to check your answer.
x(6x 2 5)(2x 1 3) 5 x(12x2 1 18x 2 10x 2 15)
Multiply 6x 2 5 and 2x 1 3.
5 x(12x2 1 8x 2 15)
Combine like terms.
5 12x3 1 8x2 2 15x 3
Multiply by x.
Possible dimensions of the storage container are x, 6x 2 5, and 2x 1 3.
Exercises
Find linear expressions for the possible dimensions of each rectangular prism.
13.
14.
V = 10x3 + 65x2 +105x
V = 12x3 + 34x2 +14x
15.
16.
V = 60x3 - 68x2 -16x
V = 12x3 - 15x2 -18x
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9-1
Quadratic Graphs and Their Properties
A U-shaped graph such as the one at the right is called a parabola.
y
• A parabola can open upward or downward.
8
• A parabola that opens upward has a minimum or lowest point.
6
• A parabola that opens downward has a maximum or highest point.
• The vertex of a parabola is its minimum or maximum point.
All parabolas have a line or axis of symmetry.
4
2
x
Ź2
O
2
Problem
What is the vertex of the graph below? Is it a minimum or maximum?
y
2
O
x
Ź4 Ź2
Ź2
The graph opens downward, so you are looking for the highest point. The vertex is
(23, 2) and it is a maximum.
Exercises
Identify the vertex of each graph. Tell whether it is a minimum or a maximum.
y
1.
2.
x
O y
2
4
3.
4
Ź6 Ź4 Ź2
y O
x
Ź2
Ź2
Ź4
Ź4
Ź6
Ź6
2
x
O
2
4
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Reteaching (continued)
Quadratic Graphs and Their Properties
Any function in the form y 5 ax2 1 bx 1 c where a 2 0 is called a quadratic
function. The graph of a quadratic function is a parabola.
Problem
What is the graph of y 5 12x2 2 4?
This is a quadratic function where a 5 12, b 5 0 and c 5 24. The graph will be a
parabola. Use a table to find some points on the graph. Then use what you know
about parabolas to complete the graph.
x
1
y â x2 Ź4
2
4
(x, y)
y
2
Ź4
yâ
1
(Ź4)2 Ź4 â4
2
Ź2
yâ
1
(Ź2)2 Ź4 âŹ2
2
0
yâ
1
(0)2 Ź4 âŹ4
2
(0, Ź4)
2
yâ
1
(2)2 Ź4 âŹ2
2
(2, Ź2)
4
yâ
1
(4)2 Ź4 â4
2
x
(Ź4, 4)
2
Ź4
Ź2
(Ź2, Ź2)
(4, 4)
Exercises
Graph each function.
4. y 5 2x2 1 5
5. y 5 x2 2 4
6. y 5 2x2 2 1
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4
Name
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9-2
Date
Reteaching
Quadratic Functions
Recall that the general equation for a quadratic function is y 5 ax2 1 bx 1 c.
2b
Using this general equation, the equation for the axis of symmetry is x 5 2a .
2b
Since the vertex lies on the axis of symmetry, the x-coordinate of the vertex is 2a .
Problem
What are the equation of the axis of symmetry and the coordinates of the vertex
of the graph of y 5 3x2 1 6x 2 4?
x 5 2b
2a
x5
Equation for axis of symmetry
26
2(3)
a 5 3 and b 5 6
x 5 21
Simplify.
Now, find the value of y when x 5 21.
y 5 3x2 1 6x 2 4
y 5 3(21)2 1 6(21) 2 4
y 5 27
The equation of the axis of symmetry is x 5 21 and the coordinates of the vertex
of the graph are (21, 27).
Exercises
Find the equation of the axis of symmetry and the coordinates of the vertex of
the graph of each function.
1. y 5 x2 1 8x
2. y 5 2x2 1 12x 1 10
3. y 5 2x2 1 4x 2 8
4. y 5 2x2 2 4x 2 5
5. y 5 23x2 1 18x 2 25
6. y 5 22x2 1 2x 2 6
7. f (x) 5 6x2 2 7
8. f (x) 5 25x2 2 10x 1 1
9. f (x) 5 4x2 2 16x 2 2
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Reteaching (continued)
Quadratic Functions
You can use the axis of symmetry and the vertex to help graph a quadratic
b
equation. Use the equation x 5 22a to find the equation of the axis of symmetry.
Because the vertex lies on the axis of symmetry, this value is also the x-coordinate
of the vertex.
Problem
What is the graph of y 5 2x2 2 4x 1 1?
1. Find the equation of the axis
2. Find the vertex.
y 5 2x2 2 4x 1 1
of symmetry.
x 5 2b
2a
x5
2(24)
2(2)
x51
y 5 2(12) 2 4(1) 1 1
y 5 21
a 5 2 and b 5 24
x51
Simplify.
The vertex is (1, 21)
Simplify.
3. Graph the axis of symmetry x 5 2 and the vertex (1, 21) .
y
(Ź1, 7)
6
4. Find a couple points on the graph.
For x 5 0, y 5 2(02) 2 4(0) 1 1 or 1.
Plot (0, 1).
(0, 1)
For x 5 21, y 5 2(21)2 2 4(21) 1 1 or 7.
Ź2
Plot (21, 7) .
O
Ź2
5. Use the axis of symmetry to complete the graph.
x
2
(1, Ź1)
x â1
Exercises
Graph each function. Label the axis of symmetry and the vertex.
10. y 5 x2 2 3
11. y 5 2x2 2 4x 1 1
12. y 5 2x2 1 8x 1 6
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Solving Quadratic Equations
An equation in the form ax2 1 bx 1 c 5 0 where a 2 0 is called a
quadratic equation. Its related quadratic function is y 5 ax2 1 bx 1 c. If you
graph the related quadratic function, the solutions of the quadratic equation are
x-values where the graph crosses the x-axis.
A linear equation can have only one solution. However, a quadratic equation can
have 2, 1, or 0 real-number solutions.
The related function of
2x2 1 4 5 0 is
y 5 2x2 1 4. The graph of
y 5 2x2 1 4 is shown
below.
The related function of
x2 2 2x 1 1 5 0 is
y 5 x2 2 2x 1 1.
The graph of
y 5 x2 2 2x 1 1 is
shown below.
y
4
The related function of
x2 2 x 1 2 5 0 is
y 5 x2 2 x 1 2. The
graph of y 5 x2 2 x 1 2
is shown below.
y
4
2
2
x
Ź3
O
y
x
3
Ź2
O
x
Ź2
2
O
2
Ź2
The graph crosses the x-axis
where x 5 22 and x 5 2.
The equation 2x2 1 4 5 0
has two solutions, 22 and 2.
The graph touches the
x-axis where x 5 1.
The equation
x2 2 2x 1 1 5 0 has
one solution, 1.
The graph does not
touch the x-axis.
The equation
x2 2 x 1 2 5 0 has no
real-number solutions.
Exercises
Solve each equation by graphing the related function. If the equation has no
real-number solution, write no solution.
1. x2 1 3 5 0
2. x2 1 4x 1 4 5 0
3. x2 1 x 2 2 5 0
4. How many times does the graph of y 5 x2 2 4 cross the x-axis? Explain.
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9-3
Solving Quadratic Equations
You can solve a quadratic equation by taking the square root of each side of the equation.
Problem
What are the solutions of 81x2 5 49?
81x2 5 49
81x2
49
81 5 81
Divide each side by 81.
49
x2 5 81
Simplify.
"x2 5 4Å49
81
Take the square root of each side.
7
x 5 49
Simplify.
Problem
What are the solutions of x2 1 9 5 0?
x2 1 9 5 0
x2 1 9 2 9 5 0 2 9
x2 5 29
Subtract 9 from each side.
Simplify.
Since x2 cannot equal 29 in the real numbers, x2 1 9 5 0 has no real-number
solutions.
Exercises
Solve each equation by finding square roots. If the equation has no real-number
solution, write no solution. If a solution is irrational, round to the nearest tenth.
5. x2 5 100
6. x2 2 144 5 0
8. 9x2 2 16 5 0
9. 3x2 1 27 5 0
11. 64x2 2 25 5 0
12. 3x2 2 30 5 0
7. 5x2 2 125 5 0
10. 7x2 2 49 5 0
13. x2 1 7 5 0
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9-4
Date
Reteaching
Factoring to Solve Quadratic Equations
If the product of two or more numbers is 0, then one of the factors must be 0. You
can use this fact to solve quadratic equations.
Problem
What are the solutions of the equation (4a 1 12)(5a 2 20) 5 0?
Since the product is 0, either (4a 1 12) or (5a 2 20) must equal 0.
4a 1 12 5 0
or
5a 2 20 5 0
4a 1 12 2 12 5 0 2 12
or
5a 2 20 1 20 5 0 1 20
4a 5 212
or
5a 5 20
4a
212
4 5 4
or
5a
20
5 5 5
a 5 23
or
a54
The solutions are 23 and 4.
Exercises
Solve each equation.
1. b(b 1 7) 5 0
2. 8y(2y 2 12) 5 0
3. (d 2 8)(d 2 2) 5 0
4. (m 1 1)(m 2 4) 5 0
5. (2a 1 14)(3a 1 12) 5 0
6. (5p 2 10)(2p 1 20) 5 0
7. (8t 1 4)(3t 1 6) 5 0
8. (4h 2 1)(2h 1 1) 5 0
9. (8n 2 16)(5n 2 12) 5 0
10. (s 1 6)(4s 2 6) 5 0
11. (5w 2 30)(2w 2 1) 5 0
12. (3g 1 1)(2g 2 5) 5 0
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Reteaching (continued)
Factoring to Solve Quadratic Equations
If you can rewrite a quadratic equation as a product of factors that equals zero, you
can solve the equation. To solve equations in this manner, you must use all your
factoring skills.
Problem
What are the solutions of the equation x2 2 x 5 20?
First rewrite the equation so that one side equals zero.
x2 2 x 5 20
x2 2 x 2 20 5 20 2 20
Subtract 20 from each side.
x2 2 x 2 20 5 0
Simplify.
Now, factor to rewrite the equation as a product of factors equal to zero. Find two
integers whose product is 220 and whose sum is 21. The product of 4 and 25 is
220, and the sum of 4 and 25 is 21.
x2 2 x 2 20 5 0
(x 1 4)(x 2 5) 5 0
x1450
or
x2550
x14245024
or
x25155015
x 5 24
x55
or
The solutions are 24 and 5.
Exercises
Solve each equation by factoring.
13. y2 1 3y 1 2 5 0
14. a2 2 a 2 20 5 0
15. m2 2 7m 1 6 5 0
16. 2d2 1 7d 2 4 5 0
17. 6t2 1 13t 1 6 5 0
18. 5p2 1 29p 2 6 5 0
19. s2 1 9s 5 220
20. x2 2 5x 5 14
21. b2 1 7b 5 8
22. 2h2 2 9h 5 5
23. 3s2 2 13s 5 212
24. 6v2 1 13v 5 5
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9-5
Class
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Reteaching
Completing the Square
You have learned to square binomials. Notice how the coefficient of the a term is
related to the constant value in every perfect-square trinomial.
(a 1 1) 2 5 (a 1 1)(a 1 1) 5 a2 1 2a 1 1 S
22
Q2 R 5 1
(a 2 1) 2 5 (a 2 1)(a 2 1) 5 a2 2 2a 1 1 S
22 2
Q 2 R 51
(a 2 2) 2 5 (a 2 2)(a 2 2) 5 a2 2 4a 1 4 S
24 2
Q 2 R 54
(a 1 3) 2 5 (a 1 3)(a 1 3) 5 a2 1 6a 1 9 S
62
Q2 R 5 9
In each case, half the coefficient of the a term squared equals the constant term.
You can use this pattern to find the value that makes a trinomial a perfect square.
Problem
What is the value of c such that x2 2 14x 1 c is a perfect-square trinomial?
2
The coefficient of the x term is 214. Using the pattern, c 5 Q 214
2 R or 49.
So, x2 2 14x 1 49 is a perfect-square trinomial.
Exercises
Find the value of c such that each expression is a perfect-square trinomial.
1. a2 1 8a 1 c
2. x2 2 16x 1 c
3. m2 1 20m 1 c
4. p2 2 14p 1 c
5. y2 2 10y 1 c
6. b2 1 18b 1 c
7. d2 1 12d 1 c
8. n2 2 n 1 c
9. w2 1 3w 1 c
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Reteaching (continued)
Completing the Square
You can use completing the square to solve quadratic equations.
Problem
What are the solutions of the equation x2 1 2x 2 48 5 0?
First rewrite the equation so that the constant is on one side of the equation and
the other terms are on the other side.
x2 1 2x 2 48 5 0
x2 1 2x 2 48 1 48 5 0 1 48
Add 48 to each side.
x2 1 2x 5 48
Simplify.
2 2
Since Q 2 R 5 1, add 1 to each side.
x2 1 2x 1 1 5 48 1 1
Add 1 to each side.
(x 1 1)2 5 49
Simplify.
x 1 1 5 4 !49
Take the square root of each side.
x 1 1 5 47
Simplify.
x 1 1 5 27
or
x1157
x 1 1 2 1 5 27 2 1
or
x11215721
x 5 28
x56
or
The solutions are 28 and 6.
Exercises
Solve each equation by completing the square. If necessary, round to the
nearest hundredth.
10. b2 1 10b 5 75
11. y2 2 18y 5 63
12. n2 2 20n 5 275
13. a2 1 16a 5 215
14. t2 1 8t 2 9 5 0
15. h2 2 12h 2 9 5 0
16. m2 2 2m 2 8 5 0
17. s2 1 6s 1 1 5 0
18. v2 1 4v 2 2 5 0
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The Quadratic Formula and the Discriminant
If a quadratic equation is written in the form ax2 1 bx 1 c 5 0, the solutions can
be found using the following formula.
x5
2b 4 "b2 2 4ac
2a
This formula is called the quadratic formula.
Problem
What are the solutions of x2 1 7x 5 60? Use the quadratic formula.
First rewrite the equation in the form ax2 1 bx 1 c 5 0.
x2 1 7x 5 60
x2 1 7x 2 60 5 60 2 60
Subtract 60 from each side.
x2 1 7x 2 60 5 0
Simplify.
Therefore, a 5 1, b 5 7, and c 5 260.
x5
2b 4 "b2 2 4ac
2a
x5
27 4 "72 2 4(1)(260)
2(1)
x5
27 4 "289
2
x5
27 4 17
2
The two solutions are
27 2 17
27 1 17
or 212 and
or 5.
2
2
Exercises
Use the quadratic formula to solve each equation.
1. x2 2 19x 1 70 5 0
2. x2 1 32x 1 175 5 0
3. 2x2 1 37x 2 19 5 0
4. x2 2 10x 5 75
5. x2 1 x 5 132
6. 6x2 1 13x 5 28
7. 20x2 1 11x 5 3
8. 4x2 1 24x 5 235
9. 15x2 1 20 5 40x
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The Quadratic Formula and the Discriminant
In the quadratic equation, the expression under the radical sign, b2 2 4ac, is
called the discriminant. Consider the quadratic formula.
x5
2b 4 "b2 2 4ac
2a
• If b2 2 4ac is a negative number, the square root cannot be found in the real
numbers. There are no real-number solutions of the equation. The graph of
the quadratic does not cross the x-axis.
• If b2 2 4ac equals 0, x 5
2b 4 !0 2b
or 2a . There is only one solution of the
2a
equation. The vertex of the quadratic is on the x-axis.
• If b2 2 4ac is a positive number, there are two solutions of the equation,
2b 2 "b2 2 4ac
2b 1 "b2 2 4ac
and
. The graph of the quadratic intersects
2a
2a
the x-axis twice.
Problem
What is the number of solutions of x2 1 13 5 25x?
First rewrite the equation in the form ax2 1 bx 1 c 5 0.
x2 1 13 5 25x
x2 1 5x 1 13 5 0
Add 5x to each side.
Therefore, a 5 1, b 5 5, and c 5 13.
b2 2 4ac 5 52 2 4(1)(13)
5 227
Since b2 2 4ac is a negative number, there are no real-number solutions of the
equation.
Exercises
Find the number of solutions of each equation.
10. 4x2 1 12x 1 9 5 0
11. x2 2 12x 1 32 5 0
12. x2 2 10x 1 1 5 0
13. 3x2 1 6x 1 8 5 0
14. 3x2 2 5x 5 26
15. x2 1 100 5 20x
16. 5x2 2 7x 5 2
17. 9x2 1 4 5 12x
18. 3x2 1 5x 5 2
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Class
10-2
Date
Reteaching
Simplifying Radicals
You can remove perfect-square factors from a radicand.
Problem
What is the simplified form of "80n5 ?
In the radicand, factor the coefficient and the variable separately into perfect
square factors, and then simplify. Factor 80 and n5 completely and then find
paired factors.
80 5 8 ? 10 5 2 ? 2 ? 2 ? 2 ? 5
Solve
Factor 80 completely.
5 (2 ? 2)(2 ? 2) ? 5 5 (2 ? 2)2 ? 5
!80 5 "42 ? 5 5 "42 ? !5
Find pairs of factors.
Use the rule !ab 5 !a ? !b.
5 4 ? !5 5 4!5
The square root of a number squared
is the number: "a2 5 a .
Factor n5 completely.
n5 5 n ? n ? n ? n ? n
5 (n ? n) ? (n ? n) ? n 5 (n ? n)2 ? n
"n5 5 "(n ? n)2 ? !n
5
n2
? !n 5
Separate the factors.
n2 !n
Remove the perfect square.
"80n5 5 4 ? n2 !5 ? n 5 4n2 !5n
Check
Find pairs of factors.
Combine your answers.
"80n5 0 4n2 !5n
Check your solution.
"80n5
4n2 !5n
0
!5n
!5n
Divide both sides by !5n .
"16n4 0 4n2
Simplify.
4n2
5
4n2 3
Solution: The simplified form of "80n5 is 4n2 !5n .
Exercises
Simplify each radical expression.
1. "100n3
2. "120b4
3. "66t5
4. !32x
5. "525c7
6. "86t2
7. "50g3
8. "54h6
9. !35y
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Reteaching (continued)
Simplifying Radicals
Problem
What is the simplified form of
27t4
?
Å 48t2
Begin by cancelling the common factors in the numerator and denominator.
Simplify the numerator and denominator separately when the denominator is a
perfect square. Remember that the radical is not simplified if there is a radical in
the denominator. Multiply to remove the radical from the denominator.
Solve
27t3
3?3?3?t?t?t
5 %3 ? 4 ? 4 ? t ? t ? t ? t
4
Å 48t
3?3?3?t?t?t
5 %3 ? 4 ? 4 ? t ? t ? t ? t
5
5
"(3 ? 3)
"(4 ? 4)t
5
"32
"42t
Cancel the common factors.
Find pairs of factors. These are the perfectsquare factors.
3
Simplify the numerator and denominator
separately to remove the perfect-square
factors. "32 5 3 and "42t 5 4!t
4"t
5
3 ( !t)
4 !t ( !t)
5
3 !t
3 !t
3 !t
5
5
4 !t ? t
2
4t
4"t
Solution: The simplified form of
Factor the numerator and denominator
completely.
Multiply the numerator and denominator by
!t to remove !t from the denominator.
Remove the perfect-square factor from the
denominator.
27t3 3 !t
is
.
Å 48t4
4t
Exercises
Simplify each radical expression.
10.
49
Å 81
11.
18x4
Å 200
12.
28s
Å s3
13.
25a5
Å 9a7
14.
40b4
Å 12b3
15.
48
Å 6t6
16.
50z3
Å 4x2
17.
t5
Å 64
18.
32t
Å t
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Operations with Radical Expressions
You can use the Distributive Property with radical expressions.
Problem
What is the simplified form of 4 !2 2 !18?
You need to simplify the radical expressions before you know if there are any like
radicals that can be subtracted.
Solve
4!2 2 !18
Look for a common radical in 4!2 and !18. 4!2 is
factored completely, but !18 can be factored further.
!18 5 !3 ? 3 ? 2
Factor !18 completely.
5 !(3 ? 3) ? 2 5 "32 ? 2
Find pairs of factors that you can factor out. These are
perfect-square factors.
5 3!2
Remove the perfect-square factor.
4!2 2 !18 5 4!2 2 3!2
Check
Now you can see that each term in the expression
shares the common radical !2.
5 (4 2 3) !2
Use the Distributive Property to combine like radicals.
a !b 2 c!b 5 (a 2 c)!b
5 1 ? !2
Subtract.
5 !2
Simplify.
4!2 2 !18 5 !2
Check your solution.
3!2 2 !18 5 0
Subtract !2 from both sides.
3!2 5 !18
Add !18 to both sides.
3!2 5 3!2 3
Simplify !18.
Solution: The simplified form of 4 !2 2 !18 is !2.
Exercises
Simplify each sum or difference.
1. 2!5 2 4!5
2. !7 1 !7
3. 3 !6 1 2 !6
4. 5!2 1 !32
5. 3 !3 2 !75
6. 4 !54 1 2 !24
7. 10 !5 2 5!20
8. 2 !8 1 !200
9. 3 !12 1 !108
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10-3
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Operations with Radical Expressions
When you have two binomial factors that include radical expressions, treat them
like any other binomials and multiply using FOIL (First, Outer, Inner, Last).
Problem
What is the simplified form of (2 !3 1 5 !7)(3 !7 2 !3)?
Solve
Use FOIL to find the product of each pair of terms. Multiply the coefficients
and then multiply the radicals. Remove all perfect-square factors.
FIRST
LAST
( 2 Ƅ3 à5 Ƅ7 )
( 3Ƅ7 ŹƄ3 )
INNER
OUTER
First: 2!3 ? 3!7 5 6 !21
Outer: 2!3 ? (2!3) 5 22 !9 5 22 ? 3 5 26
Inner: 5!7 ? 3!7 5 15 !49 5 15 ? 7 5 105
Last: 5!7 ? (2!3) 5 25 !21
5 6!21 2 6 1 105 2 5 !21
5 (6 !21 2 5!21) 1 (26 1 105)
Group like terms.
5 (6 2 5)!21 1 99
Distributive Property
5 1!21 1 99 5 !21 1 99
Simplify.
Solution: The simplified form of (2 !3 1 5 !7)(3 !7 2 !3) is !21 1 99.
Exercises
Simplify each radical expression.
10. !4(!3 1 !5)
11. !10(!8 2 9)
12. 2 !3(2 2 !3)
13. 2!8(5 2 3!5)
14. 4 !6(!2 1 4 !3)
15. 2 !6(!11 1 7)
16. (3!7 1 !3)2
17. (1 1 !3)(1 2 !3)
18. (3 !6 1 2 !2)(!2 2 4 !6)
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11-1
Date
Reteaching
Simplifying Rational Expressions
Rational expressions may be in the form of monomials or polynomials.
Simplifying rational expressions is similar to simplifying numerical fractions
where common factors are taken out.
x22
x22
1
3x 2 6 5 3(x 2 2) 5 3
Example:
Excluded values are those that make the denominator 0. A denominator can not equal 0, so
these values are not part of the solution. Consider not only the solution, but also the original
expression to figure the excluded values.
Problem
3
What is the simplified form of 2a2? State any excluded values.
4a
Solve
Monomials: reduce numbers; cancel out like variables
2a3
2 a?a?a
a
5 2 ?? 2 ? a ? a 5 2
2
4a
a
The simplified form is 2 when a 2 0.
What is the simplified form of
Solve
x2 1 4x 1 4
? State any excluded values.
x 1 2
Polynomials: cancel out factors or groups of factors
(x 1 2)(x 1 2)
x2 1 4x 1 4
5
5x12
x12
x12
The simplified form is x 1 2 when x 2 22.
Recognizing Opposite Factors
You can find the opposite of a number by multiplying by 21. For example, the
opposite of 3 is (21)(3) 5 23.
Similarly, multiplying a polynomial by 21 results in its opposite. For example, the
opposite of x 2 2 is (21)(x 2 2) 5 2x 1 2. It can also be written as 2 2 x.
Problem
Write the opposite of (20 2 x) two ways.
Solve
Multiply by (−1) to find the opposite.
(21)(20 2 x) 5 220 1 x or x 2 20
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Simplifying Rational Expressions
Exercises
Simplify each expression. State any excluded values.
a4
1. a
2.
4x3
16x2
3.
cd2
3c2d
lm
4. 2 2
l m n
5.
64y
16y2x
6.
2x2 2 4x
x
7.
5x3 2 15x2
x23
9.
2b 1 4
4
10.
11.
3p 2 21
18
4
12. 4y 2 8
13.
7z 2 28
14z
9
14. 18 2 81a
8.
5
15. 35 2 5c
17.
16.
a12
a2 1 4a 1 4
x2 1 5x 1 6
x13
3a 1 15
15
q2
2q 1 2
1 4q 1 3
2x 2 2
18. 2 2 2x
9 2 x2
19. x 2 3
20.
2a 1 4
2
Write the opposite expression and simplify the opposite expression.
10b5
40b4
36 2 z2
22. 4z 2 24
x2 2 16
23. x 2 4
30 1 2z
24. 14 1 4z
21.
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Dividing Polynomials
A few important rules are needed for successful division of a polynomial. When
dividing by a monomial, a single term, remember to divide each polynomial term
by the monomial and reduce the fraction.
Problem
What is (8x3 2 3x2 1 16x) 4 2x2 ?
Solve
(8x3 2 3x2 1 16x) 4 2x2
5 (8x3 2 3x2 1 16x) ?
5
1
2x2
8x3
3x2
16x
2
1 2
2
2
2x
2x
2x
3
Multiply by
1
, the reciprocal of
2x2
2x2 .
Use the Distributive Property.
8
5 4x1 2 2x0 1 x
3
Division equals multiplication by the reciprocal.
Subtract exponents when dividing powers
with the same base.
8
5 4x 2 2 1 x
Simplify.
Exercises
Divide.
1. (2x2 2 9x 1 18) 4 2x
2. (16x4 2 64) 4 4x3
3
3. (x5 2 3x4 1 10x3 2 4x2 2 6) 4 3x2
4. (5x3 2 25x2 2 1) 4 5x
When a polynomial (many terms) is divided by a binomial (2 terms), the
polynomial terms should be in order from highest to lowest exponent.
To make the polynomial 24 1 2x 2 16x2 1 3x3 division ready, put it in the
correct order for division, from greatest exponent to lowest. The correct order for
24 1 2x 2 16x2 1 3x3 to be division ready is 3x3 2 16x2 1 2x 2 4.
For any gaps or missing exponents, the place is held with 0. For example, x2 1 1
becomes x2 1 0x 1 1, 0x being the placeholder for the x term.
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Dividing Polynomials
Dividing a polynomial by a binomial is similar to long division.
Problem
What is (p2 2 3p 1 2) 4 (p 2 2)?
Solve
p 2 2qp2 2 3p 1 2
Write the problem as long division.
p
p 2 2qp2 2 3p 1 2
Ask how many times p goes into p2 (p times). The variables must
align by exponent, so the p goes above 23p since both match.
p
p 2 2qp2 2 3p 1 2
p2 2 2p0000
21p 1 2
Multiply p times (p 2 2). Subtract the product p2 2 2p. Bring
down 2.
p21
p 2 2qp2 2 3p 1 2
p2 2 2p0000
21p 1 2
21p 1 2
Determine how many times p goes into 21p ( 21 times). Multiply
(21) times (p 2 2) to get 21p 1 2. Subtract the product
21p 1 2. There is no remainder.
So p 2 2 goes into p2 2 3p 1 2 exactly p 2 1 times with no remainder.
Exercises
Divide.
5. (d2 1 4d 2 12) 4 (d 2 2)
6. (y2 2 4y 1 4) 4 (y 2 2)
7. (x2 2 2x 1 1) 4 (x 2 1)
8. (b2 2 b 2 20) 4 (b 2 5)
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