Download SD 9-12 Algebra

SD 9-12 Algebra
03/20/2009
Student Name:
___________________________
Class:
___________________________
Date:
___________________________
Instructions:
Read each question carefully and select the correct answer.
1.
The number of mosquitos in Kylie's
backyard varies directly as the amount of
water in a puddle. When there are 4 gallons
of water in the puddle, there are 120
mosquitos in the yard. How many
mosquitos would be in the yard if there was
1 gallon of water in the puddle?
A.
B.
C.
D.
2.
Calculate the slope of the line between the
points (- 4, 10) and (3, 8).
116 mosquitos
480 mosquitos
30 mosquitos
481 mosquitos
A.
2
B.
C.
D.
SD Algbra-Questions, Answer Key, Study Guide
Page 1 of 102
3.
Factor completely.
6.
4x2 - 36
A.
B.
C.
D.
4.
(2x + 6)(2x - 6)
2(x + 9)(x - 16)
(4x + 18)(x - 2)
4(x + 3)(x - 3)
The electrical resistance (R) of a wire varies
directly with its length (L) in meters, and
indirectly with the square of its diameter (d)
in millimeters. The equation that represents
this relationship is given by
where k is the constant for
wire made out of a specific metal. If the
resistance of a 50 meter long wire is 8.64
ohms when its diameter is 5 millimeters,
what is the resistance of a 200 meter long
wire when its diameter is 8 millimeters?
Simplify.
A.
A.
B.
C.
D.
B.
C.
13.5 ohms
14.3 ohms
15.4 ohms
17.9 ohms
D.
7.
5.
A riding stable charges $150 per month to
board a horse plus $6.75 per bale of hay that
the horse eats. What is the maximum
number of hay bales that Shiloh eats in a
month if his monthly boarding cost is at
least $225.00 and at most $255.00? Round
your answer to the nearest whole bale that
satisfies the inequality.
Find the range of the following relation r.
A.
B.
C.
D.
A.
B.
C.
D.
12 bales
11 bales
16 bales
15 bales
SD Algbra-Questions, Answer Key, Study Guide
Page 2 of 102
8.
Which function rule relates the values of the
input variable, x, to the values of the output
variable, y, in the table below?
10.
Simplify.
A.
B.
C.
3
9
324
D.
11.
A.
B.
C.
D.
y=x-4
y=x+5
y=x+7
y=x-1
Subtract the following.
A.
B.
C.
9.
D.
12.
Evaluate the expression with y = 18.
3 - 2(4y - 5)
A.
B.
C.
D.
13.
A.
B.
C.
D.
Q
R
S
T
SD Algbra-Questions, Answer Key, Study Guide
-131
-151
137
157
Two sides of a rectangular picture frame are
each 7 inches shorter than the other two
sides. The perimeter of the picture frame
measures 88 inches. What is the length of
the shortest two sides of the frame?
A.
B.
C.
D.
25.5 in.
30 in.
18.5 in.
37 in.
Page 3 of 102
14.
Salim had a square picture that he enlarged.
He increased the length by 3 inches and the
width by 1 inch so that the area of the
picture is 35 in2 . What was the length of the
picture's sides before he enlarged it?
17.
Solve for h.
A.
A.
B.
C.
D.
31 in
5 in
4 in.
7 in.
B.
C.
15.
Which of the following points, if removed
from the set, would make the set a function?
A.
B.
C.
D.
16.
(- 9, 9)
(5, - 9)
(- 5, 5)
(9, - 5)
D.
18.
9 + (4 + 7) = 4 + (9 + ?)
A.
B.
C.
D.
Solve for k.
kl - 4 = 2k
A.
Find the missing number.
19.
B.
Solve for the value of x.
A.
B.
C.
D.
C.
D.
20.
9
4
7
20
x = 36
6<x<7
x=6
5<x<6
To get a certain shade of orange, Raul needs
to mix red paint with yellow paint in the
ratio of 5:2.
Which of the following equations shows
how many quarts of red paint Raul needs to
mix with 4 quarts yellow paint?
A.
B.
C.
D.
SD Algbra-Questions, Answer Key, Study Guide
5/2 = x/4
5/2 = 4/x
5/8 = x
8/5 = x
Page 4 of 102
21.
Complete the following linear function
table.
24.
Write the absolute value function, f(x) = |-3x
- 4|, as a compound function.
A.
A.
B.
C.
D.
B.
3
4
1.5
4.5
C.
D.
22.
What is the value of x in the given
statement?
25.
A.
B.
C.
D.
23.
If f(x) = 0.5x + 3.1 and g(x) = 0.6x - 1.86,
find g(f(x)).
A.
B.
C.
D.
0.3x2 + 0.93x - 5.766
1.1x + 1.24
0.3x + 2.17
0.3x
26.
Find the area of an ellipse with a major axis
of 8 in. and a minor axis of 2 in. Express
your answer in terms of .
A.
8
B.
64
C.
4
D.
16
in.2
in.2
in.2
in.2
Solve by completing the square.
x2 - 3x - 21 = 0
A.
x = - 7 or x = 3
B.
C.
D.
SD Algbra-Questions, Answer Key, Study Guide
No Real Roots
Page 5 of 102
27.
Solve for x.
31.
What is the value of n?
5n + 15 = 100
A.
B.
C.
D.
28.
A.
B.
C.
D.
1
-128
-2
-8
What is the value of y?
32.
29.
-2
20
46
-2
A.
y = 90
B.
y = 1 1/9
C.
y=1
D.
Problem cannot be solved with the
given information.
What is the value of x?
x
0.2 = -15
A.
B.
C.
D.
30.
What is the value of y?
x = 9y
z = 10y
y + 8 - 11 = 17
A.
B.
C.
D.
n = 575
n = 425
n = 23
n = 17
n + 8 < -7
n > -1
n < -15
n > -15
n < -1
SD Algbra-Questions, Answer Key, Study Guide
Which of the following is true?
A.
7.5
75
3
3.3
Solve the inequality.
A.
B.
C.
D.
33.
B.
C.
D.
34.
Find the missing symbol.
(8 x 8) ? (7 x 7) = 15
A.
B.
C.
D.
+
x
÷
Page 6 of 102
35.
Round to the nearest cent when necessary.
39.
Which of the following is the best price?
A.
B.
C.
D.
36.
8 for $3.99
7 for $3.55
6 for $3.49
9 for $4.20
Four pancakes and three eggs at Candy's
Cafe cost $7.95. Two pancakes and three
eggs at Burger Palace cost $5.95. Which
option shows the best method for calculating
the amount that each restaurant is charging
for each pancake and each egg?
A.
-2p = $13.90
B.
6p = $13.90
C.
4p + 3e = $7.95 and 2p + 3e = $5.95
D.
4 + 3(p + e) = $7.95 and 2 + 3(p + e)
= $5.95
Which of the following best completes the
number sentence?
2 = (18 x 2) + (14 x 2)
40.
A.
B.
C.
D.
37.
(18 x 14) +
(18 x 14) x
(18 + 14) x
(18 + 14) +
A.
B.
C.
D.
Simplify
41.
A.
B.
C.
D.
38.
A
B
C
D
x equals negative y squared
y equals negative x to the fourth
42.
SD Algbra-Questions, Answer Key, Study Guide
l = 21.5; w = 10.5
l = 25; w = 7
l = 26.5; w = 5.5
l = 28; w = 4
Factor completely.
6x2 - 3x - 30
x equals negative y to the second
x equals 3y squared
P(3, 2)
P(3, 4)
P(3, 44/5)
P(3, 16)
The length of a rectangle is four more than
three times the width. The perimeter is
sixty-four meters. Find the length and the
width.
A.
B.
C.
D.
Which relation is written as a function?
A.
B.
power
C.
power
D.
Given the points O(-3, 4), Q(-2, 6), write an
equation for the line OQ and find the ycoordinate for point P (3, y) on the line OQ.
A.
B.
C.
D.
3(2x - 5)(x + 2)
3(2x - 2)(x + 5)
3(2x + 5)(x - 2)
3(2x + 2)(x - 5)
Page 7 of 102
43.
Solve the following using the quadratic
formula.
46.
Simplify.
(a + b)(3a + 2b)
2
- 2x + 3x + 4 = 0
A.
B.
A.
C.
B.
D.
C.
47.
Simplify.
D.
A.
B.
C.
D.
44.
A.
A
B
C
D
B.
C.
48.
D.
45.
A.
B.
-12
C.
D.
2
Simplify.
A.
B.
C.
D.
A
B
C
D
Find the solution set and choose the
corresponding description: 12 - (10 - 8 - 3x)
< 8 - 10 + 2x
49.
all points to the left of -12
all points to the left of -2, including -
Solve for x.
|8x + 12| - 13 = 23
A.
B.
C.
D.
SD Algbra-Questions, Answer Key, Study Guide
all points to the left of -2
all points to the left of -12, including
x = 6 or x = - 3
x=6
x=3
x = 3 or x = - 6
Page 8 of 102
SD 9-12 Algebra
Answer Key
03/20/2009
1.
C
Direct Variation
2.
D
Slope
3.
D
Factoring: Difference of Two
Squares
4.
C
Radicals:
Addition/Subtraction - A
5.
D
Inequalities - C
6.
A
Joint and Combined Variation
7.
A
Domain/Range
8.
C
Function Rules
9.
A
Exponential Functions
10.
A
Exponential Notation - F
11.
A
Polynomials: Subtraction
12.
A
Equations: Order of
Operations
13.
C
Functions: Linear
14.
C
Quadratic Equations: Real
World Problems
15.
A
Functions/Relations - A
16.
A
Equations With Two
Variables
17.
D
Literal Equations
18.
C
Missing Elements - D
19.
C
Radicals and Roots
20.
A
Ratio/Proportion - C
21.
B
Function/Pattern - C
SD Algbra-Questions, Answer Key, Study Guide
22.
D
Inequalities - A
23.
D
Functions: Quadratic
24.
B
Functions: Absolute Value
25.
C
Irrational Numbers: Pi
26.
B
Non-Linear Equations
27.
B
Rational Numbers: Equations
28.
B
Equations:
Addition/Subtraction
29.
C
Equations:
Multiplication/Division
30.
B
Inequalities - B
31.
D
Equations: Two-Step
32.
D
Solving Equations:
Substitution
33.
C
Exponential Notation - D
34.
B
Missing Elements - E
35.
D
Rates
36.
C
Properties - E
37.
A
Radicals: Simplifying
38.
B
Functions/Relations - B
39.
C
Equations: Systems
40.
D
Equations of a Line
41.
B
Number Relation Problems
42.
A
Factoring
43.
B
Quadratic Formula
44.
D
Polynomials: Addition
45.
C
Exponential Notation - E
46.
C
Polynomials: Multiplication
47.
B
Polynomials: Division
48.
C
Sets/Subsets/Solution Sets
49.
D
Absolute Value: Solve
Page 9 of 102
Study Guide
SD 9-12 Algebra
03/20/2009
Direct Variation
Variation equations are formulas that show how one quantity changes in relation to one or more other
quantities. There are four types of variation: direct, indirect (or inverse), joint, and combined.
Direct variation equations show a relationship between two quantities such that when one quantity increases, the
other also increases, and when one quantity decreases, the other also decreases. We can say that y varies
directly as x, or y is proportional to x. Direct variation formulas are of the form y = kx, where the number
represented by k does not change and is called a constant of variation.
Indirect variation equations are of the form y = k/x and show a relationship between two quantities such that
when one quantity increases, the other decreases, and vice versa.
This skill focuses on direct variation. The following is an example of a direct variation problem.
The amount of money in a paycheck, P, varies directly as the number of hours, h, that are worked. In
this case, the constant k is the hourly wage, and the formula is written P = kh. If the equation is solved
for k, the resulting equation shows that P and h are proportional to each other.
Therefore, when two variables show a direct variation relationship, they are proportional
to each other. Direct variation problems can be solved by setting up a proportion in the form below.
Example 1: The amount of fuel needed to run a textile machine varies directly as the number of hours
the machine is running. If the machine required 8 gallons of fuel to run for 24 hours, how many gallons
of fuel were needed to run the machine for 72 hours? Round your answer to the nearest tenth of a
gallon, if necessary.
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Step 1: Set up the proportion. Since the machine used 8 gallons of fuel in 24 hours, the left side of the
proportion should be 8 gallons over 24 hours. The number of gallons that the machine used in 72 hours
needs to be found, so the right side of the proportion should be g gallons over 72 hours.
Step 2: Cross-multiply across the equal sign.
Step 3: Set up the cross-multiplication equation.
Step 4: Divide both sides of the equation by 24 hours to isolate g gallons.
Step 5: Reduce the fractions on both sides of the equal sign.
Step 6: Simplify by multiplying the numbers remaining on the right side of the equal sign (8 gallons
3).
Answer: 24 gallons
Example 2: The price of jellybeans, j, varies directly as the number of pounds, p, that are purchased.
Find the equation that relates the two variables if jellybeans are $1.95 per pound.
(1) y = kx, j = kp
(2) j = 1.95p
Step 1: Remember that the formula for direct variation is: y = kx and substitute the variables from the
question into the appropriate places.
Step 2: Since the jellybeans are always $1.95 per pound, the constant, k, equals 1.95. Substitute 1.95
into the equation for k.
Answer: j = 1.95p
Activities that can help reinforce the concept of direct variation are as follows.
1. Have students solve the equation y = kx for k, and then substitute two sets of (x, y) values into
the equation and compare the values for k. If they are the same, then x and y have a direct variation
relationship.
2. Have the student think of scenarios that show a direct variation relationship. Then, make up
numbers to go with the relationships and have the students practice solving them.
Slope
Slope is the ratio of the vertical difference of two points on a line and the horizontal difference between the
same two points. Slope is also defined as "rise over run" and is found by calculating the difference in the ySD Algbra-Questions, Answer Key, Study Guide
Page 11 of 102
coordinates (rise) divided by the difference in the x-coordinates (run). Slope can be found either graphically or
algebraically.
Finding Slope Graphically:
Two points are identified on line z: A (- 3, 8) and B (3, 1).
To find the slope of line z, which passes through points A and B, follow these steps.
Step 1: Start at the left-most point (it is possible to start at either point, but it is important to be
consistent), which is point A (- 3, 8).
Step 2: Count up (positive) or down (negative) until level with the right-most point, point B (3, 1). The
count will be 7 in the downward direction, or - 7. The rise is - 7.
Step 3: Count right (positive) or left (negative) until point B is reached. The count will be 6 to the right,
or +6. The run is 6.
Step 4: Using the "rise over run" definition of slope, place - 7 on top of the fraction and 6 on the bottom
of the fraction.
Finding Slope Algebraically:
To find the slope of line z algebraically, follow these steps.
Step 1: Use the formula for the slope of a line, where m is the variable that represents slope.
Step 2:
Step 3: Substitute the given coordinate points into the slope formula and simplify the fraction.
It does not matter which method (graphically or algebraically) is used for determining slope.
Regardless, the slope of a line always remains the same.
Example 1: Find the slope of the line between Point R (2, 4) and Point S (1, 3).
SD Algbra-Questions, Answer Key, Study Guide
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Graphically:
Step 1: Start at the left-most point, which is point S (1, 3).
Step 2: Determine the rise (1).
Step 3: Determine the run (1).
Step 4: Using the "rise over run" definition of slope, place 1 on top of the fraction and 1 on the bottom of
the fraction.
Answer: m = 1
Algebraically:
Step 1: Write the formula.
Step 2:
Step 3: Simplify the fraction.
Answer: m = 1
An activity that can reinforce the concept of slope is to have students randomly plot two points on a
coordinate system and then find the slope graphically. They can check their answers by substituting the
two points into the slope formula.
Factoring: Difference of Two Squares
Consider the following equation:
3 4 = 12
The numbers 3 and 4 are said to be factors of the number 12. Factors are terms that can be multiplied
together to create another term. This concept of factoring is not reserved for numbers, but may be
extended to polynomials as well.
Definitions:
• Factoring is rewriting a mathematical expression as a product of its component factors. One way to
think of factoring is as the opposite or inverse of multiplying.
• Sum means the addition of two terms.
• Difference means the subtraction of two terms.
• Variables are letters that represent unknown numbers.
• Squares are terms whose square roots are whole numbers, zero, or variables whose powers are
SD Algbra-Questions, Answer Key, Study Guide
Page 13 of 102
divisible by two. For example, 81 is a square because the square root of 81 is 9, and x2 is a square
because the exponent (2) is divisible by two.
• A polynomial is a term or sum/difference of terms. Each term is either a number, a variable, or a
product of a number and one or more variables. Examples: 3x2 + 2x +1; 5x3 ; 4x + 1
• A monomial is a polynomial with one term. Example: 6x2
• A binomial is a polynomial with two terms. Example: 8y4 + 7y
• A trinomial is a polynomial with three terms. Example: 3x2 + 2x +1
The difference of two squares is easily factored using the rule below.
a2 - b2 = (a - b) (a + b)
Example 1: Factor completely.
x2 - 169
Step 1: This problem is a difference of two squares since x2 and 169 are perfect squares. Set up two sets
of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of the
other set.
Step 2: Take the square root of the first term (x2 ) and write x in the first space in each set of parentheses.
Step 3: Take the square root of the second term (169) and write 13 in the second space in each set of
parentheses.
Answer: (x - 13)(x + 13)
NOTE: This format will always work because if (x - 13) and (x + 13) are multiplied together using the
FOIL method, (multiply the First terms, then the Outer terms, then the Inner terms, then the Last terms),
the result is: x2 + 13x - 13x - 169. Combine like terms, or terms that have the same variable(s) with the
same exponent(s), the +13x and -13x terms add to zero. The result is x2 - 169, which is the original
problem.
NOTE: Because of the commutative property of multiplication, the order of the factors can be reversed
and an equivalent expression is made: (x - 13)(x + 13) = (x + 13)(x - 13).
Example 2: Factor completely.
169y2 - 36
Step 1: This problem is a difference of two squares since 169y2 and 36 are perfect squares. Set up two
sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of
the other set.
Step 2: Take the square root of the first term (169y2 ) and write 13y in the first space in each set of
parentheses.
Step 3: Take the square root of the second term (36) and write 6 in the second space in each set of
SD Algbra-Questions, Answer Key, Study Guide
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parentheses.
Example 3: Factor completely.
2x2 - 50
Step 1: Factor out any common factors. Since 2x2 and 50 can both be evenly divided by 2, perform the
division. NOTE: Don't forget to carry the 2 that was factored out through the entire problem.
Step 2: The remaining problem is a difference of two squares since x2 and 25 are perfect squares. Set up
two sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center
of the other set.
Step 3: Take the square root of the first term (x2 ) and write x in the first space in each set of
parentheses.
Step 4; Take the square root of the second term (25) and write 5 in the second space in each set of
parentheses.
Answer: 2(x - 5)(x + 5)
To help the student further understand this concept, write square numbers on one set of index cards.
Write squared variable terms and/or the product of numbers and variables on another set of index cards.
Have the student draw one card from each set of index cards. Ask the student to factor the differences
between each pair of cards. He or she will have to find the square roots of the terms on the cards to
correctly complete the task.
Radicals: Addition/Subtraction - A
A radical sign looks like a check mark with a line across the top. The radical sign is used to
communicate square roots. The square root of a number is the number that when squared, or multiplied
by itself, is equal to the radicand. The radicand is the number under the radical sign. For example, the
square root of 64 is equal to 8, because 8 squared (8 8) is 64, and 64 is the radicand.
Students must be able to combine (add and subtract) radicals. In order for radicals to be combined, the
radicands must be identical. The square root of 3 and the square root of 7 cannot be added together in
their present form because their radicands (in this case, 3 and 7) are not identical. When combining
radicals, the radicands themselves do not get added or subtracted.
SD Algbra-Questions, Answer Key, Study Guide
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Example 1: Add and simplify.
Step 1: Check to see if the radicands are identical. In this case, both radicands are 5, so the radicals can
be added.
Step 2: Add the coefficients (numbers in front of the radicals). The answer is the sum of the coefficients
times the radical.
Example 2: Subtract and simplify.
Step 1: Check to see if the radicands are identical. In this case, both radicands are 3, so the radicals can
be subtracted.
Step 2: Subtract the coefficients and the answer is the difference of the coefficients times the radical.
There is a silent 1 in front of the radical in the final answer.
Example 3: Simplify.
Step 1: Check to see if the radicands are identical. In this case, two of the radicands are 7, so those two
radicals can be subtracted.
Step 2: Subtract the coefficients and the answer is the difference of the coefficients times the radical,
plus the remaining radical.
Example 4: Simplify.
Step 1: Check to see if the radicands are identical. In this case, two of the radicands are 3, so these two
radicals can be added. Add the coefficients and then multiply the sum by the radical.
Step 2: Rewrite radical 16 in simplest form. Since 16 = 4 4, radical 16 can be simplified to 4.
SD Algbra-Questions, Answer Key, Study Guide
Page 16 of 102
Step 3: The final answer is the answer to step 1 minus the answer to step 2.
An activity that can help reinforce the concept of adding and subtracting radicals is to name radicals
with the names of animals. For example, name the square root of 6 "elephant," and the square root of 5
"lion." Then say, "If you have 5 elephants and you add 10 more elephants and 2 lions, how many
elephants and lions do you have altogether?" Answer: 15 elephants and 2 lions, or, 15 times the square
root of 6 plus 2 times the square root of 5. This activity keeps students from trying to combine radicands
in the answer.
Inequalities - C
An inequality is a number sentence that uses "is greater than," "is greater than or equal to," "is less than," "is
less than or equal to," or "is not equal to" symbols. For example, 6n > 4 is a number sentence with an inequality
symbol, and is read "six times n is greater than four." Inequalities can be identified in real world situations by
expressions such as "is less than," "is more than," "at least," and "at most." Sentences can be translated to
number sentences using the following symbols.
An inequality can be solved for a variable (a letter that represents a number) in the same way that an
equation is solved. An example of solving an inequality follows.
Example 1: Solve the inequality for x.
2x + 7 > 11
Step 1: Subtract 7 from each side of the inequality symbol to isolate the variable term, 2x, and simplify.
Step 2: Divide both sides of the inequality by 2 to completely isolate x.
Answer: x > 2
A compound inequality has more than one condition and can be identified in a number sentence by two
inequality symbols, or in a real world application by the words "and" or "or." For example, 3 < x < 9 is
a compound inequality. To read a compound inequality, start in the center and read left: x > 3, then go
back to the center and read to the right: x < 9. The compound inequality is a combination of x > 3 AND
x < 9. Solving a compound inequality is very similar to solving a single inequality.
Example 2: Solve the compound inequality for x.
4 < 3x + 1 < 7
SD Algbra-Questions, Answer Key, Study Guide
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Step 1: Subtract 1 from the left, center, and right side of the number sentence to isolate the variable term,
3x.
Step 2: Divide the left, center, and right side of the number sentence by 3 to isolate x.
Answer: 1 < x < 2. This answer can be interpreted to mean that x is greater than 1 and x is less than 2.
Although the above examples do not include negative numbers, students must remember to switch the
inequality sign when multiplying or dividing by a negative number.
Once the student is familiar with solving compound inequalities, he or she should be ready to solve them
in real world situations.
Example 3: The telephone company charges $21.95 per month for basic service plus $0.17 per local
call. What is the maximum number of calls that can be made in a month if a family can spend at least
$27.50 and at most $32.75 per month for telephone service? Round your answer to the nearest whole
number of calls that satisfies the inequality.
Step 1: Translate the information in the problem to a compound inequality. Let c represent the number
of calls. The family gets charged $21.95 + $0.17c each month (the service charge plus local calls).
Since the charges are at least $27.50 and at most $32.75, place the minimum amount of $27.50 to the
left of the expression that represents the charges and the maximum amount of $32.75 to the right of the
expression. Refer to the chart above to see which inequality symbol to use. Since the question uses the
words at least and at most, the needs to be used. Place the on either side of the charges expression.
Step 2: Isolate the variable term $0.17c by subtracting $21.95 from the left, center, and right of the
compound inequality. Simplify.
Step 3: Divide each term of the new compound inequality by $0.17 to isolate c.
Step 4: Simplify the inequality.
Step 5: Since the maximum number of calls must be less than or equal to 63.53, the whole number that
will satisfy the inequality is 63. Rounding up to 64 would no longer satisfy the inequality because 64 is
greater than 63.53.
Answer: 63 calls
Note: If the question had asked what the minimum number of calls would be, the answer would be 33,
since the minimum number of calls has to be greater than or equal to 32.65.
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An activity that can help reinforce the concept of inequalities is to ask the student to make a budget
using the amount of money he or she earns each month (approximate or make up an amount if
necessary). Then, make up scenarios using the budget. For example:
It costs $120 per month to insure a car, plus $2.05 per gallon of gasoline used. If you have budgeted at
least $150.00 and at most $200.00 for car expenses this month, what is the maximum number of gallons
of gas you can buy?
Joint and Combined Variation
Variation equations are formulas that show how one quantity changes in relation to one or more quantities.
There are four types of variation: direct, indirect (or inverse), joint, and combined. This skill focuses on joint
and combined variation.
Direct variation equations show a relationship between two quantities such that when one quantity
increases, the other also increases, and vise versa. We can say that y varies directly as x. Direct
variation formulas are of the form y = kx, where the number represented by k does not change and is
called a constant of variation. For example, the amount of money in a paycheck (P) varies directly as
the number of hours (h) worked. In this case, the constant k is the hourly wage, and the formula is
written P = kh.
Indirect variation formulas show that when one quantity increases, the other quantity decreases, and vise
versa. For example, when the price of an item increases, the demand decreases. Indirect variation
formulas are of the form y = k/x.
Joint variation formulas show the relationship between a quantity and the product of two other
quantities. They are of the form y = kxz. For example, the equation for the area of a triangle states that
the area is equal to one half times base times height, or A = ? bh. The number ? is the constant of
variation, and is always the same for the area of a triangle.
Combined variation formulas show how a quantity varies directly with one or more quantities and
indirectly with one or more different quantities. They are of the form y = (kx)/z. For example, the sales
(S) of a company vary directly as the amount spent on advertising (a), and indirectly as the price (p) of
the item sold. The formula is written as S = (ka)/p. When the amount spent on advertising increases,
sales increase because sales and advertising vary directly. When the price increases, sales decrease,
because price and sales vary indirectly.
To solve variation problems, follow these steps.
Step 1: Substitute the quantities given for the letters they represent.
Step 2: Solve for the constant k.
Step 3: Set up the equation again.
Step 4: Substitute given quantities and the value of k that was just determined.
Step 5: Solve for the quantity needed.
Example 1: The surface area (S) of a cube varies jointly with the length (l) and width (w) of one of the
faces. The equation that represents this relationship is given by
S = klw
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where k is a constant that needs to be determined. If the surface area of a cube is 150 square inches
when the length and width of each face are 5 inches, what are the length and width of each face when
the surface area is 216 square inches?
Step 1: Write the values for the variables that are known (do not use the values for a surface area of 216
because the constant k is not known).
Step 2: Substitute the values into the formula and solve for k. Multiply the terms on the right side of the
equal sign to get 25k. Isolate the k by dividing each side of the equation by 25. k = 6.
Step 3: Now that the constant is known, it is possible to determine the length and width of a cube that
has a surface area of 216 in.2 . Write the values for the variables that are known for a surface area of
216.
Step 4: Substitute the values into the variation equation and solve for l and w. Multiply the right side of
the equation to get 6l2 . Since length and width are the same for a cube, (l)(w) can be written l2 or w2 .
Divide each side of the equation by 6 to isolate the l2 . To remove the square from l2 , take the square
root of each side of the equation. l = 6 and w = 6 (because l and w are the same).
Answer: length = 6 inches and width = 6 inches
Example 2: The amount of monthly sales (S) for a company that makes a certain type of chair varies
directly with the amount spent on advertising each month (a) and indirectly with the price (p) of the
chair. The equation that represents this relationship is given by
where k is a constant to be determined. If the sales for July were $40,000.00 when
advertising costs were $100.00 and the price of the chair was $50.00, what were the sales for August if
advertising costs were increased to $135.00 and the price of the chair was decreased to $45.00?
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Step 1: Write the values for the variables that are known (do not use the values for advertising cost of
$135 and price of $45 because the constant k is not known).
Step 2: Substitute the values into the formula and solve for k. Multiply the terms on the top of the
fraction to get 100k. Divide 100k by 50 to further simplify the right side of the equation and get 2k.
Isolate the k by dividing both sides of the equation by 2, yielding k = 20,000.
Step 3: Now that the constant is known, it is possible to determine the sales for advertising costs of $135
and price of $45. Write the values for the variables that are known for these constraints.
Step 4: Substitute the values into the variation equation and solve for S. Multiply the numbers on the top
of the fraction to get 2,7000,000. Divide 2,700,000 by 45 to continue to simplify the equation and
determine that the sales were $60,000.00
Answer: $60,000.00
An activity that can help reinforce the concept of variation is to describe situations involving joint and
combined variation, and then have the student write the equations relating the quantities. For example,
"The volume of a cylinder varies jointly as the square of the radius of the cylinder and its height." The
student would write: V = kr2 h.
Domain/Range
A relation is a collection of (x, y) coordinate points. For example:
The domain of the relation is the set of x-values in the collection of coordinate points that make up the
relation, and the range is the set of y-values. The domain and range of a relation are written as sets of
numbers in ascending order within brackets and each number is only written once. For example:
This skill focuses on giving students practice determining the
domain and range of relations.
Example 1: Find the domain of the following relation r.
Solution: The domain is the set of all x-values of the coordinate points that make up the relation.
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Answer:
Example 2: Find the range of the following relation r.
Solution: The range is the set of all y-values of the coordinate points that make up the relation. NOTE:
Although the - 4 is used as a y-value twice, it only needs to be written once in the range.
Answer:
An activity that can help reinforce the concept of domain and range is to show students a collection of
coordinate points and ask them to name the member of the domain of each coordinate point (x-value)
and the member of the range of each coordinate point (y-value).
Function Rules
A function rule is an equation relating two variables. Variables are letters that represent unknown
numbers, such as x and y.
In a function, the value of the output variable, y, depends on the value of the input variable, x. The
function rule takes a given value of x, and performs an operation (addition, subtraction, multiplication,
or division) on it to make it equal to y.
[i]
[i]
A table of related values for input-output variables that can be generated from a function rule is called a
function table. It is important for students to understand what a function table represents. To create the
function table below, x-values were chosen and placed in the table. Next, the x-values were substituted
into the function rule and a y-value was generated for each x-value. The final function table appears on
the right without the function rule.
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It is possible to determine the function rule from a table of values for related input-output variables. The
first two examples deal with simple function rules that use addition and subtraction only.
Example 1: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: Determine what needs to be added to or subtracted from - 3 to make it equal to 3. The solution is,
"add 6," because - 3 + 6 = 3.
Step 2: Add 6 to each of the remaining x-values, and verify that they match the values listed for y.
Step 3: Write the function rule: y = x + 6.
Example 2: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: Determine what needs to be added to or subtracted from 1 to make it equal to - 4. The solution is,
"subtract 5," because 1 - 5 = - 4.
Step 2: Subtract 5 from each of the remaining x-values, and verify that they match the values listed for y.
Step 3: Write the function rule: y = x - 5.
Function rules can also be of the form y = ax + b, or y = ax. Rules of this form require the student to use
multiplication, as well as addition and subtraction.
Example 3: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
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Step 1: First, the student should test the y values to see if they are multiples of the x values. If the
function is of the form y = ax, then the student should be able to multiply all of the values in the x
column by the same number, and arrive at the values in the y column. Multiplying - 4 by 3 yields - 12.
Step 2: Multiply the remaining x values by 3, and verify that they match the values listed for y.
Step 3: Write the function rule: y = 3x.
Example 4: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: First, the student should test the y values to see if they are multiples of the x values. If the
function is of the form y = ax, then the student should be able to multiply all of the values in the x
column by the same number, and arrive at the values in the y column. If the student is unable to
determine a single factor that yields the output values, he or she should continue to step 2. Since there
isn't a factor that yields - 19 when multiplied by - 4, the student should continue to step 2.
Step 2: If there is an input value equal to 0, the student should always begin with this value. This line in
the table will help the student to determine the value that is being added or subtracted since any factor
multiplied by 0 equals zero. In this case, 3 is being subtracted from x because 0 - 3 = - 3. Therefore, the
function must be of the form y = ax - 3.
Step 3: The student must now determine what value of a satisfies the function. Determining this value is
made easier by performing the inverse of the operation found in step 2 on a chosen y value. Adding 3 to
-19 yields - 16. Since - 4, (the corresponding x value) must be multiply by 4 in order to yield -16, the
factor 4 is the correct value for a.
Step 4: Multiply each input value by 4 and subtract 3 in order to verify that they match the values listed
for y.
Step 5: Write the function rule: y = 4x - 3.
Exponential Functions
An exponential function is an equation that has a variable in the exponent. The functions below are all
exponential functions, where f(x) represents the letter y and is dependent on the value of x.
When x-values are substituted into the equation and the right side of the equal sign is simplified, a list of
f(x) (or y) values is obtained. This process gives you a set of (x, y) coordinate pairs that can be plotted
(or graphed) on a coordinate plane.
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Step 1: Make a table of values with the following column headings.
Step 2: Choose values to enter into the x column
(remember to choose the smallest possible values to keep computations as simple as possible). It is
generally necessary for the values to contain both negative and positive numbers and the number zero.
As few as three values could be used, but this example will use these five values: -2, -1, 0, 1, 2. Enter
the values into the table.
Step 3: Substitute each x-value into the equation in the center column for x and simplify. This will
determine the y-values for the coordinate points.
Step 4: Write the (x, y) pairs generated in Step 3
into the (x, y) column, using the values from the x column and the y-values from the center column.
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Step 5: Plot the points on a coordinate plane to
graph the exponential function.
Step 6: Compare this graph to the solution choices and choose
the correct answer, which is choice D.
Answer: D.
Step 1: Make a table of values and choose x-values to substitute into the equation.
Step 2: Substitute the values in the x column into
the equation in the center column to determine the corresponding y-values. Write the coordinate points
in the (x, y) column.
Step 3: Plot the points on a coordinate plane to
graph the exponential function.
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Step 4: Compare this graph to the curves in the question and
choose the correct answer.
Answer: W
Comparing Graphs of Exponential Functions:
The standard form for an exponential function is shown below.
For purposes of comparing graphs of exponential functions, two more
variables need to be added, such that the standard form becomes the form shown below.
The value of c determines whether the graph shifts upward or
downward and the value of d determines whether the graph shifts right or left. See the table below.
Example 3:
Step 1:
Step 2: Look at the table above to determine that the graph of
the exponential function should move upward 3 units and to the right 4 units.
Step 3: Compare the actual graphs below to see if these predictions were right.
An activity that will help reinforce the skill of graphing exponential functions is to explore exponential
functions with the student using a graphing calculator. Make up a list of exponential functions like the
one below.
Follow the steps to graph each function using a graphing calculator. Notice the way changing a sign or
adding a term alters the graph.
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Step 1: Follow the instructions in your graphing calculator manual to set the viewing window to the
following values: x min = -10, x max = 10, xscl = 1; y min = -10, y max = 10, y scl = 1. Exit the
window.
Step 2: Press the y = key to open the equation editor. Delete any equations that are listed. Enter the
following keystrokes for the first exponential equation on the list: 5^x. Press Enter.
Step 3: Press the Graph key to display the graph of the function.
**NOTE: This activity could be completed using graph paper and tables of values to graph the
exponential functions.
Exponential Notation - F
In the expression 32 , the number 2 is called an exponent and the number 3 is called a base. The exponent
determines the number of times the base is multiplied by itself. For example: 32 = (3)(3) = 9.
Rational exponents are exponents that are in fraction form. The following are examples of expressions
with rational exponents.
Expressions with rational exponents can be written in radical form. The numerator is the number of
times the base, which is placed under the radical sign, is multiplied by itself. The denominator is the root
of the radical expression. A root is the inverse of an exponent. The examples below show how
expressions with rational exponents are written in radical form.
A root of 2 is called a square root and is the inverse of the exponent 2. A root of 3 is called a cube root
and is the inverse of the exponent 3. The following examples show how to simplify expressions with
rational exponents in two ways.
Simplifying by converting to radical form:
Example 1: Simplify.
Step 1: Rewrite the rational expression in radical form.
Step 2: Simplify: 8 to the first power equals 8.
Step 3: Factor 8 into (2)(2)(2).
Step 4: Since there are three factors of 2 under the cube root, the radical expression will simplify to 2.
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Answer: 2
Simplifying using exponent laws:
Step 1: Rewrite the expression and factor 8 into (2)(2)(2).
Step 2: Write (2)(2)(2) in exponential form.
Step 3: Multiply 3 and 1/3 (remember that when an exponent is taken to another power, the two powers
are multiplied) to get 1. Simplify 2 to the first power to 2.
Answer: 2
Example 2: Simplify.
Step 1: Rewrite the expression as a radical expression.
Step 2: Simplify 322 : 32 32 = 1,024.
Step 3: Factor 1,024 into (4)(4)(4)(4)(4). Always try to factor the number under the radical into the
same number of factors as the root.
Step 4: Since this is a 5th root and there are five 4s under the radical, the radical expression will simplify
to 4.
Answer: 4
Activities that can help reinforce the concept of rational exponents are as follows.
1. Take 30 index cards, and make three piles of ten cards each. Write expressions with rational
exponents on the first set of ten cards. On the second set, write the radical form for each expression you
wrote on the first ten cards. On the third set, write the simplified answer to the ten expressions. Shuffle
all of the cards and lay them face down on a table. Each player takes turns flipping three cards. If all
three cards go together, the player has made a match and gets to try again. If not, it is the next player's
turn.
2. Name expressions with rational roots, and ask the student to factor the base into the same number of
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identical factors as the root. For example, if the root is 1/6, the student should factor the base into 6
identical factors, so that when the factors are rewritten in exponential form, and the exponent is
multiplied by the root, the product is 1.
Polynomials: Subtraction
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a monomial.
The sum or difference of two or more monomials is called a polynomial.
Here is an example of a polynomial:
Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are
monomials that have the same variable or variables for which the variable or variables have the same
exponent.
To subtract polynomials, first write the polynomials as one long polynomial. Then distribute the
subtraction sign through the second polynomial. Finally, combine like terms. Practice by subtracting the
following polynomials.
Example 1:
Step 1: Set up the two polynomials as one long polynomial. Since the problem is to subtract one
polynomial from another, the second polynomial in the problem must be written first.
Step 2: Distribute the subtraction sign through the second polynomial. This involves changing the sign
of each term in the second polynomial.
Step 3: Rewrite the polynomial after changing the signs in the second polynomial.
Step 4: Combine like terms.
Answer: 5p + 9
Example 2: Subtract four times a number decreased by ten from eight times the same number less six.
Step 1: "Four times a number decreased by ten" can be written (4x - 10).
Step 2: "Eight times the same number less six" can be written (8x - 6).
Step 3: Now the problem reads: Subtract (4x - 10) from (8x - 6).
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Step 4: Set up the polynomials as one long polynomial.
Step 5: Distribute the subtraction sign through the second polynomial. This involves changing the sign
of each term in the second polynomial.
Step 6: Rewrite the entire polynomial after changing the signs in the second polynomial.
Step 7: Combine like terms.
(4x - 10) subtracted from (8x - 6) equals 4x + 4.
Answer: 4x + 4
Example 3: Find area of the shaded region.
Step 1: Determine the area of the large rectangle by multiplying the length (2x + 7) by the width (3x).
This involves multiplying each term in (2x + 7) by 3x.
Step 2: Determine the area of the small rectangle by multiplying the length (2x - 2) by the width (x).
This involves multiplying each term in (2x - 2) by x.
Step 3: Now subtract the area of the small rectangle from the area of the large rectangle. Remember to
put the second polynomial in parentheses since this is subtraction.
Step 4: Distribute the subtraction sign through the second polynomial. This involves changing the sign
of each term in the second polynomial.
Step 5: Combine like terms.
Answer:
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Equations: Order of Operations
An equation is a statement in which two numbers or two expressions are set equal to each other. For example,
5 + 3 = 8 and 16 = 3c + 4 are equations.
When solving equations, find the value of the variable by getting the variable alone on one side of the
equal sign. To do this, undo any operations on the variable by using the inverse operation. Any operation
done on one side of the equal sign must be done on the other side of the equal sign in order to keep the
statement true.
If a number has been added to the variable, subtract the number from both sides of the equation.
m + 3 = 5
- 3 - 3
m = 2
If a number has been subtracted from the variable, add the number to both sides of the equation.
b - 7 = 9
+ 7 + 7
b = 16
If a variable has been multiplied by a nonzero number, divide both sides by the number.
If a variable has been divided by a number, multiply both sides by the number.
When solving 2-step equations, we must first undo the addition or subtraction using the inverse
operation, then undo the multiplication or division:
Example 1: Solve the equation for t.
5(t - 4) = t + 12
Step 1: Multiply 5 times the terms inside the parenthesis.
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Step 2: Add 20 to both sides of the equation
Step 3: Subtract t from both sides of the equation
Step 4: Divide both sides of the equation by 4.
Answer: t = 8
Example 2: Evaluate the expression for c = 3:
2(c + 4) + 2(15)
(1) 2(3 + 4) + 2(15)
(2) 2(7) + 2(15)
(3) 14 + 30
(4) 44
Step 1: Substitute 3 in place of 'c' in the expression.
Step 2: Add the numbers in parentheses.
Step 3: Rewrite the equation after performing all multiplications in order from left to right.
Step 4: Add 14 and 30 to get 44.
Answer: 44
Functions: Linear
This skill requires students to solve real world problems involving two linear functions. A linear function is a
function whose graph forms a non-vertical straight line.
Story problems are often very difficult for students to master. It may be beneficial first to confirm that
the student is comfortable with solving two linear equations outside of a real world context. To do this,
students must review how to solve linear equations. In some variable equations, students are given the
value of one variable. Letters expressing unknown values in equations are called variables. To find the
value of the second variable, substitution must be used.
If y = 4x, and x = 9, substitute the given value of x in the equation, y = 4x. The result is y = 4(9).
Calculate the right side of the equation to get y = 36. Now the values of both variables are known: x = 9
and y = 36.
To check the answer, set up the equation y = 4x as if x was an unknown variable: 36 = 4x. Divide each
side of the equation by 4. The result is x = 9, so the values for x and y are both correct.
Here is another situation where the value of a variable is substituted into an equation to solve for x.
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Example 1:
Solve for x.
x = 3y + 9
y=-2
Solution: Substitute - 2 in place of y in the first equation. Then, solve for x.
x = 3(- 2) + 9
x=-6+9
x=3
Answer: x = 3
This concept can be used to solve two equations that include the same two variables.
Example 2:
Solve for x and y in the following equations.
y = 2x - 3
3x = y + 10
Solution:
If y = 2x - 3 and 3x = y + 10, there is not a number to substitute in for either of the variables. So, an
expression must be substituted for a variable. In this case, the y variable is already isolated. It is equal
to 2x - 3. This can be substituted in for y in the second equation.
(1) 3x = y + 10
3x = (2x - 3) + 10
(2) 3x = 2x + (- 3 + 10)
3x = 2x + 7
(3) 3x = 2x + 7
-2x -2x
x=7
(4) y = 2x - 3
y = 2(7) - 3
(5) y = 14 - 3
y = 11
Step 1: Substitute 2x - 3 for y in the second equation.
Step 2: Simplify the equation by removing the parentheses and combining the like terms - 3 and 10.
Step 3: Isolate the variable by subtracting 2x from each side and combining like terms.
Step 4: Since there is now a value for one of the variables, it can be substituted back into either of the
original equations. In this example, the first equation was used.
Step 5: Calculate the right side of the equation to get y = 11.
Answer: x = 7 and y = 11
Several questions from this skill relate to measurement concepts. The formula for perimeter is
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occasionally needed to find solutions to story problems. Perimeter is the measurement around a figure.
Example 3:
Find the perimeter for the figure below.
Solution:To calculate the perimeter of a figure, add the lengths of all the sides of the figure. For
example, this figure has four sides measuring 3 inches, 7 inches, 3 inches, and 7 inches.
P = 7 + 3 + 7 + 3 = 20
Answer: The perimeter of this figure is 20 inches.
Story problems that involve perimeter and two linear equations can be solved by writing two equations,
then using the substitution method that was reviewed earlier.
Example 4:
Emilio built a square sandbox for his dog with a perimeter of 15 m. Since his dog has grown, he wants
to rebuild the sandbox so that each side is 2 m longer than the original sandbox. Before he buys the
materials to build the new sandbox, he needs to know the perimeter. What is the perimeter of the new
sandbox?
Solution:
To solve this problem it is probably best to start with drawing a picture. Because we do not know the
length of the sides of the old enclosure, they can be assigned a variable, x. The new enclosure must have
sides that are 2 m longer than the original, so the sides can be written as x + 2.
(1) For the old enclosure, x + x + x + x = 15, or 4x = 15
For the new enclosure, (x + 2) + (x + 2) + (x + 2) + (x + 2) = y, or 4(x + 2) = y.
(2) 4x = 15, so x = 3.75
(3) 4(3.75 + 2) = y
(4) y = 23
Step 1: An equation can be written for the perimeter of the new and old enclosures based on the
information in the problem. For the new enclosure, we do not know the perimeter so we give it a
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different variable, y.
Step 2: The x must be isolated in the first equation so that it can be substituted into the second.
Step 3: This value can be placed in the second equation for x.
Step 4: Calculate.
The perimeter of the new enclosure will be 23 m.
Answer: 23 m
Another common application of finding the solution to two equations in solving real world problems
involves money.
Example 5: Ariel wants to go snowboarding this weekend. Ice Mountain charges $24 for a lift ticket
and $8 per hour of snowboarding. Flurry Ridge charges $29 for a lift ticket and $6 per hour of
snowboarding. Ariel must get a lift ticket to snowboard and she wants to snowboard for 5 hours. Which
ski area would be cheaper for Ariel's snowboarding trip?
(1) For Ice Mountain, the cost = 24 + 8h.
For Flurry Ridge, the cost = 29 + 6h.
(2) For Ice Mountain, 24 + 8(5) = 64
For Flurry Ridge, 29 + 6(5) = 59.
Step 1: Set up two equations, one for each snowboarding location. We can use the variable h for the
number of hours.
Step 2: Since Ariel wants to snowboard for 5 hours, we can substitute 5 in for h in each of the equations
and solve.
Answer: Ice Mountain will cost her $64 and Flurry Ridge costs her only $59, so Flurry Ridge is
cheaper.
To help students understand the application of linear equations in solving real world problems, students
can find the costs per unit of time or distance of different services with the help of a telephone book, the
newspaper, or the internet. If specific cost information is not listed, call the company for information.
For example, a car rental company may charge a base fee of $35 per day and $0.59 per mile or a
catering company may charge a rental fee of $100 for materials and then $60 per hour for waitstaff
services. Then, have students calculate which would be the cheapest and which would be the most
expensive for a specified distance a rental car may drive or amount of time a catered party may last.
Quadratic Equations: Real World Problems
A quadratic equation is a second-order polynomial equation in a single variable. Second-order means that one
of the terms contains the variable to the power of two and no other exponents are greater than two. Students
will apply quadratic equations to evaluate real world problems in this question set.
Real world problems are often difficult for students to master. It may be beneficial first to confirm that
the student is comfortable with quadratic equations outside of a real world context. To do this, students
must first review the basic principles of quadratic equations, including how to distribute terms in order
to obtain the equation in the proper form, how to factor, and eventually, how to solve for the variable.
Review of Definitions:
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• A variable is an unknown number in an equation and is usually represented by a letter such as x.
• A coefficient is a number representing a fixed, or constant, value that is multiplied with a variable.
• A polynomial is a term or sum of terms. Each term is either a number or the product of a number and
one or more variables. For example, 2x2 + 3 is a polynomial with two terms, 2x2 and 3.
• A binomial is a polynomial with two terms, and a trinomial is a polynomial with three terms.
• Factoring is the breaking up of quantities into products of their component factors. One way to think
of factoring is as the opposite, or inverse, of multiplying.
Quadratic functions are written in the form:
where f(x) is read "f of x." The letters a, b, and c are constants and a 0.
When solving quadratic equations, replace the y or f(x) with zero. It is important to remember that all
quadratic equations have two answers. However, the answers could be imaginary (contain a square root
of a negative number), positive, or negative. When dealing with real world problems, the positive
answer is typically the only one that would make sense in the context and would therefore be the only
correct answer.
Below are a few examples of quadratic equations:
Sometimes, quadratic equations come in a form where they have already been factored. Some examples
of these equations are shown below.
(x + 5)(x - 1) = 7
or
2x(x + 6) = 54
In order to find the solution to quadratic equations, they have to be set equal to zero. To get the
equations in this form, multiply out the terms to eliminate the parentheses and subtract appropriately
until the polynomial is equal to zero. Once an equation is written in the proper form, the next step
towards solving for the variable is factoring the equation.
Example 1:
Factor completely and solve for y.
y2 + 4y - 2 = 3
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Step 1: Subtract 3 from both sides of the equation so that the equation is equal to zero.
Step 2: Determine the factors. Since the coefficient of the y2 term is equal to 1, focus on the last term,
in this case, - 5. If factors of - 5 can be found that add up to the coefficient of the "y" term (in this
example, 4), the trinomial can be factored. The factors of - 5 are 5 and - 1 or - 5 and 1. When we add
the first pair of numbers, 5 and - 1, we get 4, the number we are seeking.
Step 3: Check your answer. To check the result, multiply the binomial factors together. Binomials are
multiplied using the FOIL method. FOIL is the abbreviation for First, Outside, Inside, and Last.
• Multiply the First terms of each binomial (y y = y2 )
• Multiply the Outside terms of each binomial (y -1 = -1y)
• Multiply the Inside terms of each binomial (5 y = 5y)
• Multiply the Last terms of each binomial (5 - 1 = - 5)
• Combine like terms to simplify the polynomial.
Step 4: Set each factor equal to zero. When we multiply any value by zero the result is zero. Therefore,
if either factor is equal to zero, the entire expression is equal to zero.
Step 5: Solve each of these equations by isolating the variables.
Answer: y = - 5 and y = 1
A special type of polynomial expresses the difference of two perfect squares. Polynomials of this type
are easily factored once the pattern is remembered.
Since each term in the polynomial is a perfect square, the square root of each term (in this case x and 6,
respectively) will be used in the following way. The original polynomial is factored as (x + 6)(x - 6).
Notice that if these terms are multiplied together using the appropriate FOIL method, the original
polynomial is formed. Polynomials that are in the "difference of squares" form, that is, a2 - b2 , can
always be factored as the sum of the square roots, (a + b), multiplied by the difference of the square
roots, (a - b).
Using Quadratic Equations in Real World Problems:
When solving real world problems, clues from the text of the problem must be used to set up the
equations. Quadratic equations are often used to find solutions to story problems. The story problems
in this set also occasionally require the use of formulas for the area of a rectangle, the area of a triangle,
or the Pythagorean Theorem.
Area is the measure of the interior of a two-dimensional region and is calculated in square units. The
formulas used to find the area of a rectangle and a triangle are shown below.
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Area of a rectangle = length x width, or A = lw. A square is a rectangle with four equal sides, so the
formula for the area of a square is often shown as Area of a square = side2 , or A = s2 because the length
is equal to the width for all squares.
Area of a triangle =1/2 x base x height, or A = (1/2)bh. The height of a triangle is the length of the
line segment that extends from the vertex opposite the base and is perpendicular to the base, as shown
below.
The Pythagorean theorem is used to find the lengths of the sides and hypotenuse of a right triangle. The
Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of
the squares of the legs of that triangle. A right triangle is a triangle containing a 90º angle, the
hypotenuse is the side opposite the right angle (longest side of the triangle), and the legs are the two
other sides.
Using the diagram above, it is important to remember that a and b always denote legs of a right triangle
and c always denotes the hypotenuse of a right triangle.
Sometimes, the length or width are unknown in an area problem or the leg is unknown in a right triangle
problem. Examples of these types of problems are provided below.
Example 2:
Colleen is swimming in the ocean. She swam 4 miles south and then turned and swam 3 miles west. If
she turns and swims straight back to the point where she started swimming, how many more miles does
she need to swim?
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Step 1: Determine the values of a, b, and c. In this case, a and b are known and c is unknown.
Step 2: Substitute the values of a and b into the Pythagorean theorem.
Step 3: Square 3 (3 3 = 9) and square 4 (4 4 = 16).
Step 4: Add 9 and 16 to get 25.
Step 5: Take the square root of each side of the equation. The square root of 25 is 5 and the square root
of c2 is c.
Answer: She will swim 5 miles.
Example 3:
Find the height of a triangular sail that has an area of 60 m2 and a base of 8 m.
Step 1:
Step 2:
Step 3:
Step 4:
Begin with the formula for the area of a triangle.
Substitute in 60 for the area and 8 for the base.
Simplify the equation by multiplying 1/2 and 8 to get 4.
Isolate the height by dividing each side of the equation by 4.
Answer: The height of the sail is 15 m.
When working with story problems, use a variable for the unknown value and clues within the wording
to write the equations that will need to be solved. For example, if a problem reads "the length is 4 m
longer than the width," then the length can be represented by w + 4. After an entire equation is formed,
it can then be solved.
Example 4:
The width of a rectangular cake pan is 3 inches shorter than the length of the pan. The area of the cake
pan is 108 in.2 What is the length of the cake pan?
Solution:
(1)
(2) x(x - 3) = 108
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(3) (x)(x) - (x)(3) = 108
x2 - 3x = 108
(4) x2 - 3x - 108 = 0
(5) (x - 12)(x + 9) = 0
(6) x - 12 = 0 and x + 9 = 0
(7) x = 12 and x = - 9
Step 1: Drawing a picture of the cake pan with the information in the problem may help in solving the
problem. The length is the unknown value, so it can be represented by the variable x. The width is 3
inches shorter than the length, so it can be written as x - 3.
Step 2: Since this problem deals with the area, length, and width of a rectangle, we must use the
formula for the area of a rectangle, Area = length x width, and substitute in x for the length, x - 3 for the
width, and 108 in.2 for the area.
Step 3: Distribute the x and simplify.
Step 4: Subtract 108 from both sides to put it in the proper form.
Step 5: Factor the equation.
Step 6: Set each polynomial equal to zero.
Step 7: Isolate the variable in each problem.
Answer: The answer x = - 9 does not make sense because the length of a cake pan cannot be a negative
value. That answer is discarded and the correct answer must be x = 12. The length of the cake pan is 12
in.
To help students understand the application of quadratic equations in solving real world problems, help
the student find the missing dimensions of various objects around the house when given the area and
information about one of the dimensions. For example: Measure a picture frame, door, etc., calculate
the area of the object, and create a relationship between the dimensions (the length of the door is twice
the width). Then have the student find the missing dimension when given the area and the relationship
between the dimensions. An activity to see how the Pythagorean theorem works is to lean a board or
ladder against a wall to create a right triangle. Measure two dimensions of the triangle and use the
Pythagorean theorem to determine the length of the third side of the triangle.
Functions/Relations - A
A relation is a set of ordered pairs that represent a relationship between the elements of the two sets. A function
is a special type of relation, where each element of the first set (x-values) corresponds to an unique element of
the second set (y-values). The first set of numbers is commonly known as the input and the second set as the
output. The input, or x-values, are entered into the equation. Once evaluated, the result is the output, or yvalues. In other words, in order for a relation to be a function, for each x-value there can be no more than one
value of y. Some examples of relations are given below, with input values in A mapped to output values in B.
Relations 1 and 2 are functions, while relation 3 is not a function. The input value - 1 in relation 3 is
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matched to more than one output value (3 and 5), so the relation is not a function.
Example 1:
Which of the following relations is not a function?
Solution:
If there is a value of x resulting in more than one value of y, the relation is not a function. This only
occurs in the third set of numbers with (9, - 1) and (9, 4). Therefore, set C is not a function.
Answer: Set C is not a function.
Example 2:
Which of the following points, if removed from the set, would make the set a function?
Solution:
The ordered pairs (- 4, 5) and (- 4, 4) have the same x values but different y values. Therefore, if either
point is removed from the set, the remaining ordered pairs will represent a function.
Answer: Remove either (- 4, 5) or (- 4, 4).
Equations With Two Variables
This study guide will focus on solving equations that contain two variables.
Remember:
Variables are letters or symbols that represent numbers that are unknown.
Expressions are variables or combinations of variables, numbers, and symbols that represent a
mathematical relationship. Expressions do not have equal signs, but can be evaluated or simplified.
Example: y - 9
Equations are expressions that contain equal signs. They can be solved, but not evaluated.
Example: 5 + x = 13
Like terms are those terms that have the same variable(s) in common, or no variable. Like terms can be
combined.
Examples: 2 and 3 are like terms, 2x and 3x are like terms, but 2 and 3x are not like terms.
Solving Equations
To solve an equation, it is necessary to "undo" what was done to the variable in question. Another way
to think about this is to do the order of operations in reverse.
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P = Parenthesis, E = Exponents, M = Multiplication, D = Division, A = Addition, S = Subtraction.
**In the normal direction, first evaluate multiplication and division from left to right (whichever comes
first), then evaluate addition and subtraction from left to right (whichever comes first).
**To "undo" what was done, reverse the direction.
•Addition "undoes" subtraction.
•Subtraction "undoes" addition.
•Multiplication "undoes" division.
•Division "undoes" multiplication.
Addition/Subtraction Properties - Adding or subtracting the same real number to each side of an
equation will result in an equivalent equation.
Example 1: Solve 16 = 5 + x - 8 for x.
Step 1: Begin to isolate the variable, x, by adding 8 to both sides of the equation. Remember to only add
and subtract like terms, so the 8 must be added to the - 8 and the 16.
Step 2: Subtract 5 from both sides of the equation to completely isolate the variable, x.
Answer: x = 19
Multiplication/Division Properties - Multiplying or dividing each side of an equation by the same
(nonzero) number will result in an equivalent equation.
Example 2: Solve the following equation for t.
Step 1: Multiply both sides of the equation by 17 to eliminate the fraction.
Step 2: Divide both sides of the equation by 4 to isolate the variable, t.
Answer: t =136.
Solving Equations With Two Variables
In order to solve equations that contain two variables, the student will need to solve for one variable in
terms of the other. This means that the answer may not be entirely numeric.
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The same properties as described above should be used when solving for a given variable in a twovariable equation. Students should treat the second variable as if it were a number.
Example 3: Solve 30w - 9v = 6 for w.
Step 1: Begin to isolate w by adding 9v to both sides of the equation. Remember to only combine like
terms.
Step 2: Completely isolate the variable, w, by dividing by 30 on both sides of the equation.
Step 3: Simplify the solution. Since 6, 9, and 30 can all be divided by 3, divide the numerator and
denominator of the fraction by 3.
Example 4: Solve the following equation for r.
Step 1: Begin to isolate r by adding 7 to both sides of the equation. Remember to only combine like
terms. 10s + 7 cannot be combined because they are not like terms.
Step 2: Completely isolate r by multiplying both sides of the equation by 7.
Answer: r = 7(10s +7)
Example 5: Solve the following equation for a.
5(a - b) = 14b
Step 1: Questions of this type involve using the distributive property. The distributive property states
that for all numbers a, b, and c, a(b + c) = ab + ac. Therefore, begin by multiplying 5 by a and by b.
Step 2: Next, begin to isolate a by adding 5b to both sides of the equation. 14b and 5b are considered
like terms so they can be added together to get 19b.
Step 3: Completely isolate the variable, a, by dividing both sides of the equation by 5.
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Example 6: Solve the following equation for x.
8xy - 9x = 20
Step 1: Questions of this type involve using the distributive property to write x as a factor. Begin by
factoring out the x as a common factor.
Step 2: Isolate the variable, x, by dividing both sides of the equation by (8y - 9).
Literal Equations
This study guide will focus on finding solutions to literal equations. Students will be given an equation, usually
a formula, and be asked to solve the equation for a particular variable. This skill is especially useful when
students need to find missing values in science experiments and technical fields.
Literal equations are comprised of literals or multiple letters, also called variables. These literals can
represent known constants or numbers, or be extra variables, letters that represent a range of values to be
determined. So a literal equation uses a bunch of letters to represent a specific relation. They are
usually referred to as formulas.
It may be beneficial to confirm that the student is comfortable with solving linear equations, equations
containing one or two variables that do not contain exponents, before beginning to solve literal
equations. A quick review is provided below.
Solving Linear Equations
To solve a linear equation, it is necessary to "undo" what was done to the variable in question. Another
way to think about this is to do the order of operations in reverse.
P = Parenthesis, E = Exponents, M = Multiplication, D = Division, A = Addition, S = Subtraction.
**In the normal direction, evaluate multiplication and division from left to right (whichever comes first).
Evaluate addition and subtraction from left to right (whichever comes first).
**To "undo" what was done, reverse the direction.
•Addition "undoes" subtraction.
•Subtraction "undoes" addition.
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•Multiplication "undoes" division.
•Division "undoes" multiplication.
Example 1:
Solve for x.
x+5=7
Step 1: It is necessary to get x by itself on one side of the equation and everything else on the other side.
Undo the "+ 5" by subtracting 5 from both sides of the equation.
Step 2: x is now isolated and the other side of the equal sign is simplified.
Answer: x = 2
Solving Literal Equations
The methods used for solving literal equations are the same as those used in solving linear equations,
except fewer simplifications can be made since the variables cannot be combined. The goal is to solve
the formula for a given variable in terms of the other variables.
Example 2:
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Step 1: Square both sides to eliminate the radical.
Step 3: Take
the square root of both sides to eliminate the squared term on the side of the equation with the x-terms.
Step 5: Divide both
sides by -1 to eliminate the negative.
Example 3: Solve for w.
P = 2w + 2l.
Step 1: To isolate the w, subtract 2l from both sides of the equation to get P - 2l = 2w.
Step 2: Then divide both sides of the equation by 2.
A good way to practice this skill is to look up formulas used in different fields (physics, electronics,
chemistry, etc.) and practice solving each one for a different variable. Some formulas may be more
complicated than others, so try to stick to those that have the four basic operations, addition, subtraction,
multiplication, and division.
Missing Elements - D
Equations are presented with either a number or operational symbol omitted. Students must solve the equation
by determining what is missing.
It may be useful to develop a series of mathematical equations with either a number or operational symbol
missing. Help the student replace the missing elements.
Example 1: 6 ? 4 = 10
The missing element is the plus symbol (+), we know this because 6 + 4 = 10.
Example 2: Find the operational symbol from the multiple choices below.
99 ? 3 = 33
A. +
B. C. x
D. ÷
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Answer: D, since 99 ÷ 3 = 33.
Example 3: Find the missing number.
4 x (5 + 2) = 20 + ?
(1) 5 + 2 = 7
(2) 4 x 7 = 20 + ?
(3) 4 x 7 = 28
(4) 28 = 20 + ?
Step 1: First determine the value within the parentheses.
Step 2: Rewrite the equation with the new value in place of the parentheses.
Step 3: Perform the operations on the opposite side of the value we are finding. 4 x 7 = 28
Step 4: Rewrite the equation with the new value in place of 4 x 7.
Answer: 8
Example 4: Find the missing number.
5 x (? + 3) = (2 x 5) + (3 x 5)
(1) 2 x 5 = 10 and 3 x 5 = 15
(2) 5 x (? + 3) = 10 + 15
(3) 5 x (? + 3) = 25
(4) 5 x (2 + 3) = 25
Step 1: Perform operations within the parentheses.
Step 2: Rewrite the equation with the new values in place of the parentheses.
Step 3: Add 10 and 15 to determine the value on one side of the equation.
Step 4: We can now find the missing value. We know that 5 x 5 = 25, so we know the value within parentheses
must equal 5. So, since 2 + 3 = 5, the missing value is 2.
Answer: 2
Radicals and Roots
Mastering roots and radicals is an essential step toward learning advanced mathematics concepts. A radical
sign looks like a check mark with a line attached to the top. The radical sign is used to communicate square
roots.
The following rules are required to perform operations with roots and radicals.
1. If x multiplied by x equals y, then x is a square root of y. For example, 6 multiplied by 6 is 36, so 6 is
a square root of 36. In fact, 36 is called a perfect square because its square root, 6, is a whole number.
Most algebra text books contain a table of perfect squares.
2. -3 and 3 are both square roots of 9 because -3 x -3 = 9 and 3 x 3 = 9. 3 is referred to as the principal
square root because it is the positive square root of 9.
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3. To find the simplest radical form of a radical expression, factor the number under the radical sign (the
radicand). The square root of 45 could be factored to be the square root of 9 multiplied by the square
root of 5. The square root of 9 multiplied by the square root of 5 can be simplified further by finding the
square root of 9. The result is 3 (the square root of 9) multiplied by the square root of 5.
Example 1: Find the equivalent form.
Solution: Multiply the numbers under the radical symbols.
Example 2: What symbol would best replace the ? in the given statement?
A. =
B. <
C. >
There are two methods that can be used to solve this problem. Each method is shown and explained
below.
Solution Method 1:
Step 1: Simplify the square root of 12 by making it the square root of 6 x 2.
Step 2: Further simplify the square root of 6 x 2 by making it the square root of 3 x 2 x 2. (If you
multiply 3 x 2 x 2, you will get 12.)
Step 3: The square root of 3 x 2 x 2 becomes 2 times the square root of 3, because the square root of 2 x
2 is 2 and the 3 must remain under the square root symbol.
Step 4: Two times the square root of 3 can also be written as the square root of 3 plus the square root of
3.
Step 5: Now, we can make a comparison. We know that the larger a number is, the larger that number's
square root will be. We can determine that the square root of 5 plus the square root of 7 will be greater
than the square root of 3 plus the square root of 3 because 5 and 7 are both larger than 3.
Answer: C
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Solution Method 2:
Step 1: Estimate the square root of 5, the square root of 7, and the square root of 12. This estimation can
be done using a calculator.
Step 2: Add together the 2.24 and the 2.65 to get 4.89.
Step 3: Replace the question mark with the > symbol because 4.89 is greater than 3.46.
Answer: C
Example 3: Solve for the value of x.
Step 1: The square root of 36 is 6 because 6 x 6 = 36.
Step 2: Subtract 6 from each side of the equation to isolate the square root of x.
Step 3: The square root of x is equal to 8. We can replace the x with 64, since the square root of 64 is 8.
Step 4: Since the square root of 64 equals 8, the value of x is 64.
Answer: x = 64
Ratio/Proportion - C
A ratio is a comparison of two numbers expressed as a quotient. Ratios can be written in three ways: a fraction
(3/5), a ratio (3:5), or a phrase (3 to 5). Like fractions, ratios refer to a specific comparison. The ratios 3/5, 3:5,
and 3 to 5 (as in "the ratio of cellos to violins was 3 to 5") all express the same ratio or comparison. A
proportion reflects the equivalency of two ratios. The ratio 3/5 expresses the same proportion as the ratio 15/25.
To understand how ratios operate, students need to understand equivalent fractions. Fractions represent
portions or parts. For every fraction, there is a corresponding portion. The fraction 1/2 communicates
one portion out of two, but this specific portion can also be communicated by the fractions 2/4, 3/6,
8/16, 10/20, etc. All of these fractions are equal to 1/2 because the relationship between the numerator
and denominator in 1/2 is the same relationship between the numerators and denominators in 2/4, 3/6,
8/16, and 10/20. Ratios and proportions operate in a similar manner. The ratio 2:5 communicates a
specific portion. The ratio 4:10 communicates the same portion because 4:10 reduces to 2:5.
Example 1: Sandra has 15 lollipops and 25 jellybeans. What is the ratio of lollipops to jellybeans?
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Answer: There are 15 lollipops to 25 jellybeans, so the ratio of lollipops to jellybeans is 15:25.
Proportions occur when two ratios are equal. In a proportion the cross products of the terms are equal.
Example 2: Is the following proportion True or False?
1/3 = 3/9
The cross products are both equal to 9, so the proportion is TRUE. If the cross products were not equal,
the proportion would be FALSE.
Sometimes you must find the value of a variable in a proportion. To solve the proportion, you must find
the value of the variable that makes both ratios equal.
Example 3: What is the value of a?
9/12 = a/48
Step 1: Find the cross products. Multiply 48 by 9 and 12 by 'a'.
Step 2: 48 x 9 = 432 and 12 x a = 12a. Rewrite the equation with the new products.
Step 3: Divide each side of the equation by 12 to isolate the variable 'a'.
Step 4: Divide 432 by 12 to get a = 36.
Answer: a = 36
Example 4: Apples are 13 for $3.35. What will 32 apples cost? (Round answer to the nearest
hundredth).
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Step 1: Write the appropriate proportion. Let b represent the cost of 32 apples.
Step 2: Write the cross products. Multiply b by 13 and multiply 3.35 by 32.
Step 3: Rewrite the equation with the new values.
Step 4: Divide each side of the equation by 13 to isolate the b.
Step 5: 107.20 ÷ 13 = 8.246153846 We only need to write 3 decimal places because we are going to
round to the hundredths place (the second decimal place). The three dots above represent that the
decimal number continues on past that point.
Step 6: Round 8.246 to the hundredth place. The 6 tells us to round the 4 up to a 5. 8.246 approximately
equals 8.25.
Answer: $8.25
Take a look at the solution. Does it make sense that 32 apples would cost $8.25? We know that 32
apples would cost more than twice as much as 13 apples (13 x 2 = 26), but less that three times as much
(13 x 3 = 39). The solution is more than twice as much as $3.35 (2 x $3.35 = $6.70), but less than three
times as much (3 x $3.35 = 10.05), so our answer makes sense.
Example 5: Rachel scored 25% of the team points. If the team scored 40 points, how many did Rachel
score?
Step 1: Write a proportion to represent the problem.
Step 2: Cross multiply. This involves multiplying the bottom of one ratio by the top of the other and vice
versa. Perform the multiplication.
Step 3: Divide both sides of the equation by 100.
Answer: 10 points
Function/Pattern - C
Students must identify patterns in numbers (such as missing elements) and shapes, and then draw conclusions
based upon the identified pattern.
It may be helpful to develop a series of number patterns and have the student identify the pattern and determine
the correct equation. For example, the pattern
5, 7, 9, 11
This sequence follows a pattern of adding 2 to an odd number.
3+2=5
5+2=7
7+2=9
9 + 2 = 11
To practice critical thinking skills, have the student not only find the patterns, but also analyze the
patterns. For instance, have him or her identify the patterns of 3 lists of numbers. Then, ask the student
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which pattern will need the fewest additional numbers to reach a specific number. For example, consider
the following 3 patterns. Which pattern will need the fewest additional numbers to reach 95?
(a) 71, 74, 78, 81, 85, ...
(b) 2, 5, 11, 23, 47, ...
(c) 5, 10, 15, 20, 25, ...
For (a), the pattern is to add three and then to add four. It will take 3 more numbers to reach 95. For (b),
the pattern is to double the number and then add 1. It will take 1 number to reach 95. For (c), the pattern
is to add five. It will take 14 numbers to reach 95. Therefore, (a) would be the correct answer.
Example: Which of the following equations can be used to find the numbers in this table?
A. B = A + 9
B. B = 2A + 1
C. B = 3A - 6
D. B = A + 11
Solution: To get from 7 to 15, you must either add 8 or multiply 7 by 2 then add one. To get from 8 to
17, you must either add 9 or multiply 8 by 2 then add one. Since the pattern "multiply by 2 then add
one" is common in both of the first two rows, check to see if this pattern works for the rest of the rows in
the table. It does work for the rest of the table, so the answer is B.
Inequalities - A
An inequality is a number sentence that uses "is greater than", "is less than", or "is not equal to" symbols. For
example, 6n > 4 is a number sentence with an inequality symbol.
It may be useful to review the inequality symbols.
Example 1: Solve for y: 8y > 40
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Get the variable being solved for (y) on one side of the inequality and the whole number on the other.
To do this, divide both sides by 8.
The correct answer is that y is greater than 5.
Inequalities can be represented as a value on a number line. The following number line represents the
inequality
Example 2: Which inequality represents the value shown on the number line below?
A. n < 3
B. n > 3
C. n = 3
The answer is A. n < 3 because the dot on the number line is open.
Functions: Quadratic
A quadratic function is a function that contains polynomial expressions for which the highest power of the
unknown variable is two.
Quadratic functions are written in the form:
or
f(x) is read "f of x."
Here are a few examples of quadratic functions:
Since the value of f(x) (or y) depends on the value of x, the dependent variable is f(x) and the
independent variable is x.
Domain and Range:
The domain of a function is the set of values that can be substituted for the independent variable (x) of a
function. When dealing with the coordinate pairs of a function, the domain is the set of all x values of
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the function.
Example 1: Find the domain of the following function.
Since the domain deals with all of the x values of a function, we need to determine all of the x values of
this function. The domain of the function is:
The domain of a function is generally written in numerical order from smallest to largest.
Example 2: Find the domain of
Since any real number that is substituted in place of the independent variable (x) will result in another
real number, the domain of this function is: D = The set of all real numbers.
The range of a function is the set of values that can result from the substitutions for the independent
variable. When dealing with the coordinate pairs of a function, the range is the set of all y values of the
function.
Example 3: Find the range of the following function.
Since the range deals with all of the y-values of a function, we need to determine all of the y values of
this function. The range of the function is
The range of a function is generally written in numerical order from smallest to largest. If a y-value is
repeated in an ordered pair, it is only listed once in the range.
Example 4: Find the range of
Since any real number that is substituted in place of the independent variable (x) will result in another
real number (g(x)), the range of this function is: R = The set of all real numbers.
Evaluating Quadratic Functions:
In mathematics, evaluate means to substitute values in for the variables and calculate a result. So, to
evaluate a quadratic function, we need to substitute a value in for the independent variable (x) and
calculate the result.
Example 5: Evaluate the quadratic function at f(3).
Step 1: The directions told us to evaluate the quadratic function at f(3), so we substitute 3 in place of all
of the x's in the quadratic function.
Step 2: Following the order of operations, we perform operations on exponents first, then we multiply.
Step 3: Once again, we follow the order of operations which states that multiplying comes before adding
and subtracting.
Step 4: The final step in the order of operations states that adding and subtracting are completed by
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working from the left to the right. First we will add 18 and 9. Then we subtract 8 from that sum. We
now know that f(3) = 19.
It is possible to evaluate a function and still have a variable in the answer.
Example 6:
Step 1: Substitute (x + 4) in place of all of the x's in the quadratic function.
Step 2: Applying the order of operations, we must do all operations inside the parentheses. So we add 4
and 3.
Step 3: To take (x + 7) to the second power, we must multiply (x + 7) by itself.
Step 4: In order to multiply (x + 7) by itself, we use the FOIL method. Multiply the "first" terms together
(x times x). Then multiply the "outer" terms (x times 7). Next, multiply the "inner" terms (7 times x).
Finally, multiply the "last" terms (7 times 7).
Step 5: Since 7x and 7x are like terms, we can add them together to get 14x. Also, 49 and 5 are like
terms, so we can add them together to get 54.
Compositions of Quadratic Functions:
The composition of functions is the operation of first applying one function, then applying the other.
Compositions of functions are denoted in two ways. One of those ways is
which can be read,
"the composite of g and f of x." The other way is g(f(x)), which can be read, "g of f of x."
Example 7:
Step 1: Since we are trying to find "g of f of x," we can substitute
in place of f(x) in g(f(x))
to get
Then, we can substitute that term in place of x in the g(x) = 6x function.
Step 2: Use the distributive property to multiply each term in the parentheses by 6.
Example 8: Find f(g(x)) if f(x) = 2x + 6 and g(x) = 7x - 2.
(1) f(7x - 2) = 2(7x - 2) + 6
(2) f(7x - 2) = 14x - 4 + 6
(3) f(g(x)) = 14x + 2
Step 1: Substitute (7x - 2) in place of g(x) to get f(g(x)) = f(7x - 2). Then substitute (7x - 2) in place of
all x's in the f(x) = 2x + 6 function.
Step 2: Use the distributive property to multiply 7x and -2 by 2.
Step 3: Since -4 and 6 are like terms, they can be added together to make 2. So, f(g(x)) = 14x + 2.
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Compositions of functions can be evaluated if we are given the value of the variable.
Example 9:
Step 1: To determine the value of f(g(4)), we must first evaluate g(4). We substitute 4 in place of x
in
Step 2: Following the order of operations we square 4 (4 x 4 =16) and add one to the
answer. g(4) = 17.
Step 3: Now we substitute 17 in place of g(4) to get f(g(4)) = f(17). Then substitute 17 in place of x in
f(x) = 2x - 3.
Step 4: Following the order of operations we multiply 2 and 17 (34), then subtract 3.
Comparing Graphs of Quadratic Functions:
A parabola is the shape of the graph of
The standard form for the equation of a parabola is
(Please remember that f(x)
and y can be substituted for each other in quadratic functions.) When changes are made to an equation
such as adding a constant, changes in the position of the graph of the equation will take place. If the
value of a < 0, the graph of the function will reflect itself across the x-axis. The value of h determines
whether the graph will shift right or left and the value of k determines whether the graph will shift
upward or downward. See the chart below.
Example 10:
(1)
(2) h = -4, k = 7,
and a = -1
(3) By looking at the chart, we can determine that the graph of should shift to the left 4 units
(h = -4), upward 7 units (k = 7), and should reflect itself across the x-axis (a = -1).
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(4) Compare the actual graphs.
Functions: Absolute Value
The absolute value of a number is the positive distance between the number and 0 on a number line. Absolute
value is denoted by one straight vertical line on each side of a number or expression. Here are some examples of
absolute value notation.
(1) |-6| = 6
(2) |2| = 2
(3) |0| = 0
(4) |x - 3|
(5) |2x + 6|
The absolute value of the first three examples can be determined because they are all integers. Remember the
absolute value of a number is the positive distance between the number and zero on a number line.
An absolute value function is a function which contains at least one absolute value expression. An absolute
value function can be evaluated if the value of the variable is known by substituting the value in for the variable
and calculating the result.
Evaluating Absolute Value Functions:
Example 1: Evaluate y = |x - 7| + 3, for x = 9.
(1) y = |9 - 7| + 3
(2) y = |2| + 3
(3) y = 2 + 3
(4) y = 5
Step 1: Substitute 9 in place of x in the absolute value function.
Step 2: Before the absolute value can be figured, 7 must be subtracted from 9. This is a rule: All
operations inside an absolute value symbol must be completed before the absolute value can be taken.
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Step 3: The distance between 0 and 2 is 2 units, so the absolute value of 2 is 2.
Step 4: The final step is to add 2 and 3 to get 5.
Example 2: Evaluate y = 3|x - 2| + |2x + 4|, for x = -3.
(1) y = 3|(-3) - 2| + |2(-3) + 4|
(2) y = 3|-5| + |-2|
(3) y = 3(5) + 2
(4) y = 15 + 2
(5) y = 17
Step 1: Substitute -3 in place of x in the absolute value function.
Step 2: Compute the values inside both absolute value symbols first.
-3 - 2 = -5
and
2(-3) + 4 = -6 + 4 = -2
Step 3: The distance between 0 and -5 is 5 units, so the absolute value of -5 is 5. The distance between 2 and 0 is 2 units, so the absolute value of -2 is 2.
Step 4: Three times five is 15.
Step 5: Add 15 and 2 to get 17.
Writing an Absolute Value Function as a Compound Function:
Every absolute value function can be written as a compound function. A compound function is a
function that is made up of two or more functions. It is very similar to a compound word in language
arts. In order to write an absolute value function as a compound function, it is necessary to remember
that
An example of this rule would be:
When x = 3, |3| = 3; since
When x = -3, |-3| = -(-3) = 3; since -3 < 0
Example 3: Write the absolute value function, f(x) = |2x + 6|, as a compound function.
Step 1: In order to write an absolute value function as a compound function, it is necessary to determine
the value of x that makes the absolute value function equal to 0. To do this, we set the expression inside
the absolute value symbol equal to zero.
Step 2: Solve 2x + 6 = 0. Subtract 6 from each side of the equal sign to isolate the 2x on one side of the
equal sign.
Step 3: Divide by 2 on each side of the equal sign to isolate the x on one side of the equal sign. If -3 is
substituted into f(x) = |2x + 6|, f(x) = 0.
Step 4:
so -x would be -(2x + 6) = -2x - 6. We use the conditions
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x would be 2x + 6,
because -3 is the value that
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makes f(x) = |2x + 6| equal zero.
Finding the Vertex of an Absolute Value Function:
A vertex is a point at which two line segments, lines, or rays meet to form an angle. The graph of every
absolute value function has a vertex. The function f(x) = |x| is graphed below and the vertex is
marked.Remember, the terms f(x) and y can be substituted for each other, so f(x) = |x| can be written as
y = |x|.
The standard form for an absolute value function is y = a|bx + c| + d. It is not necessary to graph an
absolute value function to determine the vertex of the function. If the absolute value function is written
in standard form, we can use the standard form to determine the vertex. The vertex of an absolute value
function is always written as a coordinate point. The x-coordinate of the vertex is where bx + c = 0. The
y-coordinate of the vertex is the value of d.
Example 4: Find the vertex of y = |x - 3| + 2.
(1) x - 3 = 0
(2) x = 3
(3) d = 2
(4) vertex is (3, 2)
Step 1: Set the expression inside the absolute value symbols (bx + c) equal to zero and solve to
determine the x-coordinate of the vertex.
Step 2: Add 3 to each side of the equal sign to isolate the x and determine that x = 3. The x-coordinate of
the vertex is 3.
Step 3: Since the 2 in y = |x - 3| + 2 is in the same place as the d in the standard form for an absolute
value function, d = 2. The y-coordinate of the vertex is 2.
Step 4: Put the two coordinates together to make an ordered pair (x, y). The vertex of y = |x - 3| + 2 is
(3, 2).
Example 5: Find the vertex of y = -2|4x + 8| - 6.
(1) 4x + 8 = 0
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(2) 4x = -8
(3) x = -2
(4) d = -6
(5) vertex is (-2, -6)
Step 1: Set the expression inside the absolute value symbols (bx + c) equal to zero and solve to
determine the x-coordinate of the vertex.
Step 2: Subtract 8 from each side of the equal sign to isolate the 4x on one side of the equal sign.
Step 3: Divide each side of the equation by 4 to get the x by itself on one side of the equal sign. Since
x = -2, the x-coordinate of the vertex is -2.
Step 4: Since the -6 in y = -2|4x + 8| - 6 is in the same place as the d in the standard form for an absolute
value function, d = -6. The y-coordinate of the vertex is -6.
Step 5: Put the two coordinates together to make an ordered pair (x, y). The vertex of y = -2|4x + 8| - 6 is
(-2, -6).
Comparing Graphs of Absolute Value Functions:
The standard form for an absolute value function (y = a|bx + c| + d) is needed to compare graphs of
absolute value functions with the graph of y = |x|. The variable a in the standard form determines
whether the graph opens up or opens down. The variables b and c determine whether the graph shifts to
the right or the left and the variable d determines whether the graph shifts upward or downward. See the
chart below.
Example 6: Compare the graph of y = |x| with the graph of y = -|2x + 8| - 3.
(1) In y = -|2x + 8| - 3, a = -1, b = 2, c = 8, and d = -3. Refer back to the standard form of an absolute
value function to see the placements of a, b, c, and d.
(2) By looking at the chart above, we can see that the graph should open downward (a = -1, which is less
than 0). To determine whether the graph shifts left or right, we need to determine c ÷ b = 8 ÷ 2 = 4.
Since 4 is greater than zero, the graph will shift to the left 4 units. The last piece of information we have
is that d = -3. The chart above states that if d < 0, the graph will shift |d| units downward, so this graph
will shift 3 units downward since |-3| = 3.
(3) Compare the actual graphs of y = |x| and y = -|2x + 8| -3.
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Matching an Absolute Value Function (Equation) With its Graph:
It is not necessary to graph an absolute value function to match it with its graph. All that is needed is the
standard form for an absolute value function: y = a|bx + c| + d.
Example 7: Match the equation y = -|3x - 9| + 2 with its graph below.
(1) The first step in matching the equation of an absolute value function with its graph is to determine
the vertex of the graph. The vertex of this graph would be (3, 2) (refer to Example 4 and Example 5 for
an explanation about determining the vertex of an absolute value function).
(2) Now that the vertex is determined, we can eliminate any graph that does not have (3, 2) as its vertex.
Graphs A and D are eliminated.
(3) The next step in matching the equation to its graph is to look at the value of a in the equation.
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In y = -|3x - 9| + 2, a = -1. We learned earlier that if a < 0, the graph would reflect itself over the x-axis
(in other words, it will open downward).
(4) Since a = -1 in the equation we are discussing, we can eliminate any graph that does not open
downward. Graph B is not the right graph (D would also be out now if it had not already been
disqualified in step 2).
(5) The only graph left is graph C and it is the correct answer. If there were still two (or more) graphs
left, it would be necessary to substitute some values in for x to determine the value of y and see which of
the graphs contained the calculated points.
Irrational Numbers: Pi
An irrational number is a number that cannot be written as a fraction. The decimal equivalents of
irrational numbers do not terminate (end) and never repeat. For example, 0.10110111011110...and
3.14159265... are two decimals that never repeat and never end.
~
Determining the Circumference of a Circle:
Circumference is the measurement of the distance around a circle. The circumference of a circle is found
by using the following formula:
In this formula, r represents the radius of the circle. The radius of a circle is the length of the segment
from the center of a circle to any point on the circle. The radius is one-half the length of the diameter
(the distance across a circle) of a circle. Pi is the ratio of the circumference of any circle to its diameter
and is approximately equal to 3.14159. Pi is represented by the symbol
Example 1: Find the circumference of a circle with a diameter of 18 inches.
Step 1: Since the radius of a circle is one-half the diameter, divide 18 by 2 to get the value of the
radius(9).
Step 2: Substitute 9 in place of r in the formula for the circumference of a circle.
Step 3: Multiply 9 by 2 (all three terms are multiplied and it does not matter which order numbers are
multiplied).
The circumference is 18pi inches.
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Example 2: Find the diameter of a circle whose circumference is 87 feet.
Step 1: Substitute 87 in place of C because 87 is the circumference.
Step 2: To isolate the r on one side of the equation, divide each side of the equation by
Step 3: Since
the radius is half the length of the diameter, multiply the radius by 2 to get the length of the diameter.
Remember, what is done to one side of an equation must be done to the other side in order for the
equation to balance, so multiply the left side of the equation by 2 also. The 2 in the numerator cancels
with the 2 in the denominator on the left side of the equation.
Step 4: The diameter of the circle equals 87/pi feet.
Finding the Area of a Circle:
The area of a figure is the measure, in square units, of the interior region of a two-dimensional figure.
The area of a circle is found by using the following formula:
In this formula, r represents the radius of the circle.
Example 3: Find the area of the circle whose radius is 10.2 inches.
Step 1: Substitute 10.2 in place of r because the radius of the circle is 10.2.
Step 2: Square 10.2 (10.2 x 10.2) and multiply the answer by
The area of the circle whose radius is 10.2 is 104.04pi square inches.
Example 4: Find the area of the circle whose circumference is 35 inches.
Step 1: Before the area of the circle can be determined, the radius of the circle must be calculated. Since
the circumference is known, the formula for the circumference of a circle can be used to determine the
radius of the circle. Substitute 35 in place of C in the circumference formula.
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Step 2: To isolate the r on one side of the equal sign, divide each side of the equation by
Step 3:
Divide 35 by 2 and the radius is
Step 4: Now that the radius of the circle is known, it can be
substituted in place of r in the formula for the area of a circle
Step 5:
Step 6: Square 17.5 (17.5 x 17.5) to get 306.25.
Step 7: The pi in the numerator and one pi in the
denominator divide out and leave one pi in the denominator.
The area of the circle, whose circumference is 35 inches, is 306.25/pi square inches.
Finding the Area of an Ellipse:
An ellipse is a closed two-dimensional plane figure that is oval in shape. Every ellipse has two axes. The
two axes lie on the symmetry lines and intersect at the center O of the ellipse. One of the axes is the
major axis. The major axis contains the foci, has two vertices of the ellipse as its endpoints, and is
always longer.
The other axis of an ellipse is called the minor axis. The minor axis does not contain the foci and has
two co-vertices of the ellipse as its endpoints.
The formula for an ellipse with axes of lengths 2a and 2b follows.
The length of the major axis is 2a and the length of the minor axis is 2b.
Example 5: Find the area of an ellipse whose minor axis is 23 feet and major axis is 35 feet.
Step 1: The lengths of the axes can be used to determine a and b, so the formula for the area of an ellipse
can be used. Since the major axis is 35 feet long, 2a = 35. The minor axis is 23 feet long, so 2b = 23.
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Step 2: Divide each side of each equation by 2 to isolate the variable in each equation.
Step 3: Divide 35 by 2 to determine that a = 17.5. Divide 23 by 2 to determine that b = 11.5.
Step 4: Substitute the values of a and b into the formula for the area of an ellipse.
Step 5: Multiply 17.5 by 11.5 (numbers can be multiplied in any order) and multiply that product by
The area of the ellipse is A = 201.25pi square feet.
Example 6: Find the minor axes of an ellipse whose major axis is 25 inches and whose area is 55 square
inches.
Step 1: Since the area of the ellipse and the length of the major axis are known, the formula for the area
of an ellipse can be used to determine the length of the minor axis. The length of the major axis (25) can
be used to find the value of a, so 2a = 25. Divide each side of the equation by 2 to isolate the variable.
25 ÷ 2 = 12.5, so a = 12.5.
Step 2: Substitute the value of the area (55) and the value of a (12.5) into the formula for the area of an
ellipse. We can use this new equation to solve for b (which is 1/2 the length of the minor axis).
Step 3: To isolate b on one side of the equal sign, divide each side of the equation by
Step 4:
Divide 55 by 12.5 to determine that b =
Step 5: Since the length of the minor axis is 2b and we
know the value of b, we need to multiply each side of the equation by 2 to determine the length of the
minor axis.
Step 6: Multiply b by 2 to get 2b.
The length of the minor axis of the ellipse is 8.8/pi inches.
Finding the Volume of a Cone:
A cone is a 3-dimensional figure with one curved surface, one flat surface (usually circular), one curved
edge, and one vertex. The volume is the number of cubic units it takes to fill a figure. The formula for
the volume of a cone is:
In this formula, r represents the radius of the circle (base) and h represents the height of the cone. (See
the diagram of a cone below.)
Example 7: Find the volume of a cone that has a radius of the base equal to 8 inches and height equal to
12 inches.
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Step 1: Since the radius r and the height h are known, substitute the values into the formula for the
volume of a cone.
Step 2: Following the order of operations, square the 8 first (8 x 8 = 64).
Step 3: Multiply 64 and 12, then multiply that product by 1/3. (64 x 12 = 768; 768 x 1/3 = 256)
The volume of the cone is V = 256pi cubic inches.
Example 8: Find the height of a cone that has a radius of the base equal to 10 inches and volume equal
to 820 cubic inches.
Step 1: Since the volume of the cone and the radius of the base are known, the formula for the volume of
a cone can be used to determine the height of the cone. Substitute the value of the volume and the value
of the radius into the formula for the volume of a cone.
Step 2: Following the order of operations, square the 10 (10 x 10 = 100).
Step 3: To begin to isolate the h on one side of the equal sign, divide each side of the equation by
Divide 820 by 100 to get 8.2.
Step 4: Multiply each side of the equation by the inverse of 1/3, which is 3/1, to finish isolating the h on
one side of the equation. The 3 in the numerator divides out with the 3 in the denominator.
Step 5: Multiply 8.2 by 3 get 24.6.
The height of the cone is 24.6/pi inches.
Finding the Volume of a Sphere:
A sphere is a 3-dimensional figure made up of all points in space that are equally distant from a given
point called the center. The best example of a sphere is a ball.
The formula for the volume of a sphere is:
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In this formula, r represents the radius of the sphere.
Example 9: Find the volume of a sphere whose radius is 11 inches.
Step 1: Substitute the value of the radius (11) into the formula for the volume of a sphere.
Step 2: Following the order of operations, calculate
Step 3: Multiply 1331 by
4/3. To do this, first multiply 1331 by 4, then divide the product by 3 (1331 x 4 = 5324; 5324 ÷ 3 ~
1774.67).
The volume of the sphere is V = 1.774.67pi cubic inches.
Example 10: Find the radius of the sphere whose volume is 1,432 cubic feet.
Step 1: Since the volume of the sphere is known (1432), the formula for the volume of a sphere can be
used to find the radius of the sphere. Substitute the value of the volume of the sphere into the formula
for the volume of a sphere.
Step 2: Multiply both sides of the equation by the reciprocal of 4/3, which is 3/4, to begin isolating the r.
To multiply 1432 by 3/4, multiply 1432 by 3 then divide that product by 4 (1432 x 3 = 4296; 4296 ÷ 4 =
1074).
Step 3: Divide both sides of the equation by pi to isolate the Step 4: Take the cube root of each side of
the equation. This will isolate the r on one side of the equal sign.
Step 5: Once the r has been isolated, the equation can be written beginning with the r. It is acceptable to
leave the radical sign in the answer.
Non-Linear Equations
A non-linear equation is an equation whose graph is not a straight line.
An example of a linear equation (graph is a straight line) is y = x + 3 and an example of a non-linear
equation is
These two equations are graphed below.
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A basic rule to follow to determine whether an equation is linear or non-linear is that non-linear
equations have variables that have powers other than one and linear equations have powers equal only to
one.
Simplifying Square Roots:
Before we can begin solving non-linear equations, we must discuss how to simplify square roots. When
a number is multiplied by itself the product is the square of the number. A square root of a number is a
factor that when multiplied by itself equals the number. For example: 2 x 2 = 4, so 4 is the square of 2
also, since 4 = 2 x 2, 2 is a square root of 4. Another square root of 4 is -2 because -2 x -2 = 4. Numbers
that have a rational number as their square root are called perfect squares. Examples of perfect squares
are: 9 (square root is 3), 16 (square root is 4), 25 (square root is 5) and 9/25 (square root is 3/5).
The notation for a square root is this symbol:
To simplify a square root, we first determine two factors that multiply to make the whole number. One
of these two factors should be a perfect square, preferably the largest perfect square that is a factor of the
number. Then we take the square root of the perfect square factor and place that number in front of the
radical symbol.
Example 1: Simplify.
Step 1: Rewrite the problem as the square root of two factors of 72. Remember, one of the factors
should be the largest perfect square that is a factor of 72. In this case 36 and 2 were used because 36
times 2 equals 72 and 36 is a perfect square.
Step 2: Now the problem can be rewritten as the square root of 36 times the square root of 2.
Step 3: Determine the square root of 36 (which is 6) and multiply it by radical 2. Since 2 does not have a
factor that is a perfect square, radical 2 does not change.
Example 2: Another way to simplify
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Step 1: It is possible to simplify a square root if the largest perfect square factor is not known. Once
again, rewrite the problem as the square root of two factors of 72. Make sure one of these two factors is
a perfect square. In this case 9 and 8 were used because 9 is a perfect square and 9 times 8 equals 72.
Step 2: Since 8 has a factor that is a perfect square (4), the problem must be rewritten as the product of
these three factors of 72. 9 x 4 x 2 = 72.
Step 3: Since 9 and 4 are perfect squares and 2 does not have a perfect square factor, the problem can be
rewritten as the square root of 9 times the square root of 4 times the square root of 2.
Step 4: Determine the square root of 9 (which is 3), multiply it by the square root of 4 (which is 2), and
multiply them both by radical 2.
Step 5: Finally, multiply 3 and 2 to get 6. The 6 is multiplied by radical 2 to obtain the final answer:
Solving Non-Linear Equations:
Non-Linear equations can be solved in much the same way as linear equations. The goal of solving a
non-linear equation is to isolate the variable on one side of the equal sign.
Example 3: Solve the following equation.
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Answer:
Example 4: Solve the following equation.
Using the Pythagorean Theorem to Solve Non-Linear Equations:
The Pythagorean Theorem can be used to determine the length of a missing side of a right triangle. A
right triangle is a triangle that has one right (90º ) angle. A 90º angle is marked in a triangle with a box
in the angle. The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle is
equal to the square of the hypotenuse. The hypotenuse of a right triangle is the side of the triangle
opposite the right angle. The other two sides of the triangle are called the legs.
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The Pythagorean Theorem states:
In the Pythagorean Theorem, 'c' represents the length of the hypotenuse and 'a' and 'b' represent the
lengths of the legs of the right triangle. The Pythagorean Theorem only works for right triangles.
Example 5: Use the Pythagorean Theorem to solve for x.
Step 1: Write the Pythagorean Theorem. Then determine the values of a, b, and c. Remember, c always
represents the length of the hypotenuse. In this triangle, the length of the hypotenuse is 8, so c = 8. It
does not matter whether a or b is assigned the value of x or the value of 6.
Step 2: Substitute the values of a, b, and c into the Pythagorean Theorem.
Step 3: Following the order of operations, square the 8 (8 x 8 = 64) and the 6 (6 x 6 = 36).
Step 4: Subtract 36 from each side of the equation.
Step 5: Take the square root of each side of the equation.
Step 6: Simplify radical 28. In this case, the variable represents the length of the side of a triangle, so the
answer cannot be negative.
Answer:
Solving an Equation by Completing the Square:
Completing the square is a method of solving a quadratic equation in order to express the equation as a
single squared term. The method of completing the square is used when an equation cannot be factored.
Completing the square involves adding the square of one term to the equation and solving the equation
for the value of the variable. The theorem for completing the square states:
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The following is a detailed example of how to complete the square.
Example 6: Solve the equation by completing the square.
Step 1: Add 7 to each side of the equation to put the equation in the form
Step 2: Since 8 is
in the same place as b, b = 8. Add the square of 1/2 of b to each side of the equation. One-half of 8
equals 4, so we are actually adding 4 squared to each side of the equation.
Step 3: We need to fill in the final form in the theorem for completing the square. One-half of b equals
4, so inside the parentheses we have (x + 4). Then we place an exponent of 2 outside the parentheses.
Finally, 4 squared equals 16 (4 x 4).
Step 4: First, add 7 and 16 to get 23. Then take the square root of each side of the equation. Taking the
square root of a term is the opposite of squaring a term, so we get x + 4 on one side of the equal sign.
Remember, every number has a positive and a negative square root, so we write
Step 5: Subtract
4 from each side of the equation to isolate the x on one side of the equal sign. The new expression
cannot be simplified.
Step 6: The solution to the equation is
Characteristics that Describe the Graph of a Non-Linear Equation:
A quadratic equation is any equation in the form
The graph of a quadratic equation is
always a parabola. The vertex of a parabola can be found by putting the quadratic equation in vertex
form.
Once a quadratic equation is in vertex form, the vertex is the coordinate point (h, k). If a > 0, then the
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graph opens up and has a minimum. If a < 0, then the graph opens down and has a maximum.
The roots of a quadratic equation are the solutions of the quadratic equation when y = 0. The roots are
the points where the graph intersects the x-axis; therefore, y = 0. The axis of symmetry is the line that
passes through the vertex and splits the graph directly in half such that each side is the mirror image of
the other. This line is represented by the equation x = h.
Example 7: What are the characteristics of the graph of
Step 1: Rewrite the equation in vertex form. The simplest way to do this is by completing the square.
Step 1A: Subtract 4 from each side of the equation.
Step 1B: Factor -1 out of the right side of the equation to make the term positive.
Step 1C: Add 1/2 of the b term (-6) squared to the right side of the equation and subtract 1/2 of the b
term from the left side of the equation (we subtract on the left side of the equation because we factored 1 out of the equation in Step 1B and we need to multiply any number by -1 before we can add or
subtract it from the left side of the equation). b = -6, so we are actually adding -3 squared to the right
side of the equation and subtracting -3 squared from the left side of the equation.
Step 1D: Complete the theorem for completing the square.
Step 2: Determine whether the graph opens up or down. Now that the equation is in vertex form, we can
determine that a = -1. Since a < 0, the graph opens down.
Step 3: Determine the vertex of the parabola. The vertex is the point (h, k). h = 3 and k = 13, so the
vertex is (3, 13).
Step 4: Determine the axis of symmetry. The axis of symmetry is the line x = h. The axis of symmetry is
x = 3.
Step 5: Determine the roots of the equation. The roots of the equation are the values of x when y = 0.
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Step 5A: Substitute 0 in place of y.
Step 5B: Divide each side of the equation by -1.
Step 5C: Take the square root of each side of the equation.
Step 5D: The roots of the equation are
If the 13 had been negative, the equation would have had no real roots.
(1) opens down
(2) vertex is (3, 13)
(3) axis of symmetry is: x = 3
(4) roots are
Rational Numbers: Equations
Rational numbers are numbers that can be written as fractions. The following numbers are rational
numbers because they can all be written as fractions.
Adding and subtracting rational numbers includes the calculation of whole numbers, fractions, decimals,
and integers. It is common for students learning to solve equations with rational numbers to confuse
operational symbols with integer signs.
Example 1: Before solving, n - - 7 = - 9, the equation should be simplified to n + 7 = - 9. Solve for n in
n + 7 = - 9.
Solution: To solve for n, you need to first isolate n on one side of the equal (=) sign by subtracting 7
from both sides.
Answer: n = - 16
Example 2: Solve for x.
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Step 1: Multiply both sides of the equation by the reciprocal of -2/3, which is -3/2. This will isolate the x
on one side of the equal sign.
Step 2: Make the -6 into a fraction by giving it a denominator of 1 and multiply the two fractions.
Remember, multiply numerator by numerator (6 x -3 = -18) and denominator by denominator (1 x 2 =2).
Step 3: Reduce -18/2 to lowest terms by dividing -18 and 2 by 2.
Answer: -9
Equations: Addition/Subtraction
Equations are number sentences which contain equal signs.
Example: n + 5 = 9
Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified.
Example: y - 6
Variables are letters or symbols that represent numbers that are unknown.
The following are examples of how to solve equations.
Example 1: Solve: n + 6 = 20
Step 1: Write the equation.
Step 2: Subtract 6 from both sides of the equation.
Answer: n = 14
Example 2: Solve for x.
16.6 = 5.5 + x - 8.2
Step 1: Begin to isolate the variable on one side of the equation by adding 8.2 to both sides.
Step 2: Subtract 5.5 from both sides to completely isolate the variable.
Answer: x = 19.3
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Equations: Multiplication/Division
Equations are number sentences which contain equal signs:
Example: n + 5 = 9
Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified:
Example: y - 6
A common mistake among students first learning to multiply and divide equations is the failure to use
inverse operations. For instance, in the equation 3x = 8, the student might multiply 3 to both sides of the
equal sign. The correct procedure is to use the inverse operation and divide 3 from both sides (x = 8/3).
By applying the inverse operation, the student can isolate the x.
Example 1: Solve: 5n = 75
Step 1: In the equation, solve for the value of n.
Step 2: 5n is the same as 5 x n. Divide both sides of the equation by 5.
Step 3: 75 ÷ 5 = 15
Answer: n = 15
Example 2: Solve for t.
Step 1: Multiply both sides of the equation by 16 to eliminate the fraction.
Step 2: Divide both sides if the equation by 4 to isolate the variable t.
Answer: t = 128.
Inequalities - B
An inequality is a number sentence that uses "is greater than", "is less than", or "is not equal to" symbols. For
example, 6n > 4 is a number sentence with an inequality symbol.
It may be useful to review the inequality symbols.
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Example 1: Solve for y.
8y > 40
Get the variable being solved for (y) on one side of the inequality and the whole number on the other.
To do this, divide both sides by 8.
The correct answer is that y is greater than 5.
Inequalities can be represented as a value on a number line. The following number line represents the
inequality
Example 2: Which inequality represents the value shown on the number line below?
A. n < 3
B. n > 3
C. n = 3
The answer is A. n < 3 because the dot on the number line is open.
Example 3: Solve the following inequality.
-3 < -2x - 3 < 23
Step 1: We can solve by separating the inequality into two inequalities as shown below.
-3 < -2x - 3 and -2x - 3 < 23
Step 2: Solve both inequalities.
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The answer is -13 < x < 0.
NOTE: -13 < x < 0 is the same as 0 > x > -13.
Equations: Two-Step
Two-step equations require students to perform two operations before solving the equation.
For example, the equation 3x - 3 = 18 requires adding 3 to both sides of the equation and dividing both sides of
the equation by 3.
Example 1: Find the value of n in the equation 6n - 9 = 9
Step 1: In the equation, solve for the value of n by first getting n alone. Add +9 to both sides of the
equation.
Step 2: Divide both sides of the equation by 6.
Answer: n = 3
Example 2: Solve for t.
(1/8)t + 3 = 12
Step 1: To get the variable alone, subtract 3 from both sides of the equation.
Step 2: Isolate the 't' by multiplying by the reciprocal of 1/8, which is 8/1.
Answer: t = 72
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Solving Equations: Substitution
Letters expressing unknown values in equations are called variables. In these two variable equations, students
are given the value of one variable. To find the value of the second variable, substitution must be used.
If y = 4x, and x = 9, substitute the given value of x, in the equation, y = 4x. The result is y = 4(9). Calculate the
right side of the equation to get y = 36. Now the values of both variables are known: x = 9 and y = 36.
To check the answer, set up the equation y = 4x as if x was an unknown variable: 36 = 4x. Divide each side of
the equation by 4. The result is x = 9, so the values for x and y are both correct.
Example 1: Solve for x.
x = 3y + 9
y = -2
Solution: Substitute -2 in place of 'y' in the equation. Then, solve for x.
x = 3(-2) + 9
x = -6 + 9
x=3
When we are not given enough information, the problem cannot be solved.
Example 2: Solve for t.
t = s -12
r=s+3
Solution: Since we do not know the value of either 's' or 'r', we cannot determine the value of t.
Exponential Notation - D
An exponent is a number that represents the number of times the base is used as a factor. For example, the
number 2 with an exponent of 3 is equal to 2 x 2 x 2. The result of 2 to the 3rd power is 8.
Have the student find the equivalent whole number forms of these exponential numbers:
Scientific Notation:
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Scientific notation is based upon exponential properties and is used to communicate very large or very
small numbers. Scientific notation deals with significant digits. The most significant digit in a number is
the first non-zero digit in the number (reading from left to right). To write a number using scientific
notation, place the decimal point to the right of the most significant digit and count the digits between
the new placement of the decimal point and the old placement of the decimal point. The number of
places that the decimal point moved will be represented by a power of 10.
Example 1: Write 123,000,000 using scientific notation.
Step 1: Determine where the decimal point is in the number to be written using scientific notation.
Step 2: Place the decimal point to the right of the most significant digit and count the number of places
the decimal point was moved.
Step 3: Write all of the significant digits (with the decimal in the new postion) and multiply by 10 to a
power. The power on the ten is the number of places that the decimal point was moved. The power is
positive because the decimal point was moved from the right to the left.
Answer:
If the decimal point had been moved from the left to the right, the power of the ten would have been
negative. To take a number out of scientific notation, move the decimal point the same number of places
as the exponent in the power of ten. If the exponent is positive, move the decimal point to the right and
if the exponent is negative, move the decimal point to the left.
Example 2: Write the expression in standard form.
Step 1: Rewrite the expression showing the decimal point for the number being multiplied by the power
of ten.
Step 2: Since the exponent is -4, move the decimal point to the left four places.
Step 3: Fill in the places between the decimal point and the significant digit with zeroes. The negative
sign that was out in front of the number remains in front of the number and it is negative.
Answer: -0.0007
Evaluating Negative Exponents:
There is a rule for evaluating negative exponents.
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Example 3: Evaluate
Step 1: Use the rule for negative exponents to rewrite the expression. The exponent is -3, so we move
the (-2) to the bottom of the fraction and make the exponent 3.
Step 2: Multiply -2 by itself 3 times to evaluate
Step 3: (-2)(-2)(-2) = -8, so the fraction is now 1
over -8
Step 4: Rewrite the fraction so that the negative sign is even with the fraction bar.
Answer:
Missing Elements - E
Equations will be presented with either a number or operational symbol omitted. Students must solve the
equation by determining the missing part.
It may be useful to review the following examples with the student. Help him or her determine whether the
missing element is an operational symbol, number, or expression. When the nature of the missing element is
determined, make a list of possible answers. When you have the list completed, test each element to determine
the correct solution.
Example 1: Find the missing symbol.
13 ? 14 = -1
Answer: The missing element is an operational symbol. Possible answers, therefore, are symbols for addition,
subtraction, multiplication, or division. The answer is -1, so we can rule out addition and multiplication
immediately. The subtraction symbol fits because 13 - 14 = -1.
Example 2: What is the value of N given that A = 87?
10 x N = A - 7
(1) 10 x N = 87 - 7
(2) 10 x N = 80
(3) 10 ÷ 10 x N = 80 ÷ 10
(4) N = 8
Step 1:
Step 2:
Step 3:
Step 4:
Substitute the value of A into the equation.
Calculate 87 - 7.
Divide each side of the equation by 10 to isolate the missing element on one side of the equation.
The value of N is 8.
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Rates
Rates refer to the price per unit of a particular commodity. Pricing tactics often cause confusion for today's
consumer. Products are sold at 3 for one amount or 5 for another. Calculating rates allows students to figure out
the exact cost of the items.
Example: Apples are 5 for $1.25 and bananas are 6 for $1.80, which fruit is less expensive if you only buy one?
(1) $1.25 ÷ 5 = $0.25 --> cost of one apple
(2) $1.80 ÷ 6 = $0.30 --> cost of one banana
(3) $0.25 < $0.30
Step 1: Determine the cost of one apple by dividing 1.25 by 5. One apple costs $0.25.
Step 2: Determine the cost of one banana by dividing 1.80 by 6. One banana costs $0.30.
Step 3: Compare the cost of the two pieces of fruit. $0.25 is less than $0.30.
Answer: One apple is less expensive than one banana.
Properties - E
Mathematical properties are, in essence, statements about the ways in which certain combinations of numbers
relate to each other. The student should be familiar with the following properties of multiplication and addition:
multiplication by 0, associative, reciprocal, commutative, and order of operations.
Please review the following rules with the student:
1. Multiplication by 0: the product of any integer and 0 equals 0.
-3 x 0 = 0
3x0=0
2. Associative Property of Addition: (a + b) + c = a + (b + c).
(1 + 2) + 3 = 1 + (2 + 3)
3. Associative Property of Multiplication: (a x b) x c = a x (b x c).
(1 x 2) x 3 = 1 x (2 x 3)
4. Reciprocals: two numbers are reciprocals if their product equals 1.
5. Commutative Property of Addition: a + b = b + a
1+2=2+1
6. Commutative Property of Multiplication: a x b = b x a
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1x2=2x1
7. Order of Operations:
A. When calculations for a given expression or equation require both addition and multiplication, the
rule is to multiply first and add second.
(3)(2) + 3 = ?
6+3=?
6+3=9
B. The number outside the parentheses is multiplied with each number within the parentheses:
x(y + z) = xy + xz.
(1) 3(x + y)
(2) 3(x) + 3(y)
(3) 3x + 3y
C. If a given expression contains both parentheses and brackets, calculations should be completed
working from the innermost parentheses or bracket outward.
The following are sample questions using the above properties.
Example 1: Which answer best completes the number sentence?
5 = (5 x 4) + (5 x 6)
A.
B.
C.
D.
x (4 + 6)
+ (4 + 6)
x (20 + 30)
+ (4 x 6)
Answer: A (because of rule 7A)
Example 2: Which answer best completes this number sentence?
0.32 + 0.45 + -62 = ?
A.
B.
C.
D.
-0.15
0.32
0.15
0.45
Answer: C
Example 3: Which one of the following best completes the number sentence?
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(2.3 + 3.1) x 5.6 = ?
A.
B.
C.
D.
3.5 x 5.6
7.2 x 5.6
9.1 x 5.6
5.4 x 5.6
Answer: D (because of rule 7B)
Example 4: What is the value of n in the following statement?
13 x (3.4 x 0) = n
A.
B.
C.
D.
44.2
0
13
3.4
Answer: B (because of rule 1)
Radicals: Simplifying
A radical sign looks like a check mark with a line across the top. The radical sign is used to communicate
square roots. The numbers -3 and 3 are the square roots of 9 because both -3 and 3 squared equal 9. The
definition of a square root is: if x squared = y, then x is a square root of y. Simplifying radicals assesses the
ability to perform operations with radicals.
In order to simplify radicals, it is important to understand the product property of square roots. We
already know the square root of 81 is 9. We can also determine this by the following method:
Simplify the square roots below:
When simplifying with variables, we can do the following:
Also, the same method can be used to simplify radicals taking the 3rd, 4th, 5th,...etc. roots.
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Functions/Relations - B
A relation can be expressed as a set of ordered pairs such as
type of relation.
A function is a special
There are a few ways to determine if a relation is a function. One way is to look at all of the ordered
pairs of a relation. If no two of these ordered pairs have the same x - term (abscissa), then the relation is
a function. Another method involves graphing the ordered pairs of a relation on a coordinate graph. If
no vertical line crosses this graph at more than one point, then the relation is a function.
A linear function is a function whose graph is a line or subset of a line which is not vertical. A special
type of linear function is a constant function, whose graph is a horizontal line or subset of a horizontal
line.
Using these guidelines, find the value for q that will make the following relation not a function:
The answer is 4 because if g = 4, then the second ordered pair would be (16, 7). The two ordered pairs
would have the same x - term, and for that reason the relation would not be a function.
Equations: Systems
A system of equations contains at least two equations that may be linear, non-linear, or a combination of the
two types. A graphical interpretation of the solution of a system of equations is that point (or points) where the
graphs of the equations intersect. One method of finding the solution(s) of a system of equations involves
adding the two equations together.
Example 1:
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Step 1: Write the equations in a vertical format, aligning the x-terms, y-terms, equal signs, and constant
terms.
Step 2: The objective is to add the corresponding parts of the two equations together and eliminate either
the x- or y-term. Multiplying each term in the top equation by 4 will create the necessary conditions for
eliminating the y-term.
Step 3: Add like terms in the two equations, and notice the resulting y-term will have a coefficient of
zero and be eliminated.
Step 4: Solve the resulting equation for x. In this case, that means divide both sides of the equation by
14. This results with x = 5.
Step 5: To solve for y, substitute the value of x (5) in either of the original equations. The second
equation was chosen for this example.
Step 6: Subtract 10 from both sides of the equation.
Step 7: Solve the resulting equation for y. In this case, that means divide both sides of the equation by 4. This results with y = -2.
As with all other equations, substitute the values of x and y into the original equations to ensure they are
correct solutions.
Both values check out. Therefore, the solution to the system of equations are x = 5 and y = -2. This can
be interpreted as the ordered pair (5, -2) of the point of intersection of the graphs of these two equations.
Equations of a Line
Every line on any coordinate graph has a corresponding equation which describes every point on the line.
Every linear equation (equation of the line) contains a slope. The slope of a line is the same between any two
points on the line.
Before you can find the equation of a line, you must first be able to find the slope of a line when given
two coordinate points on the line. These two points are named:
The formula for the
slope of a line follows.
Example 1: Find the slope of the line between Point R (2, 4) and Point S (1, 3).
Step 1: Substitute the given coordinate points into the formula.
Step 2: Simplify the fraction.
Answer: The slope of the line is 1.
The Point-Slope form for the equation of a line:
Example 2: Use the following points to find the equation of the line.
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Point T (7, -3)
Point U (-4, 6)
Step 1: Solve for the slope of the line between Point T and Point U.
Step 2: Use one of the coordinate points and the slope and substitute them into the Point-Slope form for
the equation of a line.
Step 3: Simplify both sides of the equation.
Step 4: Subtract 3 from both sides of the equation.
The equation of the line that passes through (7, -3) and (-4, 6) is y = -9/11x + 30/11.
Number Relation Problems
Number relation problems involve sentences that must be translated into equations.
To translate sentences, review the following phrase/number equivalences:
If x is equal to 4, then x = 4
If the sum of 2x and 2y is 23, then 2x + 2y = 23
If 3 is less than twice y, then 2y > 3
If 5x decreased by y is more than 2y, then 5x - y > 2y
If the result of 7 more than 3 times x is y, then 3x + 7 = y
Example: One number is 2 less than another number. If twice the larger number is decreased by 3 times
the smaller number, the result is 20.
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Step 1: Translate the first sentence in the problem into the first equation.
Let x and y represent the unknown numbers, with y representing the larger number and x representing
the smaller number.
Step 2: Translate the second sentence in the problem into the second equation.
Use x and y as in the first equation, substituting them into their respective locations in the equation.
Step 3: Substitute x + 2 for y in the second equation.
Step 4: Distribute the 2 across (x + 2).
Step 5: Combine the similar terms.
Step 6: Subtract 4 from both sides.
Step 7: Divide each side by -1 to find the value of x.
Step 8: Using x = -16, substitute for x in the second equation.
Step 9: Multiply -3 and -16.
Step 10: Subtract 48 from both sides.
Step 11: Divide each side by 2.
Answer: x = -16 and y = -14
As always, check the solutions by substituting both equations with x = -16 and y = -14.
Factoring
Consider the following equation:
3 x 4 = 12
The numbers 3 and 4 are said to be factors of the number 12. This concept of factoring is not reserved for
numbers, but may be extended to polynomials as well.
Factoring is the breaking up of quantities into products of their component factors. One way to think of
factoring is as the opposite or inverse of multiplying.
A polynomial is a term or sum of terms. Each term is either a number or a product of a number and one or more
variables.
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A monomial is a polynomial with one term.
A binomial is a polynomial with two terms.
A trinomial is a polynomial with three terms.
Consider the following polynomial:
A typical question on factoring will include a polynomial like the one above. Notice that 4y is a
common factor of each term of the polynomial.
Step 1: Factor out the 4y by dividing each term of the trinomial by 4y.
Step 2: The trinomial in parentheses can be factored further. Since the coefficient of the "y squared"
term is equal to 1, focus on the last term, in this case, -5. If factors of -5 can be found that ADD up to the
coefficient of the middle term (in this case, 4) the trinomial can be factored. Two factors of -5 are
5 and -1, and when ADDED together the result is equal to the coefficient of the middle term, 4. Notice
how these numbers are put together to construct the fully factored trinomial:
(4y)(y + 5)(y - 1)
To check the result, use the rules for multiplying polynomials and you should have the original
polynomial when finished.
Sometimes factoring must be done by grouping. The polynomial given below may appear impossible to
factor at first, but if you examine the steps you will see a method to use with polynomials of this type.
Step 1: (5x) is a common term to both 15x and 20xy. Factor it out of only those two terms.
Step 2: (6n) is a common term of both 18nx and 24ny. Factor it out of only those two terms.
Step 3: Notice that the quantity (3x + 4y) is a common factor of 5x and 6n. The expression is rewritten
to indicate this, and the polynomial is completely factored.
A special type of polynomial expresses the difference of two perfect squares. Polynomials of this type
are factored easily once the rule is remembered.
Since each term in the polynomial is a perfect square, the square root of each term (in this case x and 6
respectively) will be used in the following way. The original polynomial is factored as (x + 6)(x - 6).
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Notice that if these terms are multiplied together, the original polynomial is formed. Polynomials that
are in the "difference of squares" form may always be factored as the sum of the square roots times the
difference of the square roots.
Quadratic Formula
A quadratic equation is a polynomial equation in which the highest power of the unknown variable is two.
An example of a quadratic equation is below.
The format of a quadratic equation is
Quadratic equations can be solved by factoring,
graphing, or by using the quadratic formula. The quadratic formula is as follows:
It can be found in any algebra textbook. This formula should be memorized.
To apply the formula to a quadratic equation, use the quadratic equation format given above as a
guideline.
Example 1: Solve the quadratic equation.
Step 1: Determine the values of a, b, and c and substitute them into the quadratic formula. a = 1, b = 6,
and c = -91
Step 2: Determine the value under the radical symbol. 6 squared is 36 and -91 times -4 equals 364.
36 + 364 = 400
Step 3: The square root of 400 is 20 (20 x 20 = 400).
Step 4: Split the remaining problem into two problems: (-6 + 20) ÷ 2 and (-6 - 20) ÷ 2 and solve the two
problems.
The answers are x = 7 and x = -13.
Example 2: Solve the quadratic equation.
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Step 1: Write the equation.
Step 2:
This will put the equation in standard form.
Step 3: Determine the values of a, b, and c.
Step 4: Substitute the values of a, b, and c into the quadratic formula.
Step 5: Determine the value under the radical sign. The square root of 0 is 0.
Step 6: Solve for x.
Answer: x = -2
The discriminant is the portion of the quadratic equation under the radical sign
The
discriminant properties below will give you vital information about quadratic equations.
1. If the discriminant is a perfect square, then the quadratic equation can be factored.
2. If the discriminant is greater than 0, then the equation has two real solutions.
3. If the discriminant is less than 0, then the equation has no real solutions.
4. If the discriminant is equal to 0, then the equation has one real solution.
Example 3: How many solutions does the following quadratic equation have?
Step 1: Determine the values of a, b, and c.
Step 2: Substitute the values for a, b, and c into
Step 3: Simplify the discriminant.
Since the discriminant is greater than zero, there are two real solutions.
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If the discriminant is a perfect square, then the solutions are rational.
If the discriminant is not a perfect square, then the solutions are irrational.
Polynomials: Addition
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a
monomial. The sum of two or more monomials is called a polynomial. Here is an example of a
polynomial:
Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are
monomials that have the same variable or variables for which the variable or variables have the same
exponent.
To add polynomials, combine similar terms.
Example 1:
Step 1: Set the two polynomials up as one long polynomial.
Step 2: Combine the like terms.
Step 3: Add the results of combining the like terms to determine the answer.
The sum is
Example 2:
Step 1: The area of the large rectangle would be the area of the shaded region added to the area of the
small rectangle. Since we know both areas, we simply add them together.
Step 2: Collect like terms so they can be added.
Step 3: Add together any like terms that were collected to determine the final answer.
The area of the large rectangle is
Sometimes it is necessary to use the distributive property before we can combine like terms.
Example 3:
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Step 1: Multiply each term of the first polynomial by 3. Then multiply each term of the second
polynomial by -4.
Step 2: Group like terms together.
Step 3: Combine like terms.
Answer:
Example 4: Solve for a, b, and c.
Step 1: Group the similar terms on the left side of the equation together.
Step 2: Now, group like terms from both sides of the equal sign together.
Step 3: Solve for a, b, and c.
Answer: a = -4, b = -2, and c = 1
Exponential Notation - E
Exponents communicate the number of times a base number is used as a factor. The base number 5 to the 3rd
power (with an exponent of 3) translates to 5 x 5 x 5. (5 to the 3rd power is not 5 x 3.) The result of 5 to the 3rd
power is 125. To perform operations with exponents, all exponential properties must be understood.
To simplify an expression or to find the missing term in a simplified expression, apply the following
exponential properties which are listed below.
Product of Powers:
When multiplying two (or more) numbers with the same base that have exponents, the base remains the
same and the exponents are added.
Power to a Power:
When taking a number with an exponent to another power, the base remains the same and the exponents
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are multiplied.
The Power of Zero:
Any number taken to the power of zero (except zero) equals 1.
Negative Exponents:
There is a rule for evaluating negative exponents.
Example 1: Simplify.
Step 1: Separate the expression into products of whole numbers times products of variables with like
bases. A variable or number that is written without an exponent automatically has an exponent of 1.
Step 2: First, multiply the whole numbers to get 6. Use the "Product of Powers" rule to evaluate the
terms in the second set of parentheses and the terms in the third set of parentheses.
Step 3: Add the exponents to complete the expression.
Answer:
Example 2: Simplify.
Step 1: Separate the expression into products of each of its terms. The exponent (2) is given to each term
inside the parentheses.
Step 2: -2 to the second power is (-2)(-2). Use the "Power to a Power" rule to evaluate the rest of the
expression.
Step 3: Multiply the powers to determine the exponents on the variables. The final expression can be
written with the parentheses still in it, or without the parentheses. It is more acceptable to remove the
parentheses.
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Polynomials: Multiplication
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a
monomial. The sum of two or more monomials is called a polynomial. Here is an example of a
polynomial:
A binomial is a polynomial with exactly two monomial terms. 3x + 4 is a binomial. A trinomial is a
polynomial with exactly three terms. 4xy - 3x +6y is a trinomial.
Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are
monomials that have the same variable or variables for which the variable or variables have the same
exponent.
To multiply monomials and polynomials, the exponential properties must be followed. These properties
apply to all real numbers with positive exponents.
Exponential property #1 can be used to obtain results for problems such as:
Exponential properties #2 and #3 can be used to obtain results such as:
To multiply a monomial and a polynomial, multiply the monomial by each term of the polynomial. In
other words, distribute the monomial.
To multiply two polynomials, each term of the first polynomial must be multiplied by each term of the
second polynomial. Distribute each term of the first polynomial across the second polynomial.
Example 1: Which choice below represents the following trinomial:
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We know that when multiplying two polynomials, each term of the first polynomial must be multiplied
by each term of the second polynomial. Let's look at our solutions.
Answer: B
Example 2: Given these measurements, find the area of the rectangle.
Step 1: Area = Length x Width. Determine the length and width of the rectangle.
Step 2: Multiply the length and width to determine the area of the rectangle.
Step 3: Use FOIL to multiply the two binomials. Remember, multiply the first terms in each binomial,
then multiply the outer terms, next multiply the inner terms, and finally multiply the last terms.
Step 4: The final step in solving this problem is to add the like terms. 3n + 7n = 10n
Answer:
Polynomials: Division
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a
monomial. The sum of two or more monomials is called a polynomial. Here is an example of a
polynomial:
A binomial is a polynomial with exactly two monomial terms. 3x + 4 is a binomial. A trinomial is a
polynomial with exactly three terms. 4xy - 3x +6y is a trinomial.
Before dividing polynomials, recall the following properties associated with exponents:
Example 1: Divide.
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Step 1: Divide the whole numbers: 12 ÷ -3 = -4.
Step 2: Use the properties above to divide the variables. Begin with the x-variables. x-cubed divided by
x equals x-squared.
Step 3: Now divide the y-variables. y divided by y equals y to the power of zero. Any number taken to
the power of zero equals 1.
Step 4: Finally, multiply the quotients back together.
The answer is
Dividing a Polynomial by a Monomial:
To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. Then,
combine the similar terms.
Example 2: Divide.
Step 1: Divide 3m by 3, to get m.
Step 2: Divide -9n by 3, to get -3n.
Step 3: Combine the terms.
Answer: m - 3n
Dividing a Polynomial by a Polynomial:
Dividing one polynomial by another is very similar to long division.
Example 3:
Step 1: Write the problem as a long division problem. The binomial belongs on the outside of the
division symbol because it is the term we are dividing by.
Step 2: Now, we can begin dividing.
So, 2x belongs above the 8x.
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Step 3: The next step is to multiply 2x by (3x + 1).
Subtract that product from
Now, bring the + 8 straight
down beside the 6x.
Step 4: (3x)(2) = 6x, so we place the 2 above the 8 in the answer.
Step 5: Multiply 2 by (3x + 1) to get 6x + 2. Subtract (6x + 2) from (6x + 8). There is a remainder of 6,
so we write the remainder as a fraction with the binomial as the denominator.
Answer:
Sets/Subsets/Solution Sets
A set is a collection of numbers. A subset is a collection of numbers that all belong to a larger set. Every set is
also a subset of itself. If Set C is a subset of Set D, then every element of Set C is also in Set D. Set C is a part
of (a subset of) Set D. A solution set of an open sentence is the set of all elements which make the sentence true
when replaced for the unknown.
Sets are indicated by braces,
Example 1: Find the solution.
2x + 3 = 3(x + 3)
Step 1: Distribute the 3 to the numbers in the parentheses. This involves multiplying each term in the
parentheses by 3.
Step 2: We want the variable on one side by itself. Subtract 2x from each side of the equation to get the
variable on one side of the equation.
Step 3: Subtract 9 from each side of the equation to isolate the variable on one side of the equation.
Step 4: Rewrite -6 = 1x with the variable first. x = -6
Answer: x = -6
Example 2: Choose the corresponding number line: y < -2 or y > 3.
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The answer is B.
We are also able to determine sets and subsets from Venn diagrams. Venn diagrams are drawings that
show relationships among sets of items. Venn diagrams are usually sets of overlapping circles or ovals.
The relationships between the sets can be represented by the following symbols.
Example 3: Given the sets A, B, and C expressed in the diagram, which option is NOT true?
Step 1: Determine the
numbers that are in each set.
Step 2: Use the sets to determine which of the four answer choices
is NOT true.
Answer: Choice D is NOT true.
Absolute Value: Solve
The absolute value of a real number is the distance the real number x is from 0 on a number line. The absolute
value of a real number is denoted by placing the real number within two vertical lines: |x|. In other words, |-5|
("the absolute value of -5") is 5 because -5 is 5 units from 0 on a number line.
There are two major principles of absolute value:
1. The absolute value of a negative number is always a positive number.
2. The absolute value of 0 is 0.
One way to find the absolute value of single digit numbers such as |-5|, is to draw a number line and plot
-5 and 0 on the number line.
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Notice that -5 is 5 points from 0, so the absolute value of |-5| is 5.
The student must also be able to find the absolute value of single digit numbers, solve absolute value
equations, evaluate absolute value expressions, and recognize absolute value expressions on number
lines.
The following example demonstrates how to solve equations with absolute value.
Example 1: Solve: |2x - 3| = 17
Step 1: Set the equation up to equal both 17 and -17. This is because 2x - 3 can equal either 17 or -17 to
make the original equation true.
Step 2: Solve the equation 2x - 3 = 17. Add 3 to both sides of the equation to get 2x = 20. Then divide
each side of the equation by 2 to get x = 10.
Step 3: Solve the equation 2x - 3 = -17. Add 3 to both sides of the equation to get 2x = -14. Then divide
each side of the equation by 2 to get x = -7.
The solutions for |2x - 3| = 17 are x = 10 or x = -7. However, to be sure, always check solutions by
substituting both solutions (10 and -7) for x (in |2x - 3| = 17). For x = 10 (shown in Step 4) and for
x = -7 (shown in Step 5):
Example 2: Evaluate expression |x - 3| when x = -2.
(1) |-2 - 3|
(2) |-5| = 5
Step 1: Substitute -2 in the absolute value expression for x.
Step 2: The expression |-2 - 3| becomes |-5|. The absolute value of |-5| is 5 because -5 is 5 points from 0
on a number line.
It may be useful to review the inequality symbols.
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Inequalities can be represented as a value on a number line. The following number line represents the
inequality
Example 3: Solve and graph the inequality. |3x + 5| < 10
Step 1: Set up the inequality as being less than 10 and greater than -10. (The second inequality should
have the "is greater than" symbol because: when we make the 10 negative, we turn the inequality
symbol the opposite direction. In this case, we change it from "is less than" to "is greater than.")
Step 2: Solve the inequality 3x + 5 < 10. Subtract 5 from both sides of the inequality, then divide both
sides of the inequality by 3.
Step 3: Solve the inequality 3x + 5 > -10. Subtract 5 from both sides of the inequality, then divide both
sides of the inequality by 3.
The answer is: x < 5/3 and x > -5, which can also be written as -5 < x < 5/3.
Now, graph the inequality. Place an open dot on the -5 and an open dot on 5/3 because the inequality is
strictly less than 5/3 and strictly greater than -5 (x cannot equal 5/3 or -5). Shade the portion of the graph
that is greater than -5 and less than 5/3.
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