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ESCC Mathematics Tutorials
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Mathematics Tutorials
These pages are intended to aide in the preparation for the Mathematics Placement test.
They are not intended to be a substitute for any mathematics course.
Arithmetic Tutorials
Algebra I Tutorials
Algebra II Tutorials
Word Problems
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Arithmetic Tutorials
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Arithmetic Tutorials
Whole Numbers
Sets of Numbers
Properties of Real Numbers
Addition of Whole Numbers
Subtraction of Whole Numbers
Multiplication of Whole Numbers
Division of Whole Numbers
Order of Operations
Exponential Notation
Prime Numbers and Factoring
Roman Numerals
Fractions
Least Common Multiple
Greatest Common Factors
Reducing Fractions
Addition of Fractions and Mixed Numbers
Subtraction of Fractions and Mixed Numbers
Multiplication of Fractions and Mixed Numbers
Division of Fractions and Mixed Numbers
Converting between Mixed Numbers and Improper Fractions
Decimals
Introduction to Decimals
Addition of Decimals
Subtraction of Decimals
Multiplication of Decimals
Division of Decimals
Converting Fractions to Decimals
Scientific Notation
Ratio and Proportions
Introduction to Ratios
Introduction to Rates
Introduction to Proportions
Percents
Introduction to Percents
Solving Percent Equations
Signed Numbers
Negative Numbers
Addition and Subtraction of Signed Numbers
Multiplication and Division of Signed Numbers
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Algebra I Tutorials
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Algebra I Tutorials
Linear Equations and Inequalities
What is a Variable?
Evaluating Algebraic Expressions
Combining Like Terms in Algebraic Expressions
Properties of Equalities
Algebraic Equations
Solving First Degree(Linear) Equations
Literal Equations and Formulas
Linear Inequalities
Exponents
Multiplication with Exponents
Division with Exponents
Polynomials
What is a Polynomial?
Evaluating a Polynomial
Addition of Polynomials
Subtraction of Polynomials
Multiplication of Polynomials
Special Products
Division of Polynomials
Factoring Polynomials
Common Factors
Factoring using Common Factors Factoring by Grouping
Solving Equations by Factoring
Factoring the difference of two squares
Factoring the sum or difference of cubes
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Algebra II Tutorials
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Algebra II Tutorials
Rational Expressions
Introduction to Rational Expressions
Equivalent Rational Expressions
Simplifying Rational Expressions
Addition and Subtraction of Rational Expressions
Multiplication and Division of Rational Expressions
Complex Fractions
Equations involving Fractions
Linear Equations
Graphing Linear Equations
Cartesian Coordinate System
Graphs of Linear Equations
Intercepts Slope of the Line
Systems of Linear Equations
Introduction to Systems of Linear Equations
Solving by Substitution
Solving by Elimination
Quadratic Equations
Introduction to Quadratic Equations
Solving Quadratic Equations
Factoring
Extracting the Root
Completing the Square
Quadratic Formula
Complex Numbers
Introduction to Complex Numbers
Addition and Subtraction of Complex Numbers
Multiplication of Complex Numbers
Division of Complex Numbers
Complex Solutions to Quadratic Equations
Roots and Radicals
Common Roots and Radicals
Properties of Radicals
Addition and Subtraction of Radicals
Multiplication of Radicals
Division of Radicals
Equations involving Radicals
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Word Problems
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Word Problems
Translating Word Problems to Algebra
Number Problems
Age Problems
Coin Problems
Work Problems
Mixture Problems
Distance Problems
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Mathematical Numbers
Natural Numbers
Natural numbers, also known as counting numbers, are the numbers beginning with 1,
with each successive number greater than its predecessor by 1. If the set of natural
numbers is denoted by N, then
N = { 1, 2, 3, ......}
Whole Numbers
Whole numbers are the numbers beginning with 0, with each successive number greater
than its predecessor by 1. It combines the set of natural numbers and the number 0. If the
set of whole numbers is denoted by W, then
W= { 0, 1, 2, 3, .......}
Rational and Irrational Numbers
Rational numbers are the numbers that can be represented as the quotient of two integers
p and q, where q is not equal to zero. If the set of rational numbers is denoted by Q , then
Q = { all x, where x = p / q , p and q are integers, q is not zero}
Rational numbers can be represented as:
(1) Integers: (4 / 2) = 2, (12 / 4) = 3
(2) Fractions: 3 / 4, 13 / 3
(3) Terminating Decimals: (3 / 4) = 0.75, (6 / 5) = 1.2
(4) Repeating Decimals: (13 / 3) = 4.333... (4 / 11) = .363636...
Conversely, irrational numbers are the numbers that cannot be represented as the quotient
of two integers, i.e., irrational numbers cannot be rational numbers and vice-versa. If the
set of irrational numbers is denoted by H, then
H = { all x, where there exists no integers p and q such that x = p / q, q is
not zero }
Typical examples of irrational numbers are the numbers π and e, as well as the principal
roots of rational numbers. They can be expressed as non-repeating decimals, i.e., the
numbers after the decimal point do not repeat their pattern.
Real Numbers
Real numbers are the numbers that are either rational or irrational, i.e., the set of real
numbers is the union of the sets Q and H. If the set of real numbers is denoted by R , then
R=Q∪H
Since Q and H are mutually exclusive sets, any member of R is also a member of only
one of the sets Q and H. Therefore; a real number is either rational or irrational (but not
both). If a real number is rational, it can be expressed as an integer, as the quotient of two
integers, and a terminating or repeating decimal can represent it; otherwise, it is irrational
and cannot be represented in the above formats.
Complex Numbers
Complex numbers are the numbers with the format a + b i, where a and b are real
numbers and
i² = - 1. If we denote the set of complex numbers by C, then
C = { a + b i , where a and b are real numbers, i² = -1 }
If in the number x = a + b i, b is set to zero, then x = a, where a is a real number. Thus, all
real numbers are complex numbers, i.e., the set of complex numbers includes the set of
real numbers.
Real Number System
The real number system is comprised of the set of real numbers and the arithmetic
operations of addition and multiplication (subtraction, division and other operations are
derived from these two). The rules and relationships that govern the real number system
are the basis for most algebraic manipulations.
Properties of Real Numbers
All real numbers have the following properties:
(1) Reflexive Property
For any real number a, a = a.
Example: 3 = 3, y = y, x + z = x + z (x, y and z are real numbers)
(2) Symmetric Property
For any real numbers a and b, if a = b, then b = a.
Example: If 3 = 1 + 2, then 1 + 2 = 3
(3) Transitive Property
For any real numbers a, b and c, if a = b and b = c, then a = c.
Example: If 2 + 3 = 5 and 5 = 1 + 4, then 2 + 3 = 1 + 4.
(4) Substitution Property
For any real numbers a and b, if a = b, then a may be replaced by b, and b may be
replaced by a, in any mathematical statement without changing the meaning of the
statement.
Example: If a = 3 and a + b = 5, then 3 + b = 5.
(5) Trichotomy Property
For any real numbers a and b, one and only one of the following conditions holds:
(1) a is greater than b ( a > b)
(2) a is equal to b ( a = b)
(3) a is less than b ( a < b)
Example: 3 < 4 , 4 + 2 = 6 , 7 > 5
Absolute Values
The absolute value of a real number is the distance between its corresponding point on
the number line and the number 0. The absolute value of the real number a is denoted by
|a|.
From the diagram, it is clear that the absolute value of nonnegative numbers is the
number itself, while the absolute value of negative integers is the negative of the number.
Thus, the absolute value of a real number can be defined as follows:
For all real numbers a,
(1) If a >= 0, then |a| = a.
(2) If a < 0, then |a| = -a.
Examples:
|2|=2
| -4.5 | = 4.5
|0|=0
Addition of Whole Numbers
Addition is the process of finding the total of two or more numbers.
We first learn addition through counting
*** + **** = *******
3
4
7
We can also look at addition on our old friend the number line.
each of these line segments is 3 units long (length does not depend upon position
7(total)
3
+
4
Therefore, we also have 3 + 4 = 7
There a few special properties of addition that we need to be aware of.
Addition Property of Zero
Zero added to any number does not change the number
4+0=4
0+1=1
Commutative Property of Addition
Two numbers can be added in any order, the sum is the same
4 + 8 = 12
8 + 4 = 12
Associative Property of Addition
Grouping of addition in any order does not change the sum
(4 + 2) + 3 = 6 + 3 = 9
4 + (2 + 3) = 4 + 5 = 9
The number line and other assorted aides are fine to learn with but the basic addition of 2
one-digit numbers must be memorized. The addition of larger numbers is basically the
repeated usage of the basic addition facts.
Addition Table
+
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
3
4
5
6
7
8
9
10
2
2
3
4
5
6
7
8
9
10
11
3
3
4
5
6
7
8
9
10
11
12
4
4
5
6
7
8
9
10
11
12
13
5
5
6
7
8
9
10
11
12
13
14
6
6
7
8
9
10
11
12
13
14
15
7
7
8
9
10
11
12
13
14
15
16
8
8
9
10
11
12
13
14
15
16
17
9
9
10
11
12
13
14
15
16
17
18
To use this table to add two numbers, find the first number to be added in the top row
(put your finger there). Find the second number to be added in the first column (put
another finger there). Now, bring your fingers together (first finger straight down and
second finger straight across). The place where your fingers meet is the sum.
Addition of larger numbers
The easiest way to add larger numbers is to arrange the numbers vertically, keeping
the digits of the same place value in the same column.
Add 321 + 6472
321
+6472
6793
Addition with carry
If one of the columns in an addition sums to a value greater than 9 then you must
perform a carry. Write down the one’s digit of the sum and carry the ten’s digit into the
next column to the left.
Add 645 + 476
645
+476
5 + 6 = 11
1
645
+476
1
1
645
+476
1121
1 + 4 + 7 = 12
1 + 6 + 4 = 11
When do we use Addition?
There are several types of problems that require the use of addition. One of the
major clues to the use of addition is the major key words leading to addition.
Addition Key Words
Added to
More than
The sum of
Increased by
The total of
Plus
3 added to 5
7 more than 5
the sum of 3 and 9
4 increased by 6
the total of 3 and 8
5 plus 10
3 + 5
5 + 7
3 + 9
4 + 6
3 + 8
5 + 10
Subtraction of Whole Numbers
Subtraction is the process of finding the difference between two numbers.
We learn subtraction (as with addition) by counting.
********
8
Minuend
-
********
3
Subtrahend
=
*****
5
Difference
We can also show subtraction on the familiar number line.
8
3
5
You can readily see that addition and subtraction are related
Subtrahend +Difference = Minuend
3
+
5
= 8
You can use this fact to check you subtraction with addition.
Subtraction of Larger Numbers
To perform subtraction on larger numbers by arranging the numbers vertically ( as in
addition). Then subtract the numbers in each column.
Subtract 8955 – 2432
8955
-2432
6523
Subtraction with borrowing
If during the course of perform a subtraction on a large number, you are attempting to
subtract a large number from a smaller number you must use borrowing.
Subtract 692 – 378
692
-378 you can not subtract 8 from 2 so we need to borrow 1 ten from the 9 tens in the
tens column leaving 8 tens and 12 ones.
81
692
-378 12 – 8 = 4
4
81
692
-378 8 – 7 = 1
14
81
692
-378 6 – 3 = 3
314
When do we use Subtraction? There are many key words that lead us to perform
subtraction.
Subtraction Key Words
Minus
Less
Less than
The difference between
Decreased by
8 minus 5
9 less 3
2 less than 7
the difference between 8 and 2
5 decreased by 1
8–5
9–3
7–2
8–2
5–1
Multiplying Whole Numbers
Multiplication is basically repeated additions.
3 × 2 = 2 + 2 + 2 = 6
6 × 8 = 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 48
The numbers that are multiplied are called factors (6 and 8) and the result is the product
(48).
There are three basic ways to represent multiplication
a × b
a ⋅ b all mean the same thing (a multiplied by b)
a(b)
As is addition the best way to learn multiplication is to memorize the basic facts.
Multiplication Table
0
1
2
3
4
5
6
7
8
9
0
0
0
0
0
0
0
0
0
0
0
1
2
3
4
5
6
7
8
9
0
2
4
6
8
10
12
14
16
18
0
3
6
9
12
15
18
21
24
27
0
4
8
12
16
20
24
28
32
36
0
5
10
15
20
25
30
35
40
45
0
6
12
18
24
30
36
42
48
54
0
7
14
21
28
35
42
49
56
63
0
8
16
24
32
40
48
56
64
72
0
9
18
27
36
45
54
63
72
81
To use the table, place one finger on the top row on the first factor and place
another finger on the second factor on the first column. Bring the finger together
and you have the product.
×
0
1
2
3
4
5
6
7
8
9
Properties of Multiplication
There are several useful properties of multiplication that will help us in our computations
Multiplication Property of Zero
The product of any number and 0 is 0
0 × 4 = 0
7 × 0 = 0
Multiplication Property of One
The product of any number and one is the number
1 × 5 = 5
6 × 1 = 6
Commutative Property of Multiplication
Two numbers can be multiplied in either order and the product is unchanged.
4 × 3 = 3 × 4 = 12
Associative Property of Multiplication
Grouping the numbers in a multiplication problem in any order gives the same
result
(4 × 2) × 3 = 8 × 3 = 24
4 × (2 × 3) = 4 × 6 = 24
Multiplying Larger Numbers
Multiplying large numbers involves the repeated usage of basic one-digit
multiplication facts.
Multiply 37 × 4
37
× 4
above the ten’s column
2
37
4
×
8
×
4 × 7=28 write the 8 in the one’s column and carry the 8
3 × 4=12, add the carry digit – 12 + 2 = 14
37
4
148
Multiply 47 × 23
47
× 23
47 × 3 = 141
47
× 23
141
20 × 47 = 940
47
× 23
141
940
1081
141 + 940 = 1081
Multiply 439 × 206
439
× 203
3 × 439 = 2634
439
× 203
2634
0 × 439 = 000
439
× 203
2634
0000
439
× 203
2634
00000
87800
90434
200 x 439 = 87800
2634 + 0000 + 87800 = 90434
When do we use multiplication?
There are key words that indicate the use of multiplication
Multiplication Key Word
Times
7 times 3
7×3
The product of The product of 6 and 9 6 × 9
Multiplied by
8 multiplied by 2
8×2
Division of Whole Numbers
Division is used to separate objects into groups of equal size.
Division is the inverse of multiplication (3 × 4 = 12 and 12 ÷ 3 = 4 and 12 ÷ 4 = 3)
We right division in two different ways 12 ÷ 4 is the same as 4 12
6
Look at the division 4 24 , we refer to 4 as the divisor, 24 as the dividend and 6 as the
quotient. In general, we have
quotient
divisor dividend
6
Also, we can see the relationship between division and multiplication 4 24
because 4 × 6 = 24
Important Division Rules
Any number, except zero, divided by itself equals 1
1
1
88
2 2
Any number divided by 1 is the number itself
8
27
18
1 27
Zero divided by any number is zero
0
0
7 0 101 0
Division by zero is not allowed
?
0 8 , there is no number whose product with 8 is 0
Dividing single digits into larger numbers
Divide 3192 ÷ 4
4 divides into 31 - 7 times since 4 * 7 = 28
4 3192
7
4 3192
− 28
39
79
4 3192
− 28
39
− 36
3
subtract 28 from 31, bring down the 9
7
4 3192
− 28
39
− 36
3
7
4 3192
− 28
39
− 36
32
− 32
0
4 divides into 39 - 9 times since 4*9=36,
subtract 39-36
bring down the 2,
4 divides into 32-8 times since 4*8=32,
subtract 32 – 32 = 0
Dividing by single digit with a remainder
Divide 3 14
4
3 14
3*4 = 12
4
3 14
− 12
2
4 r2
3 14
− 12
2
subtract 14 – 12 = 2
so the result is 4 with a remainder of 2 (3*4
+ 2=14)
Divide larger numbers
Divide
34 1598
34 1598
4
34 1598
4
34 1598
− 136
238
47
34 1598
− 136
238
− 238
Think about 3 * 5 = 15 but 5 * 34 = 170,
which is larger than 159, so use 4
4*34 = 136
subtract 159 – 136 = 23
bring down the 8
7*34 = 238
subtract 238-238= 0
So the solution is 37
since 47 * 34 = 1598
When do we use Division?
There are a couple of key words that indicate the use of division.
The quotient of
Divided by
Division Key Words
The quotient of 9 and 3
6 divided by 2
9÷3
6÷2
Order of Operations
Many times, in math classes, the problems involve more than one operation in the
same problem. We need a system to determine the order in which we perform our
operations. There is a hierarchy of operations that keep us from being confused by the
messy problems.
The Order of Operations can be remembered by learning the phrase Please Excuse
My Dear Aunt Sally.
Parenthesis – Exponents – Multiplication & Division – Addition & Subtraction (4
levels)
Ex.
4+5*6
from left to right we see addition and multiplication
(multiplication first priority)
4+30
now we can add
34
Ex.
55 – 2*10 + 43
(multiplication first)
55 – 20 + 43
priority)
35 + 43
78
subtraction – multiplication – addition
now left to right (subtraction and division equal
Ex.
(3 + 4)2
72
49
parenthesis and exponent (p first)
now exponent
Exponential Notation
Repeated multiplication of the same number can be written in two different ways
3*3*3*3 or 34
exponent
The exponent shows how many time 3 is multiplied by itself. 34 is in a format called
exponential notation.
Examples of exponential notation
6 = 61
6*6 = 62
6*6*6 = 63
etc.
six to the first power (usually don’t write the 1)
six squared or six to the second power
six cubed or six to the third power
3*3*3*3*5*5*5 = 34*53
Place values are actually powers of 10
Ten
Hundred
Thousand
Ten-Thousand
Hundred-Thousand
Million
101
102
103
104
105
106
Factoring Numbers
We can divide whole numbers into two categories (prime and composite). Prime numbers
are numbers that are only divisible by 1 and itself such as 3, 5, 11, 13. Composite
numbers are numbers that are products of prime numbers such as 6, 15, 20.
One of the major things that we need to do with whole numbers is to factor the composite
numbers into their prime parts, called factoring.
Ex.
10 = 2*5
20 = 2*2*5
Ex.
Factor 105
Start with the small primes and check for divisibility
2 does not work since 2 does not divide 105 evenly
but 3 works
105 = 3*35 now factor 35 as 5*7 so we get
105 = 3*5*7
Ex.
Factor 129
2 won’t work but 3 does
129 = 3*43, 43 is prime so
129 = 3 *43
Ex.
Factor 400
400 = 2*200 FACTOR 200
400 = 2*2*100 FACTOR 100
400 = 2*2*2*50 FACTOR 50
400 = 2*2*2*5*5 DONE
Roman Numerals
Prior to the development of our number system, there have been many other
civilizations who have had their own unique way of handling mathematics and
arithmetic. The one system that has survived to this day and is still in wide use is the
Roman Numeral System.
The Roman Numeral System
A few major things to realize about the Roman Numeral System
There is no zero
It uses what we think of as letters (I, V, X, L, C, M)
Placing a lower value to the left of a higher value subtracts
Placing a lower value to the right of a higher value adds
I=1
V=5
X = 10
L = 50
C = 100
D = 500
M = 1000
Notice the significance of 5 in this system (like 10 in our system)
See a pattern
1=I
2 = II
3 = III
4 = IV
5 – 1 you add through three higher then for 4 you subtract
6 = VI
7 = VII
8 = VIII
9 = IX
10 – 1
LX = 60
XL = 40
50 + 10
50 – 10
Starting on the left you build the value
MCMLXVI
M = 1000
CM = 900
1000-100
LX = 60
50 + 10
VI = 6
5+1
1000 + 900 + 60 + 6 = 1966 you see these type of things on movie dates.
Let’s go from our system to Roman
2121
2000 = MM
100 = C
20 =XX
1=I
MMCXXI
Least Common Multiple and Greatest Common Factor
When working with a group of two or more numbers, we sometimes have to find two
specials numbers, the least common multiple and the greatest common factor.
Least Common Multiple
The Least Common Multiple is the smallest number that is a multiple of each number in
the group. The least common multiple of 2 and 3 is 6 since it is the smallest number both
2 and 3 divide evenly
Finding the least common multiple.
Factor each number
Write down all the factors of the first number
Add in the factors from the other numbers that are not in the LCM already
Multiply all the numbers together
Ex.
Find LCM for 30 and 45
30 = 2*3*5
45 = 3*3*5
LCM – start with 30 and write down 2*3*5
Look at 45, it has 2-3’s and a 5, the LCM has a 5 but only one 3 so we put in the
other 3 to get
LCM = 2*3*5*3 = 90
Ex.
Find LCM for 6. 8.15
6 = 2*3
8=2*2*2
15=3*5
LCM – start with 6 and get 2*3
Go to 8 and put in 2 3’s and get 2*3*2*2
Go to 15 and put in the 5 and get 2*3*2*2*5
LCM = 2*3*2*2*5 = 120
Greatest Common Factor
The greatest common factor is the largest number that is a factor of each number in the
set of numbers. It is used in the reduction of fractions.
Finding the Greatest Common Factor
Factor each number
Look at each factor in the first number and if it occurs in the other numbers (if it
occurs in all numbers then include it in the GCF)
Ex.
Find the GCF of 8 and 12
8 = 2*2*2
12 = 2*2*3
GCF – look at first 2 in 8 it is also in 12 so get 2 so far
8 = 2*2*2
12 = 2*2*3
now go to the next 2 of 8 it is also in 12 so we get 2*2 so far
8 = 2*2*2
12 = 2*2*3
now go to the next 2 of 8 it is not in 12 so we have 2*2 = 4 as the GCF
Ex.
Find the GCF of 60 and 200
60 = 2*2*3*5
200 = 2*2*2*5*5
look at things in common
60 = 2*2*3*5
200 = 2*2*2*5*5
so the GCF is 2*2*5 = 20
Reducing Fractions
Whenever we are dealing with numbers in the terms of fractions, we like to have
them reduced to lowest terms.
The lowest terms of a fraction is the terms when the numerator and the
denominator have no factors in common (relatively prime).
Ex.
3
is in lowest terms since 3 and 4 are relatively prime
4
8
is not in lowest terms since they have 4 as a common factor
20
To reduce fractions to lowest terms
Factor numerator and denominator
Cancel out factors in common
Ex.
Reduce
8
to lowest terms
20
2⋅2⋅2
becomes
2⋅2⋅5
or
2
5
Ex.
Reduce
44
to lowest terms
100
2 ⋅ 2 ⋅ 11
becomes
2⋅2⋅5⋅5
or
11
25
Adding and Subtracting Fractions
There will be occasions where will be necessary to add or subtract numbers that are
fractions. (I know we don’t like fractions, but they are necessary).To add fractions they must
have the same denominator (bottom). If they do not have the same denominator then we must
convert each of them to a fraction with a common denominator.
Fractions with same denominators
Ex.
1 2 1+ 2 3
+ =
= , just add the numerators (tops)
4 4
4
4
Ex.
4 2 4 − 2 2
− =
= , just subtract numerators
15 15
15
15
Fractions with different denominators.
Ex.
1 1
+ , different denominators. We must find a common multiple for the denominators to
2 3
use as a common denominator. The least common multiple of 2 and 3 is 6, so we use 6 as
the common denominator. We convert each fraction to a new one with the denominator
of 6.
1 1 3 3
3
= ⋅ = multiply by 1 in the form of
2 2 3 6
3
1 1 2 2
2
= ⋅ = multiply by in the form of
3 3 2 6
2
1 1 3 2 5
+ = + =
2 3 6 6 6
Ex.
5 1
− different denominators. The Least Common Multiple of 8 and 3 is 24
8 3
5 5 3 15
= ⋅ =
8 8 3 24
1 1 8 8
= ⋅ =
3 3 8 24
therefore
5 1 15 8
7
− =
−
=
8 3 24 24 24
Multiplying and Dividing Fractions
Multiplication of fractions is very simple, just multiply numerators and denominators
Ex.
1 3 1⋅ 3 3
⋅ =
=
2 4 2⋅4 8
Ex.
2 3 6 1
⋅ =
=
3 4 12 2
Division of fractions is not much harder but has one thing important to remember. You
must invert the divisor and then multiply (Flip the last guy and multiply)
Ex.
2 3 2 4 2⋅4 8
÷ = × =
=
3 4 3 3 3⋅3 9
Ex.
15 5 15 4 60 3
÷ = × =
=
8 4 8 5 40 2
Mixed Numbers and Improper Fractions
There are two ways of expressing fractions representing numbers greater than one, mixed
number and improper fractions.
Mixed numbers are expressed as a whole number part and a fractional part in the form
B
1
A like 3
C
2
Improper Fractions are fractions whose numerator is larger than the denominator
A
where A>B
B
Mixed Numbers to Improper Fractions
To convert a mixed number to an improper fraction
Multiply denominator by whole number part
Add numerator
Place over the denominator
Ex.
Convert 3
1
to an improper fraction
2
Multiply denominator by whole number part – 3*2 = 6
Add numerator – 6 + 1 = 7
Place over the denominator ­
7
2
Ex.
Convert 5
3
to an improper fraction
4
5*4 = 20
20 + 3 = 23
so we get
23
4
Improper Fractions to Mixed Numbers
To convert improper fractions to mixed numbers, we have to remember the long
division that we learned in elementary school – division with remainders.
To convert from Improper Fractions to Mixed Numbers
Perform implied division with remainder
Write quotient as whole number part
Place remainder over divisor as fractional part
Ex.
Write
19
as a mixed number
9
2
1
9 19R1 so we get 2
9
Ex.
Write
23
as a mixed number
4
5
3
4 23R3 so we get 5
4
Introduction to Decimals
Numbers that cannot be represented as whole numbers are written as either
fractions or in decimal notation. We are familiar with the concept of decimal notation
from numerous examples in our lives, namely the use of money ($3.12 is decimal
notation for 3 dollars and 12 cents)
We can think of decimal notation as another way of writing certain special types
of fractions (those with multiples of ten in the denominator)
3
10
3
100
239
1000
Three tenths
Three hundredths
Two hundred thirty-nine
thousandths
0.3
Note: 1 zero in the denominator and 1
decimal place
0.03 Note: 2 zeroes in the denominator and 2
decimal places
0.239 Note: 3 zeroes in the denominator and 3
decimal places
We should be able to note that there are exactly three parts to a decimal number.
3.12
whole number part
decimal point
decimal part
Writing decimals numbers in words
0.03 is read as 3 hundredths since the 3 is in the second decimal place (1/100)
0.6481 is read as six thousand four hundred eighty-one ten-thousandths since the 1 is in
the fourth decimal place (1/10000)
Writing decimal numbers in standard form
Five and thirty-eight hundredths – hundredths implies a total of 2 decimal places to be
filled by the 38 so we get 5.38
Nineteen and four thousandths – thousandths implies a total of 3 decimal places to be
filled by 4 so we add two leading zeroes to make 004 and get 19.004
Rounding decimals
Sometimes we are called upon to limit the number of decimal places that can be used in a
specific application (it makes no sense to take money out to 3 places). This process is
known as Rounding.
Rounding rules
If the number to the right of the given place value is less than 5, drop that
number and all numbers to the right of it.
If the number to the right of the given place value is 5 or greater, increase
the number in the given place value by one and drop all numbers to the
right of it
Round 26.3799 to the nearest hundredth
Look at 26.3799, 7 is in the hundredths (second) place and 9>5 so increase
7 to 8 and drop the 99 and get 26.38
Round 42.0237412 to the nearest hundred thousandth
Look at 42.0237412, 4 is in the hundred thousandth (fifth) place and 1 < 5
so drop the 12 and get 42.02374
Addition of decimals
The addition of decimal numbers is almost identical to the addition of whole
numbers. The only difference is that we need to remember to line up the vertical
columns with respect to the decimal point.
Add 0.237 + 4.9 + 27.32
0.237
4.9
+ 27.32
Remember to line up on the decimal points
0 0.237
04.900
+ 27.320
Rewrite with zeroes added in appropriate place values to make things
line up properly
0 0.237
04.900
+ 27.320
32.457
7+0+0=7
3+0+2=5
2 + 9 + 3 = 14 carry 1
0 + 4 + 7 + (1) = 12 carry 1
0 + 0 + 2 + (1) = 3
Subtraction of decimals
Subtraction of decimals is almost identical to subtraction of whole numbers. The only
difference is to remember to line up the columns on the decimal point. (all rules of
borrowing in subtraction apply)
6.93
− 3.7
6.93
− 3.70
Subtract 6.93 – 3.7
Add necessary zero to line up columns
3–0=3
9–7=2
6–3=3
6.93
− 3.70
3.23
Subtract 39.047 – 7.96
39.047
− 7.960
39.047
− 7.960
7
39/ 8.0/ 9 1 47
− 7. 9 60
31. 0 87
7–0=7
We need to borrow from the 9 (changed to
8) to change the 0 to 10 so that we can
borrow from the 10(changed to 9) to
change the 4 to 14
So the result is 31.087
Multiplication of Decimal Numbers
Decimal numbers are multiplied in the same way as whole numbers, with special
consideration given to the number of decimal places in each of the factors.
(number of decimal places in first factor + number of decimal places in second
factor = number of decimal places in product)
Multiply 21.14 × 0.36
21.4
× 0.36
1284
642
7.704
1 decimal place
2 decimal places
1+2=3
3 decimal places
Multiply 0.037 × 0.08
0.037 has 3 decimal places
0.08 has 2 decimal places
37 × 8 = 296 to make this number have 5 decimal places, we need to add 2
addition zeroes 0.00297
(the idea of keeping track of decimal places was necessary in the days
when we used slide rules to perform our multiplications. The slide rule
would do the whole number multiplication for us
(3 × 8 = 297) but you had to put the decimal places in for yourself)
Multiplication by multiples of ten
To multiply by multiples of ten,
move the decimal point to the left the same number of places as there are zeroes in the
multiple of ten factor.
3.82 × 10 = 38.2
3.82 × 100 = 382.
3.82 × 1000 = 3820.
1 zero and 1 move to the left
2 zeroes and 2 moves to the left
3 zeroes and 3 moves to the left (needed to add a 0
at the right end to make the move)
Division of Decimal Numbers
To divide decimal numbers, move the decimal point in the divisor to the right
enough place to make it a whole number. Also move the decimal point in the
dividend an equal number of places.
(remember to keep the decimal point in the quotient directly above the decimal
point in the dividend)
Divide 3.25 15.275
We need to change 3.25 to 325 by moving the decimal point 2 places to the right
Therefore, we need to change 15.275 to 1527.5 by moving the decimal point 2
places to the right. So the problem becomes 325 1527.5
4.7
325 1527.5
−1300
2275
− 2275
0
note: the decimal point in 4.7 is above the decimal point in1527.5
Not all divisions of decimal numbers will come out even as in the above example.
We generally round the quotient in a decimal division instead of carrying a
remainder.
Divide 0.3 0.56 and round to two decimal places
First convert the problem to 3 5.6 , to be able to round to two decimal places we
must have 3 decimal places in the quotient so we will arrange to have 3 decimal
places in the dividend 3 5.600
1.866
3 5.600
−3
2 6
−2 4
20
−18
20
−18
We will round 1.866 to 1.87
Dividing by multiples of ten
To divide by multiples of ten, move the decimal point to the left the same number
of places as there are zeroes in the multiple of ten divisor (placing leading zeroes
as necessary)
34.65 ÷ 10 = 3.465
1 zero in 10 so move 1 place left
34.65 ÷ 100 = 0.3465
2 zeroes in 100 so move 2 places left
34.65 ÷ 1000 = 0.03465
3 zeroes in 1000 so move 3 places left (needed to
add a leading zero for the move)
Converting between Decimals and Fractions
Fractions to Decimals
To convert from a fraction to a decimal it is as simple as performing the implied
division as we learned in elementary school.
Terminating Decimals (division ends)
Ex.
2.5
5
becomes 2.5 since 2 5.0
2
.25
1
becomes .25 since 4 1.00
4
Non-terminating Decimals (repeating decimals)
Ex.
.3333
1
becomes .3333 since 1 3.0000 repeating 3’s forever
3
.1414
14
becomes .1414 since 99 14.00 repeating 14’s forever
99
Decimals to Fractions
Remember that decimals are actually fractions with the denominator as an
appropriate power of ten (the number of zeroes after the one is equal to the
number of places to the right of the decimal point)
Ex.
5
1
reduced to
10
2
125
1
.125 has 3 places right so it is
reduced to
1000
8
.5 has 1 place to right so it is
3.4 has 1 place right so it becomes
2
34
reduced to 3
10
5
Scientific Notation
Occasionally, we need to deal with very large or very small numbers. It is convenient to
use a system called scientific notation to represent these numbers. Scientific notation is
based on powers of ten to represent these numbers.
Large numbers
101 = 10
10 2 = 100
notice the exponent indicates the number of zeroes
10 3 = 1000
10 4 = 10,000
We can express large numbers as a number between 1 and 10 multiplied by the
appropriate power of 10
289 = 2.89 ×10 2 notice the decimal point moved 2 units to the left (same as
exponent)
36782 = 3.6782 ×10 4 4 place move and exponent of 4
Small Numbers
10-1 = 0.1
10-2 = 0.01
10-3 = 0.001
decimal point
notice the exponent indicates the number of places right of
10-4 = 0.0001
We can express small numbers as a number between 1 and 10 multiplied by the
appropriate power of 10.
0.234 = 2.34 ×10 −1 we moved the decimal point 1 unit to the right
0.000123 = 1.23 ×10 −4 we moved the decimal point 4 units right
Introduction to Ratios
In real life, numbers are usually quantities of objects and have units associated with them.
(6 balls, 12 feet and 9 cars)
A ratio is a comparison between two numbers that have the same units.
Ratios can be written in three different ways
3 feet 3
= , ratios do not have units
• As a fraction –
4 feet 4
• As two numbers separated by a colon 3 feet : 4 feet, 3 : 4
• As two numbers separated by the word to 3 feet to 4 feet, 3 to 4
Compare two boards, one 6 feet long to another of 8 feet long.
6 feet 6 3
= = , this means that the shorter board is ¾ the length of the longer
8 feet 8 4
board
Introduction to Rates
A rate is the comparison of two quantities that have different units.
Rates are written as fractions
A cyclist rode 144 miles in 10 hours. We can write a distance to time ratio for the trip
144miles 72miles
, always simplify the fraction, whenever possible
=
10hours
5hours
Unit rates
Whenever a rate has 1 as the denominator, it is referred to as a unit rate.
2 pounds of sirloin sell for $6.25,
A car travels 360 miles in 6 hours,
$6.25 $3.25
=
, or $3.25 per pound
2lbs
1lb
360miles 60miles
=
, 60 miles per hour
6hours
1hour
Introduction to Proportions
A proportion is an expression of the equality of two ratios or rates.
50miles 25miles
=
is a proportion
4gals
2gals
A proportion is true if the two fractions are equal (when written in lowest terms). The
best way to check if a proportion is true is to check the equality of the “cross products”
a c
.i.e. given = is true if ad = bc
b d
Is
2 8
= a true proportion?
3 12
2*12 = 24
3*8 = 24
24 = 24 so it is a true proportion
A proportion is false if the cross products are not equal.
Is
4 8
= a true proportion?
5 9
4*9 = 36
5*8 = 40 therefore a false proportion
Solving proportion problems
Sometimes, we do not know one of the values of a proportion. We use the above
mentioned property of cross products to solve for the missing value
Solve
9 3
=
6 n
By cross multiplying we get
thus
9 3
= is a true proportion
6 2
9n = 6(3)
9n = 18
9n ÷ 9 = 18 ÷ 9
n=2
Introduction to Percents
Percents, like fractions and decimals, are ways of writing numbers that are not
necessarily whole numbers.
Percent means “parts of 100”, so they can always be written as fraction with 100 in the
denominator.
Writing percents as a fraction
To write percents as a fraction, multiply the percent by
1
100
1
13 1
13
= ⋅
=
100 1 100 100
1
120 1
120 6
1
120% = 120 ⋅
=
⋅
=
= =1
100
1 100 100 5
5
2
2 1
50 1
50 1
16 % = 16 ⋅
=
⋅
=
=
3
3 100 3 100 300 6
13% = 13 ⋅
Sometimes it helps to change mixed numbers to improper fractions for the
calculation
Writing percents as decimals
To write a percent as a decimal, multiply the percent by 0.01 (move decimal point
two places to the right)
5% = 5 * 0.001 = 0.05
215% = 215 * 0.001 = 2.15
Writing fractions or decimals as percents
To write fractions or decimals as percents, multiply the fraction or percent by
100%
Change3/8 to a percent
3
3 100
300
⋅100% = ⋅
%=
% = 37.5%
8
8 1
8
Change 0.48 as a percent
0.48 * 100% = 48%
Table of Common Fractions and Their Percentage Equivalents
1
/2 = 50%
1
/3 = 33 1/3%
2
1
3
1
2
/4 = 25%
/5 = 20%
1
2
/6 = 16 /3%
1
/7 = 14 2/7%
/3 = 66 2/3%
/4 = 75%
/5 = 40%
5
3
4
3
4
5
6
/5 = 60%
/6 = 83 /3%
2
/7 = 28 4/7%
/7 = 42 6/7%
/7 = 71 3/7%
/8 = 12 1/2%
/5 = 80%
1
3
5
7
1
2
4
5
7
8
7
9
/9 = 22 2/9%
/8 = 62 1/2%
/7 = 85 5/7%
1
/9 = 11 1/9%
/8 = 37 1/2%
/7 = 57 1/7%
/9 = 44 4/9%
/9 = 77 7/9%
1
/10 = 10%
3
/10 = 30%
/10 = 70%
/8 = 87 1/2%
/9 = 55 5/9%
/9 = 88 8/9%
/10 = 90%
1
/12 = 8 1/3%
Table of Common Fractions and Their Decimal Equivalents or
Approximations
1
/2 = 0.5
1
/3 = 0.3333...
2
/3 = 0.6666...
1
/4 = 0.25
3
/4 = 0.75
1
/5 = 0.2
2
/5 = 0.4
1
/6 = 0.1666...
5
/6 = 0.8333...
1
3
/5 = 0.6
/7 = 0.142857142857...
2
/7 = 0.285714285714...
3
/7 = 0.428571428571...
4
/7 = 0.571428571428...
5
/7 = 0.714285714285...
6
/7 = 0.8571428571428...
1
/8 = 0.125
3
/8 = 0.375
5
/8 = 0.625
7
/8 = 0.875
1
/9 = 0.111...
2
/9 = 0.222...
4
/9 = 0.444...
5
/9 = 0.555...
7
/9 = 0.777...
8
/9 = 0.888...
7
/10 = 0.7
9
/10 = 0.9
1
/10 = 0.1
1
/12 = 0.08333...
3
/10 = 0.3
4
/5 = 0.8
Percent Equations
There are many ways to solve problems involving percents. All of them require
that you identify three parts of the problem (percent, base and amount), but most of
require that you also learn rules for when you multiply and when you divide. This is
fairly confusing.
The best way to solve the basic percent problem is to use the proportion,
percent amount
and the cross multiplication, percent * base= 100 * amount.
=
100
base
4% of 85,000 is what number?
First, we must determine where the above numbers go in the proportion. As a
amount
general rule, you can look at the ratio
by looking at the wording of the
base
problem. The number associated with the word of generally as the base and the
amount is
=
number associated with the word is generally is the amount. So,
.
base
of
4 = percent
is = what number (the variable (n))
of = 85,000
4
n
=
, by cross multiplying we get
100 85,000
4(85,000) = 100n
340000 = 100n
3400 = n
What percent of 40 is 30?
Is = 30
Of = 40
Percent = what (n)
n
30
=
100 40
40n = 3000
n = 75%
18% of what is 900?
Percent = 18
Is = 900
Of = what (n)
18 900
=
100
n
18n = 90000
n = 5000
Applications of Percents
A certain charitable organization spent $2940 for administrative costs. This is
12% of the total amount of the monies they collected. How much did they collect
in total?
We are really asking $2940 is 12% of what number?
12 2940
=
100
n
12n = 294000
n =24,500 so, they collected a total of $24,500
The fire department reports that 24 false alarms were sent in out of a total of 200
alarms. What percent of all the alarms are false?
We are really asking 24 is what percent f 200?
n
24
=
100 200
200n = 2400
n = 12% so, 12% were false alarms
An antiques dealer claims that 86% of her sales are for items that cost less than
$1000. If she sells 250 items in a given month, how many will sell for less than
$1000?
We are really asking what number is 86% of 250? (notice $1000 is not used
anywhere)
86
n
=
100 250
100n = 21500
n = 215 so, 215 sold for less than $1000
Negative Numbers
There are special numbers associated with the positive numbers that are called the
additive inverses. These are numbers that when added to a positive number will give 0 as
the result.
Ex.
4 + something = 0
something = -4, therefore 4 + -4 = 0
6.6 + -6.6 = 0
1024 + -1024 = 0
These additive inverses of positive numbers are called negative numbers. Many of us are
familiar with negative numbers when we look at our checking accounts (the US
Government).
Adding and Subtracting Signed Numbers
There are basic rules used in the addition and subtraction of signed (+ or -)
Addition of signed numbers
There are two situations to consider in the addition of two signed numbers (or the signs
the same or different)
Signs the same
If the signs of the two numbers are the same, just add the numbers and
apply the common sign
Ex.
4 + 5 = 9 Both are positive so 9 is positive
(-4) + (-5) = (-9) Both are negative so -9 is negative
Signs different
If the signs of the two numbers are different then subtract the two numbers
and use the sign of the number with the larger absolute value
Ex.
4 + (-5)
5 – 4 = 1 and since 5 > 4 we use the – from -5 to get -1
-41 + 51
51 – 41 = 10 and 51 > 41 so we get +10 (10)
Subtraction of Signed Numbers
The best way to subtract signed numbers is to not do it. You can always convert a
subtraction into an addition and work from there.
Ex.
4 – 5 convert to 4 + (-5), the conversion to to addition requires a sign
change in the number after the subtraction sign
4 + (-5) = -1 by addition rules above
Ex.
-4 – 5 convert to -4 + (-5) = -9
Ex.
4 – (-5) convert to -4 + 5 (note -5 changed to 5) so -4 + 5 = 1 by addition
rules
When it comes to larger, more complicated problems, remember you can only add
or subtract two numbers at a time, so only work with two at a time
Ex.
4 + (-5) – 6 + (-7) – 9
pretty long, but take two at a time
4 + (-5) = -1 so we get
-1 – 6 + (-7) – 9
-1 – 6 = -1 + (-6) = -7 so we get
-7 + (-7) – 9
-7 + (-7) = -14 so we now get
-14 – 9 = -14 + (-9) = -23
Multiplying and Dividing Signed Numbers
There are exactly two rules that must be followed when multiplying or
dividing two signed numbers
•
If the signs agree, the answer is positive
•
If the signs disagree, the answer is negative
Remember, the operations of multiplication and division can only be done
with 2 numbers at a time – complicated problems must be taken two
numbers at a time.
Ex.
4*(-5) = -20
opposite signs
(-4)*(-5) = 20 same signs
− 20
= −5
4
opposite signs
− 20
=4
−5
same signs
Ex.
(-4)(-5)(2)(-6)
(-4)(-5) = 20
20(2)(-6)
(20)(2) = 40
40(-6)
-240
Variables
Throughout our study of algebra, we hear about these things called variables.
What is a variable?
Variables are just symbols to represent values that are currently unknown. They stand in
the place of numbers.
Ex.
In the expression 3x, x represents a number and is called a variable
In the expression 2x -4y, x and y both represent different numbers and are
called variables
Throughout algebra, and actually real life, we encounter unknown values that can be
expressed as variables. If you go to the grocery store and find that 3 cans of cream corn
cost $1.29, the price of one can is unknown and can be represented by a variable (maybe
p for price)
Evaluating Algebraic Expressions
Algebraic expressions will have a “value” dependent upon the number that is substituted
into the variable.
Example:
Evaluate 4x + 3 if x = 1
4(1) + 3 = 7 + 3 = 10
Evaluate 4x + 3 if x = -2
4(-2) + 3 = -8 + 3 = -5
Combining Like Terms in Algebraic Expressions
Two or more terms in an algebraic expression that have the same variable part are
called like (similar) terms. (Think of apples and oranges. If an expression is made up of
3 apples and 2 oranges, they cannot be combined since apples ≠ oranges)
If an algebraic expression contains 2 or more like terms, they can be combines to
form a single term.
Examples:
4x + 7x = (4 + 7) x = 11x
13u – 7u + 2u = (13 – 7 + 2) u = 8u
6x + 3x + 7y = (6 + 3) x + 7y = 9x + 7y
Not like terms
Properties of Equations
The major thing to remember when working with equations, is when you make a
change to one side of the equation you must make an identical change to the other side.
Rule: You can add or subtract a value to both sides of an equation without
changing the solution to the equation.
Examples;
X + 3 = 7, if I add 3 to the left, I must add 3 to the right
X+3+3=7+3
X + 6 = 10
Y + 6 = 10, if I subtract 6 from the left, I must subtract 6 from the
right
Y + 6 – 6 = 10 – 6
Y=4
Rule: You can multiply both sides of the equation by a constant or divide a
constant into both sides of the equation without changing the solution to the
equation.
Examples:
-4c = 20
-4c ÷ -4 = 20 ÷ -4
c=5
2
5
x=
3
9
3 2
3 5
∗ x= ∗
2 3
2 9
15 5
x=
=
18 6
I can divide both sides by –4
I can multiply both sides by 3/2
2/3 * 3/2 =1
You can also combine both of the above rules in the same problem to simplify.
3x + 2 = 8
3x + 2 – 2 = 8 – 2
3x = 6
3x ÷ 3 = 6 ÷ 2
x=2
Subtract 2 from both sides
Divide both sides by 3
1
x−2=6
2
1
x − 2 + 2 = 6 + 2
2
1
x=8
2
1
x ∗ 2 = 8 ∗ 2
2
x = 16
Add 2 to both sides
Multiply both sides by 2
Algebraic Equations
When we connect two algebraic expressions using = , we call the combination an
equation.
There are three parts to an equation
X+3=7
Left side
equal sign
right side
An equation can be true or false, dependent upon the number substituted into the
variable.
Examples
X + 3 = 7 is true if X = 4
X + 3 = 7 is false if X = 1
Some equations are true for any value of the variable. These types of equations
are called identities.
Example:
2(3x + 7) = 6x + 14 is an identity
x = 5 2(3(5) + 7) = 6(5) + 14
44 = 44
x = 1 2(3(1) + 7) = 6(1) + 14
20 = 20
Most equations are not identities. They are true for only certain values of the
variable. These are called conditional equations.
Example:
5x – 2 = 3
5x – 2 = 3
5x – 2 = 3
is true if x = 1
is false if x = any other number
is a conditional equation
The process of determining the values that make a conditional equation true is
called solving the equation.
Solving Linear Equations
Linear Equation
An equation where the only exponent involved with the variable is 1 is called a
linear equation.
Examples:
3x + 2 = 8
2(x + 4) – x + 7 = 6 are both linear equations
The best way to look at the procedures for solving a linear equation is to undo the
operations that have been done to the variable.
Example:
Equation
3x + 2 = 8
3x + 2 – 2 = 8 – 2
3x = 6
3x ÷ 3 = 6 ÷ 2
x =2
Done to x
2 added to 3x
How undone
Subtract 2
Multiplied by 3
Divide by 3
We also need to be able to get the variables on one side and the constants on the
other sides.
Example:
3x + 2 = x – 6
3x + 2 – x = x – 6 + x
2x + 2 = 6
2x + 2 –2 = 6 – 2
2x = 4
2x ÷ 2 = 4 ÷ 2
x=2
Move x to left by subtracting x from both
sides
Subtract 2 from both sides
Divide both sides by 2
Done
Procedures for Solving Linear Equations
1.
2.
3.
4.
5.
Perform any implied operations (multiply through parenthesis, etc.)
Combine like terms on each side
Perform necessary operations to separate constants and variables
Combine like terms as necessary
Multiply or divide to make the numerical coefficient 1
Literal Equations
Literal equations are also called formulas. They form a relationship between two or more
variable, such as the formula to find the area of a circle from the radius A = π r 2 .
Occasionally, we will need to solve for the variable that is not explicitly solve for in the
equation, like trying to find r from the formula above.
To solve for a variable in a formula, treat the other variables as constants and solve as
you normally would any other equation.
Ex.
Solve A = 2L + 2W (area of a rectangle) for L
A = 2L + 2W subtract 2W from each side
A – 2W = 2L divide both sides by 2
(A-2W)/2 = L given A and W can find L
Solve C = 2 r for r (circumference of a circle)
�
C = 2 r divide both sides by 2 (remember
�
C/2 = r can find r given C
�
�
�
is just a number)
Linear Inequalities
The rules for solving linear inequalities are the same as those for linear equalities except
for one thing, if you multiply or divide the inequality by a negative number the inequality
sign switches.
The solutions to linear inequalities are intervals of numbers, not individual numbers like
in equalities.
Interval Notation
There are 2 symbols used to show the endpoints of intervals
( - endpoint of the interval not included
[ - endpoint of interval included
Ex.
(0,2) all numbers between 0 and 2 but not including 0 or 2
[0,2) all numbers between 0 and 2 including 0 but not 2
[0,2] all numbers between 0 and 2 including 0 or 2
(- ∞ , 2] all numbers less than or equal to 2
(2, ∞ ) all numbers greater than 2 but not 2
Solving Inequalities
Ex.
Solve 3x + 2 > 0
subtract 2 from each side
3x > -2
divide by 3
x > -2/3
(-2/3, ∞ )
Solve -4x – 8 < 12
add 8 to each side
-4x < 20
divide by -4(remember to switch sign)
x > -5
(-5, ∞ )
Solve 3x + 4 < 2x – 1 put x’s on one side and constants on the other
3x + 4 < 2x – 1 subtract 4 from each side
3x < 2x – 5 subtract 2x from each side
x < -5
(- ∞ , -5)
Multiplication using Exponents
Exponents are a shortcut form for repeated factors in a multiplication.
Examples;
34 = 3*3*3*3 = 81
(-2)5 = (-2)*(-2)*(-2)*(-2)*(-2) = -32
2
2 2 2 8
( )3 = ∗ ∗ =
3
3 3 3 27
n
x = x ∗ x ∗ x ∗∗∗ x
n times
Parts of the exponential number
xn
Base
Exponent
There are a few rules governing the use of exponential numbers
Rule 1
Multiplication Property of like terms
am * an = am+n
34 * 37 = 34+7 = 311
Rule 2
Power to a Power Property
(am)n = amn
(42)6 = 42*6 = 412
Rule 3
Power of a Product Property
(ab)n = anbn
(5x)2 = 52x2 = 25x2
(-x)5 = (-1)5x5 = -x5
(-x)6 = (-1)6x6 = x6
Further Examples:
(2x3)4 = 24x3*4 = 24x12 = 16x12
(-3xy2)2(2x3y3)3
(-3)2x2y2*223 x3*3y3*3
9x2y48x9 y9
72 x11 y13
Division with Exponents
an
we need to develop a few rules that will lead us to a some
am
interesting properties of exponents.
In order to evaluate
Look at
note:5 – 3 = 2
an
n−m
=
a
am
Therefore we have a rule
Examples:
56
= 5 6−4 = 5 2
4
5
64
= 60
64
64
also
= 1 therefore 6 0 = 1
4
6
From here, we get
also
a 0 = 1 Zero Power Rule
54
= 5 4 −6 = 5 −2
6
5
54
5 *5 *5 *5
1
1
1
=
=
= − 2 therefore 5 −2 = 2
6
5*5 *5 *5 *5 *5 5 *5 5
5
5
leading us to the
a −n =
1
an
Negative Power Rule
3
3 3
32
9
( ) 2 = ( )( ) = 2 =
4
4 4
16
4
:
giving us
a n an
( ) = n
b
b
Power of a Quotient Rule
Further Examples:
(
x4 5
x 4*5
x 20
)
=
(
)
=
y3
y 3*5
y 15
9 x 2 ( y 4 ) 3 9 x 2 y 12 9 x 2 y 12
1
y6
6
=
= ∗ ∗
= 1∗ 2 ∗ y = 2
(3 x 2 y 3 ) 2
9x4 y6 9 x4 y6
x
x
note: the variable ends up on the side of the fraction line as the
largest power
x 6
y4
note: the negative power causes the variable to change sides of the
fraction line
( x −3 y 2 ) −2 = x −3*−2 y 2*−2 = x 6 y −4
=
Polynomials
Polynomials are a special type of algebraic expression where the exponents are positive
integers.
Ex.
x2 + 2x -1 is a polynomial since exponents are 2 and 1
x3/2 is not a polynomial since exponent is 3/2
x is not a polynomial (radical variables have fractional exponents)
3x is a polynomial since exponent is 1
Evaluating Polynomials
Polynomials will have a “value” dependent upon the number that is substituted into the
variable.
Example:
Evaluate 4x + 3 if x = 1
4(1) + 3
7 + 3
10
Evaluate 4x + 3 if x = -2
4(-2) + 3
-8 + 3 = -5
Evaluate y3 + 3y –5 if y = -3
(-3)3 + 3(-3) – 5
-27 - 9 – 5
-40
Evaluate y3 + 3y –5 if y = 0
(0)3 + 3(0) – 5
0 + 0 –5
-5
Evaluate 3u2 + 2uv – 12 if u = -1 and v = 5
3(-1)2 + 2(-1)(5) – 12
3(1) –10 –12
-19
Adding Polynomials
Adding polynomials is basically the same as the addition of numbers with units.
You can only add terms that are similar.
Ex.
2 apples + 3 oranges + 5 apples = (2 + 5) apples + 3 oranges = 7 apples + 3
oranges (you can’t add apples and oranges)
Ex.
2x + 3x = (2 + 3)x = 5x
Ex.
5y + 3x + 8y + 4x = (5 + 8)y + (3 + 4)x = 13y + 7x
Ex.
2x2 +3x2 + 4x3 = 5x2 + 4x3 (note:x2 and x3 are not similar terms)
Subtracting Polynomials
The subtraction of polynomials is basically the same as the addition of polynomials, with
one exception (you have to deal with the subtraction sign). We accomplish this by
remembering how to subtract signed numbers – change the subtraction to addition and
change the sign of everything after the minus sign.
Ex.
(3x2 + 2x + 4) – (-2x2 + 3x + 1)
change to
(3x2 + 2x + 4) + (2x2 – 3x – 1)
combine like terms
(3x2 +2x2) + (2x – 3x) + (4 – 1)
perform operations
5x2 -1x + 3
Ex.
(3xy – 4y + 3xz) – (3y -3xy +xz)
change to
(3xy - 4y + 3xz) + (-3y + 3xy –xz)
combine like terms
(3xy +3xy) + (-4y +3y) + (3xz – xz) perform operations
6xy – y + 2xz
Multiplying Polynomials
Multiplying polynomials is mostly a trial in keeping things lined up and not missing any
parts of the problem.
Technique to multiply polynomials
Break into several simple problems using distribution
Multiply each part
Combine like terms
Ex.
use distribution of first term
(3x2 + 2)(3x – 1)
3x2(3x -1) + 2(3x – 1) distribute again
3x2(3x) -3x2(1) + 2(3x) – 2(1) Multiply terms
9x3 – 3x2 + 6x -2
Ex.
(6x +2y + 1)(2x -3y – 2)
6x(2x -3y – 2) + 2y(2x – 3y – 2) + 1(2x -3y – 2)
6x(2x) – 6x(3y) -6x(2) +2y(2x) – 2y(3y) -2y(2) +1(2x)-1(3y)-1(2)
12x2 -18xy-12x + 4xy – 6y2 -4y + 2x – 3y -2
12x2 -18xy +4xy -12x +2x -4y -3y -6y2 -2
12x2 -14xy -10xy -7y - 6y2 – 2
FOIL Method
If you are multiplying a binomial by a binomial you can use a variation of the above
technique call First Outside Inside Last (FOIL)
(ax + b)(cx + d)
First
Outside
Inside
Last
(ax)(cx) = acx2
ax(d)
b(cx)
bd
acx2 +(ad + bc)x + bd
Ex.
(x + 2)(x + 3)
First
Outside
Inside
Last
x(x)
x(3)
2(x)
2(3)
x2 +3x +2x + 6
x2 + 5x + 6
Special Products
There are a few special types of products that are handy to keep in your back
pocket for use in a hurry. These are not necessary to memorize, but they can help you
perform multiplication of polynomials rapidly, in the cases where they apply.
Special Product #1
(a + b) 2 = a 2 + 2ab + b 2
Example
(6 + y) 2 = 6 2 + 2(6)( y) + y 2 = 36 + 12 y + y 2
Special Product #2
(a − b) 2 = a 2 − 2ab + b 2
Example:
(6 − y) 2 = 6 2 − 2(6)( y) + y 2 = 36 −12 y + y 2
Special Product #3
(a + b)(a − b) = a 2 − b 2
Example:
(3 + c)(3 − c) = 3 2 − c 2
Further Examples using the Special Product Rules
(8 + 2 y) 2 = 8 2 + 2(8)(2 y) + (2 y) 2 = 64 + 32 y + 4 y 2
(4m 3 n 2 − 12 ) 2 = (4m 3 n 2 ) 2 − 2(4m 3 n 2 )( 12 ) + ( 12 ) 2 = 16 m 6 n 4 − 4m 3 n 2 +
(3x
2
)(
) ( ) − (2 y )
− 2 y 3 3x 2 + 2y 3 = 3x 2
2
3 2
= 9x 4 − 4 y
6
1
4
Dividing Polynomials
There are two types of division problems for polynomials, those that can be done as
fractions and those that require long division. If the divisor is a single value then the
division can be broken into a group of simple fractions, otherwise you must use long
division.
Division using fractions
Ex.
18 x 2 − 4 x 18 x 2 4x
=
−
= 9x − 2
2x
2x
2x
Long Division
Ex.
x + 1 x 2 + 2x + 1
x
x + 1 x + 2x + 1
− (x 2 + x)
x +1
x2 divided by x gives us x
2
multiply x(x+1) and subtract
x divided by x gives 1
x +1
x + 1 x + 2x + 1
− (x 2 + x)
x +1
− (x + 1)
0
2
Multiply 1(x + 1) and subtract
Therefore, we get ( x 2 + 2 x + 1) ÷ ( x + 1) = x + 1
Common Factors
We can expand the idea of the Greatest Common Factor, that is studied in earlier
classes into the world of algebra. We will be looking for the common factors in algebraic
expressions.
Look at 30x2 and 42x3. We can factor them individually.
30x2 = 2 * 3 * 5 * x * x
42x3 = 2 * 3 * 7 * x * x * x
Each term has a 2, a 3, and 2-x’s so they have 2 * 3 * x * x = 6x2 in common
30x2 = 6x2(5)
42x3 = 6x2(7x)
Example:
Find the common factors for 15x3 y3, 6x2 y3 and 9xy4
First look at the coefficients 15, 6, and 9
The common factor of 15, 6, and 9 is 3
Look at the x variables
The exponents on x are 3, 2, and 1 choose the lowest
number, x1 = x is the common x term
Look at the y variable
The exponents of y are3, 3, and 4 choose the lowest
number, x3 is the common term.
Therefore the common term is 3xy3
15x3y3 = 3xy3 (5x2)
6x2y3 = 3xy3 (2x)
9xy4= 3xy3 (3y3)
Factoring polynomials by common factors
A polynomial is factored when it is expressed as a product of 2 or more
polynomials.
The distributive property of integers and algebra, P(Q + R) = PQ + PR is the
greatest aide in factoring. For the purposes of factoring we generally write the property
in reverse order PQ + PR = P(Q + R). In this way we can see that P and (Q +R) are the
factors of PQ + PR.
Example:
Factor 5x + 30 by common factors
Since 5x = 5 * x and 30 = 5 * 6, 5 is the common factor
5x + 30 = (5 * x) + (5 * 30)
5x + 30 = 5(x + 30) by reverse distributive law.
Procedure to factor by common factors.
1. Find the common factor in all the terms
2. Write each term as a product using the common factor found in step 1
3. Use reverse distributive law to factor out the common factor
Example:
Factor 5y3 + 25y
The common factor in the terms is 5y
5y3 + 25y = 5y(y2) + 5y(5)
so
5y3 + 25y = 5y(y2 + 5)
Example:
Factor 18h5 + 12h4 – 21h3
The common factor is 3h3
18h5 = 3h3(6h2)
12h4 = 3h3(4h)
21h3 = 3h3(7)
so
18h5 + 12h4 – 21h3 = 3h3(6h2) + 3h3(4h) - 21h3 - 3h3(7)
18h5 + 12h4 – 21h3 = 3h3(6h2 + 4h – 7)
Example:
Factor 4a7b5c + 10a3b8 – 16a4b6 + 18a5b7
The common factor is 2a3b5, note c is not in all terms
4a7b5c = 2a3b5(2a4c)
10a3b8 = 2a3b5(5b3)
16a4b6 = 2a3b5(8ab)
18a5b7 = 2a3b5 (9a2b2)
so
4a7b5c + 10a3b8 – 16a4b6 + 18a5b7 =
2a3b5(2a4c) + 2a3b5(5b3) - 2a3b5(8ab) + 2a3b5 (9a2b2)
4a7b5c + 10a3b8 – 16a4b6 + 18a5b7 = 2a3b5(2a4c + 5b3 - 8ab + 9a2b2)
Factoring by Grouping
Certain polynomials, though they have no common factors, can be still be
factored by using common factors. We have to be able to re-group the terms so that they
are in groups that have common terms.
Look at 3xm + 3ym – 2x –2y
2 terms have x and 2 terms have y
(3xm –2x) + (3ym – 2y)
Factor x out of the first part and y out of the last part
x(3m – 2) + y(3m – 2)
note: 3m –2 is now in common
(x + y)(3m – 2)
Example:
Factor 3a + 3b – ma – mb
(3a + 3b) + (-ma –mb)
Factor 3 out of first part and –m out of second part
3(a + b) – m(a + b) note: factoring –m out of –mb leaves +b
(3 – m)(a + b)
Solving Equations by Factoring
We can use factoring to solve algebraic equations. We will use the property of
multiplication of two numbers equaling zero, (if a*b = 0 then either a = 0 or b = 0)
Solve x2 + 5x = 0
By factoring we get
x(x + 5) = 0 therefore
x = 0 or x + 5 = 0
which implies
x = 0 or x = -5
Solve 2x2 – 6x = 0
By factoring we get
2x(x – 3) = 0 therefore
2x = 0 or x – 3 = 0
which gives us
x = 0 or x = 3
Solve x2 +5x + 6 = 0
By factoring we get
(x + 3)(x + 2) = 0
therefore
x + 3 = 0 or x + 2 = 0
which leads to
x = -3 or x = -2
Factor the difference of two squares
We know from our multiplication rules that (a + b)(a – b) = a2 – b2. Applying this
rule backwards we get a factoring rule for the difference of two squares,
a2 – b2 = (a + b)(a – b)
If we can identify the problem as being the difference of two squares, we can
apply above rule.
Example:
Factor x2 – 4
x2 – 22
a = x and b = 2
(x + 2)(x – 2)
Example:
Factor 4t2 – 9s2
(2t)2 – (3s)2 a = 2t and b = 3s
(2t + 3s)(2t – 3s)
Example:
Factor u4 – 16
(u2)2 - 42
a = u2 and b = 4
2
2
(u + 4)(u – 4)
(u2 + 4)(u2 – 22)
a = u and b = 2
(u2 + 4)(u + 2)(u – 2)
Factoring the sum or difference of cubes
If a polynomial is made up of the sum of difference of two perfect cubes, we have
special rules to handle the factoring (these are not easy to memorize, I can never
remember them without looking).
a3 + b3 = (a + b)(a2 – ab + b2)
a3 - b3 = (a - b)(a2 + ab + b2)
Example:
Factor x3 + 8
note: 8 = 23
Therefore, we get
x3 + 23
a = x and b = 2
(x + 2)(x2 –2x +22)
(x + 2)(x2 –2x +4)
Example:
Factor 64u3 – 27v6
note: 64u3 = (4u)3 and 27v6 = (3v2)3
(4u)3 - (3v2)3
a = 4u and b = 3v2
2
2
(4u – 3v )((4u) + (4u)(3v2) + (3v2)2)
(4u – 3v2)(16u2 + 12uv2 +3v4)
Introduction to Rational Expressions
A rational expression is a fraction (p/q) where p and q are polynomials.
Ex.
3 2 y 7c +1
,
,
are all rational expressions
x y +1 c 2 −1
Evaluation of Rational Expressions
As with polynomials the value of a rational expression is dependent upon the
value chosen for the variable.
Ex.
Evaluate
x+2
for x = -1, x = 2, and x = 1
x −1
x = -1 gives us
−1+ 2
1
1
=
=−
−1−1 − 2
2
x = 2 gives us
2+ 2 4
= =4
2 −1 1
x = 1 gives us
1+ 2 3
= which implies there is no solution
1−1 0
It is important to realize the denominator (bottom) of a fraction can not equal to zero.
Zero in the denominator of a fraction causes the fraction to be undefined.
Ex.
For what value of x is
x −5
undefined?
3x +1
For a rational expression to be undefined, the denominator must be equal
1
to zero therefore we set 3x + 1 = 0 and get x = - . Thus we cannot
3
1
x −5
substitute - into the fraction
3
3x +1
Equivalent Rational Expressions
If two rational expressions can be reduced to the same rational expression, then
the two rational expressions are called equivalent rational expressions
Examples:
6x
2
and are equivalent fractions.
9x
3
6y
2y
and
are not equivalent
2y + 6
y+3
2y
is in reduced form
y+3
3y
2y
≠
y+3 y+3
Equivalent fractions in signed number systems
−p
p
p
=
=−
q
−q
q
p
−p
p
− p
=−
=−
=
q
q
−q −q
Examples:
1
−1
−1
=
=
y − x − ( y − x) x − y
a − b − (b − a) b − a
=
=
x − y − ( y − x) y − x
Simplifying Rational Expressions
As with the more familiar numerical fractions we can reduce rational expressions
by canceling common factors in the numerator and the denominator.
pk p
=
qk q
Examples;
Reduce
8x
12 x
Reduce
12x 2 y
9xy 2
Reduce
4x 2 −1
2x 2 + x
Warning: it is common to try factoring improperly. Remember, you can only
factor out factors that in a product, not terms in and addition or subtraction
Proper reducing
Improper reducing
Addition and Subtraction of Rational Expressions
As in numerical fractions, we can only add and subtract rational expressions
(algebraic fractions) if they have like (common) denominators.
a c a+c
+ =
b b
b
a c a−c
− =
b b
b
Examples:
2x 3x 2x + 3x 5x
+
=
=
7
7
7
7
If the two fractions do not have a common denominator, we must convert the
individual fractions, separately, to fractions with the same denominator(LCD).
Example:
Add
2x 4x
+
, the denominators are 3 and 5, so the LCD is 15
3
5
2x 2x 5 10x
4x 4x 3 12x
=
• =
and
=
• =
therefore we get
3
3 5 15
5
5 3 15
10x 12x 22x
+
=
15
15
15
Finding the Least Common Denominator
We create a LCD in algebraic fractions just like we did in arithmetic
fractions
1.
2.
3.
Factor each denominator completely
List each prime factor the greatest number of times that it appears
in either of the factored forms of the denominator.
The product of these factors is the LCD.
Examples:
7
10x
+
10x = 2*5* x
15x = 3*5* x
LCD = 2*3*5* x = 30x
8
15x
7
7 3 21
=
• =
10x 10x 3 30x
8
8 2 16
=
• =
15x 15x 2 30x
therefore
21 16
37
+
=
30x 30x 30x
18x = 2⋅3⋅3⋅ x
12xy = 2⋅ 2 ⋅3⋅ x ⋅ y
LCD = 2⋅ 2⋅3⋅3⋅ x ⋅ y = 36xy
5
7
+
18x 12xy
5
5 2y 10 y
=
•
=
18x 18x 2y 36xy
7
7
3
21
=
• =
12xy 12xy 3 36xy
therefore
10y
21
10 y + 21
+
=
36xy 36xy
36xy
3
2
a b
−
4
b 3c
LCD = a 2 b 3 c
3
3 b 2 c 3b 2 c
=
•
=
a 2 b a 2 b b 2 c a 2 b 3
c
2
2
4
4 a
4a
= 3 • 2 = 2 3
3
b c b c a
a b c
therefore
3b 2 c − 4a 2
a 2 b 3 c
do not simplify at this point 1
4
+
y+3 y+4
LCD = ( y + 3)( y + 4)
y+4
y+4
1
1
=
•
=
y + 3 y + 3 y + 4 ( y + 3)( y + 4)
y +3
4( y + 3)
4 y +12
4
4
=
•
=
=
y + 4 y + 4 y + 3 ( y + 3)( y + 4) ( y + 3)( y + 4)
therefore
y+4
4 y +12
5y +16
+
=
( y + 3)( y + 4) ( y + 3)( y + 4) ( y + 3)( y + 4)
Multiplication and Division of Rational Expressions
Multiplication of Rational Expressions
Multiplication of Rational Expressions is primarily the same as multiplication with
numerical fractions.
Examples:
Division of Rational Expressions
Division of Rational Expressions involves using the old adage “flip the last guy and
multiply” – Invert and Multiply
Examples:
4 3t
4 7t 28t 28
÷ =
• =
=
25 7t 25 3t 75t 75
Note the inverting
16a 3
8a
16a 3 25b 2 400a 3 b 2 10a 2
÷
=
•
=
=
8a
b
5b 3 25b 2
5b 3
40ab 3
Complex Fractions
A complex fraction is one in which the numerator or denominator or both contain
fractions.
Examples:
5 16
7 , x , 9 are all complex fractions
3 11 x
x x−3
To simplify a complex fraction, we express it as a simple fraction of lowest terms.
The best way to handle a complex fraction is to realize that fractions are really quotients.
Then you can use the old adage “Flip the last guy and multiply” (Invert and multiply)
Examples:
3
4 = 3 ÷ 3 = 3 • 5 = 15 = 5
3 4 5 4 3 12 4
5
3 1
+
4 3
1 1
+
3 6
Hint: confront the top and bottom separately
3 1 9
4 13
+ =
+
=
4 3 12 12 12
1 1 2 1 3 1
+ = + = =
3 6 6 6 6 2
13
12 = 13 • 2 = 13
1 12 1 6
2
1
x remember to handle top and bottom separately
1
1+
x
x−
x−
1 x 1 x 2 1 x 2 −1
= − =
− =
x 1 x
x x
x
1+
1 1 1 x 1 x +1
= + = + =
x 1 x x x
x
x 2 −1
2
x = x −1 • x = (x + 1)(x −1)x = x −1
x +1
x
x +1
x(x + 1)
x
Equations involving Algebraic Fractions
We have to occasionally solve equations involving fractions. We have a
procedure to follow to help us in this quest.
Solve equations involving fractions
1. Determine the LCD
2. Multiply both sides of the equation by the LCD (this step eliminates the
fractions)
3. Solve the resulting equation
4. Check you solution. Sometimes the solution to the resulting equation will not
be a solution to the original algebraic problem (it will cause the denominator
of the fraction to be zero)
Examples:
t 2t 55
+ =
LCD = 12
4 3 12
t 2t 55
12( + = )
4 3 12
t
2t
55
12( ) + 12( ) = 12( )
4
3
12
3t + 8t = 55
11t = 55
t =5
1 1
5 1
+ =
+ , LCD = 6y
y 2 6 y 3
1 1
5 1
6 y( + =
+ )
y 2 6 y 3
1
1
5
1
6 y( ) + 6 y( ) = 6 y( ) + 6y( )
y
2
6y
3
6 + 3y = 5 + 2 y
3y − 2 y = 5 − 6
y = −1
since y = -1 will not cause any of the denominators to become zero, it is
the solution
w
2
+2=
, LCD is w-2
w−2
w − 2
w
2
(w − 2)(
+2=
)
w−2
w − 2
w
2
(w − 2)(
) + (w − 2)2 = (w − 2)(
)
w−2
w−2
w + 2w − 4 = 2
3w = 6
w=2
But w =2 causes the denominator of the fractions to become zero therefore
it cannot be a solution. (There is no solution)
The Cartesian Coordinate System
The Cartesian Coordinate System (Rectangular Coordinate System) is the
perpendicular crossing of two number lines (axis), one horizontal (x) and one vertical (y).
Using the Cartesian Coordinate System, we can assign names to points on a plane.
The point where the two axes cross is called the origin.
Origin
We name points on the Cartesian Plane by using the ordered pair (x,y), where x
(abscissa) is the position along the horizontal (x) axis and y (ordinate) is the position
along the vertical (y) axis.
(x,y) is called an ordered pair because, in this case, order matters.
The x- axis and the y-axis divide the Cartesian Plane into 4 sections called
quadrants (QI, QII, QIII, QIV)
Points on either axis are not in any quadrant.
The graph of an equation is the set of points that are solutions to the equation
3x + 2y = 12
(4,0) is a solution
(0,6) is a solution
(1,1) is not a solution
Graphs of the Linear Equation
The graph of an equation of two variables is the set of all points that are solutions
to the equation. (Obviously, it is impossible to plot all the points that are solutions to
equation. We determine the type of graph that is associated with the specific type of
equation and plot enough points to position the graph).
The Linear Equation
An equation of the form Ax + By + C = 0 is called a linear (first degree) equation.
The graph of a linear equation is a straight line.
To graph a linear equation
1. Find three points that solve the equation (pick 3 values for x and evaluate
the corresponding values for y). (Note: Technically you can draw a line
using only two points, but we use three for accuracy in our rough
sketches)
2. Plot the3 points on a Cartesian Plane.
3. Draw a straight line through the 3 points. (If you cannot draw a straight
line between the three points, you probably made a mistake in step 1)
Examples:
Graph y = x – 2
x
-1
0
2
y= x–2
y = 1 – 2 = -3
y = 0 – 2 = -2
y=2–2=0
Plot the points
Point
(-1, -3)
(0, -2)
(2, 0)
Draw line through the points
Graph y = -2x + 4
x
0
1
2
y = -2x + 4
y = -2(0) + 4 = 4
y = -2(1) + 4 = 2
y =–2(2) + 4 = 0
Point
(0,4)
(1,2)
(2,0)
Special Lines
A linear equation with B = 0, Ax + C = 0 is represented by a vertical line
through the point
Example:
Graph 2x + 4 = 0
2x + 4 = 0
2x = -4
x = -2
A linear equation with A = 0, By + C = 0 represents a horizontal line
through the point
Example:
Graph 3y –12 = 0
3y = 12
y=4
Intercept of a Line
All straight lines cross at least one of the coordinate axes. The points where the
line crosses the axis is called the intercept.
The x-intercept is the point where line crosses the x-axis. To find the x-intercept
of a line, set y = 0 and solve the resulting equation for x. (x,0)
The y-intercept is the point where the line crosses the y-axis. To find the yintercept of a line, set x = 0 and solve the resulting equation. (0,y)
Example:
Find the x-intercept and the y-intercept of 3x + 2y = 6
x-intercept
y = 0 ⇒ 3x + 2(0) = 6
3x = 6
x = 2 (2,0)
y-intercept
x = 0 ⇒ 3(0) + 2y = 6
2y = 6
y = 3 (0,3)
Special Cases
Vertical Lines (x = K) have only a x-intercept
Example:
x=2
Horizontal Lines (y = K) have only a y-intercept
Example:
y=3
Slope of a Line
We generally use the term “slope” to refer to the steepness of a line.
We also use this term when we discuss the steepness of such physical objects as
the slope of a roof or a ski-slope.
In algebra, we define the slope of a line as the ratio of the change of Vertical
distance (rise) to the change in Horizontal distance (run) between two points
rise y 2 − y1
, where (x1,y1)and (x2,y2)
=
run x 2 − x1
are two points on the line.
slope = m =
Examples:
Find the slope of the line through the points (1,2) and (2,4)
4−2 2
m=
= =2
2 −1 1
Find the slope of the line through (0,0) and (2,-4)
− 4 − 0 − 4
m=
=
= −2
2−0
2
Properties of Slope
Simultaneous Linear Equations
The solution to a linear equation in 2 variables (x,y) is a line, with and infinite number of
solutions. If we have the need to solve two linear equations in 2 variables there are three
possible scenarios.
First, the two lines could be parallel and have no points in common
There are NO Solutions in this case
Second, the two lines meet in one point
Then there is one solution to the system
Third, the two lines are actually the same line
Then there is an infinite number of solutions
Solving Simultaneous Linear Equations (Substitution Method)
We can solve simultaneous linear equations (2 equations and 2 unknowns) by
solving one of the equations for one of the variables and substituting into the other.
Example:
2x + 3y = 8
x + 2y = 5
We can solve the second equation (x + 2y = 5) for either x or y. It is simpler in this case
to solve for x, so we get
x = 5- 2y
Now we can substitute this value for x into the first equation (2x + 3y = 8) and solve for y
2(5-2y) + 3y = 8
10 – 4y + 3y = 8
10 – y = 8
10 – y – 10 = 8 – 10
-y = -2
y = 2
Now, we know y = 2, we can substitute this back into either equation to get x
2x + 3(2) = 8
2x + 6 = 8
2x + 6 – 6 = 8 – 6
2x = 2
x =1
Check:
2(1) + 3(2) = 2 + 6 = 8 and 1 +2(2) = 1 + 4 = 5
Therefore the point (1,2) is on both lines
Example:
5x + 2y = -7
3x + y = -5
Solve the bottom equation for y and get y = -5 – 3x
Then substitute –5 – 3x in place of y in the top equation
5x + 2(-5 – 3x) = -7
5x – 10 – 6x = -7
-10 – x = -7
-x = 3
x = -3
Now we can solve for y, in the equation that says y = -5 –3x and get y = -5 – 3(-3)
= -5 + 9 = 4
So the point (-3,4) solve the system
Solving Simultaneous Linear Equations (Elimination Method)
A method for solving simultaneous linear equations is to eliminate one of the
variables by adding or subtract the equations in the system.
There are a few rules that we need to keep track of
We can multiply each of the equations by a constant (we do this to make the
coefficients of one the variables in both equations the same
We can add the equations together to form a third equation that only has one
variable
�
�
Example: (no multiplication needed)
x+y=8
x–y=4
Notice that if we add the two equations together we eliminate the y.
x+ y =8
x− y =4
2x = 12
x = 6
now we put x = 6 into either equation to solve for y
6 + y = 8
y = 2
so the point (6,2) solves this system
Example: (multiplication in only one equation)
3x + 2y = 5
9x – 4y = 5
if we multiply the top equation by 2 we get 6x + 4y = 10 and then addition will
eliminate the y
6x + 4 y = 10
9x − 4y = 5
15x = 15
x =1
now, we can substitute x = 1 into either equation to get y
9(1) – 4y = 5
9 – 4y = 5
-4y = -4
y=1
therefore, (1,1) is the solution
Example: (multiplication in both equation)
4x – 3y = 8
3x + 5y = -2
if we multiply the top by 5 and the bottom by 3 we can get the coefficients of y to
be –15 and 15 which will add to 0 and eliminate y
20x −15 y = 40
9x +15y = −6
29x = 34
34
29
now we can substitute 34/29 into the first equation to solve for y
x=
34
) − 3y = 8
29
136
− 3y = 8
29
136
− 3y = 8 −
29
232 136
− 3y =
−
29
29
96
− 3y =
29
− 32
y=
29
4(
⎛ 34 − 32 ⎞
therefore the point ⎜ ,
⎟ is the solution.
⎝ 29 29 ⎠
Don’t let fractions frighten you, they appear occasionally and we have to deal
with them
Introduction to Quadratic Equations
Many of the applications we encounter in the study of algebra involve solving
quadratic equations of the form ax 2 + bx + c = 0 .
Quadratic equations can have 0, 1 or 2 solutions in the real number system.
Test for the number of solutions of a quadratic equation
(
Define the discriminant b 2 − 4ac
(
(
If (b
)
)
)
− 4ac ) < 0 , there are no solutions in the real number system
If b 2 − 4ac > 0 , there are 2 solutions in the real number system
If b 2 − 4ac = 0 , there is 1 solution in the real number system
2
Examples:
6x2 +5x – 4 = 0
b 2 − 4ac = 52 – (4)(6)(-4) = 25-(-48) = 25 + 48 = 73
73 > 0 so there are 2 solutions
(
)
x2 –2x + 1 = 0
b 2 − 4ac = 22 – (4)(1)(1) = 4 – 4 = 0
0 = 0 so there is 1 solution
(
)
x2 + 1 = 0
b 2 − 4ac = 02 – (4)(1)(1) = 0 – 4 = -4
-4 < 0 so there are no solutions
(
)
Solving Quadratic Equations by Factoring
If a quadratic expression can be factored, then it is quite simple to solve by using
the factors and the fact that if a*b = 0 then either a = 0 or b = 0.
Examples:
x2 – 5x + 6 = 0
(x – 2)(x-3) = 0
x–2=0
or
x–3=0
x=2
or
x=3
two solutions
x2 –2x + 1 = 0
(x – 1)(x – 1) = 0
x – 1= 0
or
x=1
or
one solution
x -1 = 0
x=1
x2 = 36
x2 – 36 = 0
(x – 6)(x + 6) = 0
x–6=0
or
x+6=0
x=6
or
x = -6
two solutions
6x2 + 5x = 4
6x2 + 5x – 4 = 0
(2x –1)(3x + 4) = 0
2x –1 = 0
or
2x = 1
or
x=½
or
3x + 4 = 0
3x = -4
x =-3/4
Solving Quadratic Equations by Extracting the Roots
Some quadratic equations can be written with a perfect square of a form of the
variable on one side of the equal sign. We can use roots extraction to solve the equation
(not as bad as it might sound, dentally)
Examples:
x2 = 36
we can take the square root of both sides
x 2 = 36
x = 6 or x = -6
remember
4u2 – 9 = 0
4u2 = 9
4u 2 = 9
2u = 3 or 2u = -3
u = 3/2 or u = -3/2
25x 2 − 8 = 0
8
x2 =
25
8
8
x=
or x = −
25
25
x=
2 2
2 2
or x = −
5
5
(4 y − 5)2 − 6 = 0
(4 y − 5)2 = 6
4 y − 5 = 6 or 4 y − 5 = − 6
4 y = 6 + 5 or 4 y = − 6 + 5
y=
6 +5
− 6 +5
or y =
4
4
36 = ±6
Solving Quadratics by Completing the Square
If a quadratic equation is composed of a perfect square, then it is easy to solve by
extracting the root. Using a technique called completing the square, we can transform
any quadratic equation into one composed of a perfect square.
(Completing the square is not a popular method for solving quadratic equations,
but the technique is useful in other applications, so discussing the method is important.)
Completing the square
If we look at the pattern formed by the perfect square (a + b) 2 = a 2 + 2ab + b 2 ,
we notice that the middle term is double the product of the square root of the end terms.
In general, we will see the perfect square as (x + k) 2 = x 2 + 2xk + k 2 , we can see
1
the relationship between the last two terms. k 2 = ( (2k) 2 ) the last term is the square of
2
one-half the coefficient of x.
Using this relationship we can change any quadratic into a perfect square.
Examples:
Change x2 + 10x into a perfect square
x2+10x
½ of 10 is 5
x2 + 10x +52 – 52
Add 52 to the problem, also subtract the
same thing to maintain the integrity of the
problem
(x2 + 10x + 25) - 25
Factor the first part.
2
(x + 5) –25
Note: the + comes from the +10x and the 5
is the square root of the 25 in the
parenthesis
x2 – 2x
x – 2x + 12 –12
(x2 – 2x + 1)2 –1
(x – 1)2 – 1
2
r2 + 7r
r2 +7r + 7
2
2
½ of 7 is
-
7
2
4
(r +
7
2
) -
49
4
7
2
, you could say 3.5 but that is
harder to work with
Add 7 2 to the problem and also subtract it
2
(r2 +7r + 49 ) - 49
2
½ of 2 is 1
Add 12 to the problem and also subtract 12
Factor the first part
Note: the – comes from the –2x and the 1 is
the square root of the 1 in the parenthesis
2
4
off
Factor the first part
Note: the + is from + 7 and the
2
square root of
49
4
7
2
is the
Solving Quadratic Equations using Completing the Square
1. Write the problem with variables on one side of the equal sign and constants
on the other side
2. Divide both sides of the equation by the coefficient of the squared term
(completing the square requires there to be a 1 in the first term)
3. Complete the square on the variable side
4. Add the new constant to both sides (eliminates it from the variable side)
5. Solve the equation by extracting the root
Examples:
Solve x2 + 2x- 4 = 0
x2 + 2x- 4 = 0
x2 + 2x = 4
2
(x + 2x – 1) - 1 = 4
(x2 + 2x – 1) = 5
(x + 1)2 = 5
(x+1) = 5 or (x + 1) = - 5
x = 5 - 1 or x = - 5 -1
Solve 2y2 – 12y – 7 = 0
2y2 – 12y – 7 = 0
2y2 – 12y = 7
y2 – 6y = 7
Move 4 to other side
Complete the square
Add 1 to both sides
Factor the variable side
Extract the root
Solve equations
Add seven to both sides
Divide both sides by 2
Complete the square
2
y2 – 6y + 32 – 32=
y2 – 6y + 9 – 9 =
y2 – 6y + 9 =
y2 - 6y + 9
7+
2
= 25
2
7
2
7
2
Add 9 to both sides
9
Factor left side
Extract the root
(y – 3)2 = 25
2
y – 3 = 25 or y – 3 = ­ 25
2
y = 25 + 3 or y = - 25 +3
2
2
5 2
5 2
+ 3 or y = +3
2
2
6+5 2
6 + 5 2
y=
or y = ­
2
2
y=
Add 3 to both sides
2
Simplify
Solving Quadratic Equations using the Quadratic Formula
There are many techniques for solving quadratic equations. Completing the
square will solve any quadratic equation, but it is not very efficient. The most efficient
method for solving quadratic equations is to use the quadratic formula.
The Quadratic Formula
The standard quadratic equation has the form ax2 + bx + c = 0. The solution for
this type of equation can be found by the following formula
x=
− b ± b 2 − 4ac
2a
Examples:
Solve 3x2 – 4x + 1 = 0
a = 3, b = -4, c = 1
therefore we get
x=
− (−4) ± (−4) 2 − 4(3)(1)
2(3)
x=
4 ± 16 −12
6
x=
4± 4
6
x=
4 ± 2
6
x=
4+2
4−2
or x =
6
6
x = 1 or x =
1
3
Solve 2x2 + 3x – 1 = 0
a = 2, b = 3, c = -1
− 3 ± 3 2 − 4(2)(−1)
x=
2(2)
x=
− 3 ± 9 + 8
4
x=
− 3 ± 17
4
x=
− 3 + 17
− 3 − 17
or x =
4
4
Complex Numbers
Most of the quadratic equations you will need to solve will have solutions in the
real number system. But there are quadratic equations that do not have solutions in the
real number system.
Example:
x2 +1 = 0
x 2 = −1
x = −1
−1 has no meaning in the real number system.
To solve problems of this type, we introduce a new number system, the complex
number system.
Complex Number System
Define −1 = i, so that we get x2 = -1
We then define a number a + bi where a, b are real numbers and i is defined as
above.
We can now define the square root of negative numbers in the complex number
system.
− 9 = 9 −1 = 3i
− 121 = 121 −1 = 11i
Examples of complex numbers
5 + 3i, 2 – 7i, 8, 4i are all complex numbers
Addition and Subtraction of Complex Numbers
The addition and subtraction of complex numbers follows the same rules as we
use in algebra (combining “like” terms). Add and/or subtract the real parts and the
complex parts separately.
Examples:
(2 + 7i) + ( 3 – 4i) = (2 + 3) + (7 – 4)i = 5 + 3i
(6 + 5i) – (3 – 7i) = (6 – 3) + (5 – (-7)i = 3 + 12i
Multiplication of Complex Numbers
The multiplication of complex numbers follows the same rules of algebraic
multiplication, using either the distributive property or FOIL.
Examples:
(2 + 3i)(1 + 7i)
2(1) + 2(7i) + 3i(1) + (3i)(7i)
2 + 14i + 3i + 21i2
note: i2 = -1
2 + 14i + 3i – 21
-19 + 17i
(4 – i)(3+ 2i)
4(3) + 4(2i) – i(3) – i(2i)
12 + 8i – 3i + 2
14 + 5i
note: i2 = -1
Division of Complex Numbers
Division of complex numbers follows the same sort of rules as the multiplication
of radicals (multiplication by a conjugate to get the i out of the denominator).
(a + bi) and (a – bi) and complex conjugates
(a + bi)(a – bi) = a2 –abi + abi – (bi)2 = a2 + b2 note: real number
Examples:
1
1
2 − 3i 2 − 3i 2 3
=
⋅
=
= − i
2 + 3i 2 + 3i 2 − 3i
13
13 13
(2 + 3i)(2 − 3i) = 2 2 + 3 2 = 13
(1 + i)(3 − 4i) = 3 − 4i + 3i − 4i 2 = 7 − i
1 + 4i 1 + 4i 3 − 4i 7 − i 7
1
=
⋅
=
=
− i
3 + 4i 3 + 4i 3 − 4i
25
25 25
(3 + 4i)(3 − 4i) = 3 2 + 4 2 = 25
Complex Solutions to Quadratic Equations
If the discriminant of a quadratic equation (b2 –4ac) is less than zero, then the
quadratic equation ax2 + bx + c = 0 has solutions only in the complex number system.
Examples:
Solve x2+ x + 1 = 0, a = 1, b = 1, c = 1
By the quadratic formula
−1 ± 12 − 4(1)(1) −1± 1− 4 −1± − 3 −1 ± i 3
x=
=
=
=
2(1)
2
2
2
x=
−1
3
−1
3
+
i or x =
−
i
2
2
2
2
Solve 2x2 – 4x + 3 = 0, a = 2, b = -4, c = 3
By the quadratic formula
− (−4) ± (−4) 2 − 4(2)(3) 4 ± 16 − 24 4 ± − 8 4 ± 2i 2 2(2 ± i 2)
x=
=
=
=
=
2(2)
4
4
4
4
x=
2 ± i 2
2
x = 1+
2
2
i or x = 1 −
i
2
2
Common Roots and Radicals
Square Root
If a is a real number, we say that the real number b is the square root of a
2
if b = a. Usually written as a = b
Examples:
42 = 16 and (-4)2 = 16 all real numbers have two square roots
(+ and -). We generally refer to the positive square root as the principle
root. Therefore, we usually say 16 = 4 .
Cube Root
If a is a real number , we that the real number b is the cube(third) root of a
3
if b = a. Usually written 3 a = b .
Examples:
2 3 = 8 so we say
3
8
= 2
also notice (−2) = −8 so
3
3
− 8 = −2
Nth Root
In general if bn = a then we say n a = b . This symbolism may be new, so
lets look at the format.
n
Index
a
Radicand
Radical
Principal Roots
1 = 1
4 = 2
3
1 =1
3
8 =2
4
1 =1
4
25 = 5
27 = 3
16 = 2
5
3
1 =1
81 = 3
5
36 = 6
64 = 4
4
3
32 = 2
4
49 = 7
125 = 5
256 = 4
5
3
243 = 3
4
64 = 8
3
625 = 5
9 =3
16 = 4
81 = 9
100 = 10
216 = 6
Examples:
− 121 = −11
3
− 64 = −4
(−4) 3 = −64
− 5 − 32 = −(−2) = 2
− 25 = ?
Negative numbers can not have even roots, since any real number
to an even power is positive
Variables in the Radicand
x2 = x
4
16t 4 = 4 16 ⋅ 4 t 4 = 4t
− 3 y6 = − y2
( y 2 )3 = y 6
Approximation of Square Roots
Many numbers will not have whole number square roots. Your calculator
will help you find an approximation to the root.
7 = 2.6457573...
13 = 3.6055513...
Properties of Radicals
We know 36 = 6 but we can also evaluate it this way
36 = 4 ⋅ 9 = 4 ⋅ 9 = 2 ⋅ 3 = 6 , from here we can see a basic property of roots.
Roots of Products Property
n
ab = n a ⋅ n b
This property is very useful in simplifying radical expressions.
To simplify radical expressions
Write the expression under the radical as the product of two more terms
that can be more easily dealt with and then apply the Roots of Products Property.
Examples:
75 = 25 ⋅ 3 = 25 ⋅ 3 = 5 3
18 = 9 ⋅ 2 = 9 ⋅ 2 = 3 2
3
16 = 3 8 ⋅ 2 = 3 8 ⋅ 3 2 = 23 2
If there are variables in the expression, I find it easier to separate the
variables from the constants and deal with each part separately.
To deal with the roots of variable expression, it is easier to perform a
simple 4th grade division with remainders.
t 32 , divide 32 ÷ 3 = 10 R2 (10 becomes the power outside the
radical and 2 becomes the power inside the radical), therefore we
3
get
3
t 32 = t 10 ⋅ 3 t
2
Examples:
9x 5 = 9 x 5 = 3x 1 x 2
3
16t 4 u 5 = 3 16 ⋅ 3 t 4 ⋅ 3 u 5 = 23 2 ⋅ t 3 t ⋅ u 3 u 2 = 2tu ⋅ 3 2tu 2
3
x 7 y 9 z 16 , break this up
3
x7 = x2 ⋅ 3 x
3
y9 = y3
3
z 16 = z 5 ⋅ 3 z
and get x 2 y 3 z 5 ⋅ 3 xz
Roots of Quotients Property
n
a
=
b
n
a
n
b
, you can handle the numerator and denominator separately
Examples:
25
25 5
=
=
16
16 4
5
11
=
49
11
7
=
x2
7
49
x
2
=
11
7
=
7
x
5
5
19
19
19
=
=
10
5
32c
2c 2
32c 10
Addition and Subtraction of Radicals
We add and subtract radicals in the same way that we add and subtract algebraic
expressions, we add “like” terms. Two or more radical expressions are said to be “like”
terms if they have the same indices and radicands.
7 & 3 7 are like terms
7, 3 7 & 11 are not like terms
Examples:
3 2 + 2 2 − 2
(3 + 2 −1) 2
4 2
73 5 + 33 5 + 3 5 − 5
(7 + 3 + 1)3 5 −
5
Note: the last term is not like the others
113 5 − 5
Occasionally, we can take expression that do not look “like” and do a
transformation and make them “like”. (Just like a magic trick)
8 & 50 do not appear to be “like” terms
8 = 4⋅2 = 4 ⋅ 2 = 2 2
but
so they are really “like” terms
50 = 25 ⋅ 2 = 25 ⋅ 2 = 5 2
Examples:
3 50 + 7 8
( ) ( )
35 2 +7 2 2
15 2 + 14 2
29 2
23 16 + 54
43 2 + 33 2
3
7 2
23 16 = 23 8 ⋅ 2 = 23 8 3 2 = 2 ⋅ 23 2 = 43 2
3
54 = 3 27 ⋅ 2 = 3 27 3 2 = 33 2
4 12 + 5 8 − 50
4 12 = 4 4 ⋅ 3 = 4 ⋅ 2 3 = 8 3
8 3 +10 2 − 5 2
5 8 = 5 4 ⋅ 2 = 5 ⋅ 2 2 = 10 2
8 3 +5 2
50 = 25 ⋅ 2 = 5 2
Like Terms
An example from algebra
7 u 3 + u 27u − 5 3u 2
7 u 3 = 7 ⋅ 3 ⋅ u 3 = 7u 3u
7u 3u + 3u 3u − 5u 3
u 27u = u 9 3 u = 3u 3u
10u 3u − 5u 3
5 3u 2 = 5 3 u 2 = 5u 3
Like Terms
Multiplication of Radicals
Multiplication of radicals is fairly straight forward, using the following rule
n
a n b = n ab
Examples:
18 2 = 36 = 6
3
3c 2 d 5 • 3 9c 4 d 7
3
3⋅ 9 • 3 c2 ⋅ c4 • 3 d 5 ⋅ d 7
3
27 • 3 c 6 • 3 d 12
3c 2 d 4
(
3 5+ 2
)
3 5 + 3 2
15 + 6
( 10 + 2 )( 10 − 2 )
10 10 + 10 2 − 10 2 − 2 2
100 − 4
10 − 2
8
Note the use of the Distributive Property
( x + y )(
x + 2y
)
x x + y x + 2y x + y ⋅ 2 y
x 2 + 3y x + 2 y 2
x + 3y x + 2y 2
Division of Radicals
Division of radicals follow the same kind of rules as multiplication
n
a n a
=
,b≠0
n
b
b
Examples:
27
3
27
= 9 =3
3
=
75x 3
3x
3
32u 7
3
4uv 3
=
75x 3
= 25x 2 = 5x
3x
=3
32u 7 3 8u 6 2u 2
=
=
4uv 3
v
v3
On occasion, you will find a problem leaves you with a radical in the
2
denominator, i.e.
. It is generally easier to work with the fraction if there are no
7
radicals in the denominator. We use a process called “rationalizing the denominator” to
free the denominator of radicals.
Rationalizing the denominator
Rationalizing the denominator utilizes the following two rules (that we already
know)
Rule1
Rule 2
(
a ⋅ a = a2 = a
)(
a+ b
)
Radical free
a − b = a−b
Examples:
1.
2.
2
3
=
2
3
⋅
3
8
=
10
3
=
2 3
9
8
=
10
=
2 3
3
4
=
5
4
5
=
4
5
⋅
5
5
=
2 5
5
3.
6
3+ 5
4.
=
6
3+ 5
⋅
3− 5
3− 5
=
(
) (
)
(
6 3− 5
6 3− 5
=
= −3 3 − 5
3−5
−2
(
)
(
)
)
x
x
2 x −1
x 2 x −1
x 2 x −1
=
⋅
= 2
=
2
2
4x −1
2 x + 1 2 x + 1 2 x −1 2 ( x) −1
Note: It is not wrong to leave a radical in the denominator (the math police will not
come and arrest you, if you do), but most teachers prefer to see the answers in this
form.
Radical Equations
Radicals appear in a wide variety of applications, so we need to spend time
learning to solve radical equations.
Solving Radical Equations
There are several steps to follow in solving radical equations
i. Rewrite the equation so that the radical is on one side of the equal
sign and everything else is on the other side
ii. Square both sides of the equation (eliminates the radical)
iii. Solve remaining equation
iv. Check your solution
Examples:
x =3
2
x = 32
x=9
check: 9 = 3
1 + 2t + 5 = 4
(
2t + 5 = 3
)
2
2t + 5 = 3 2
2t + 5 = 9
2t = 4
t=2
check: 1 + 2(2 ) + 5 = 1 + 4 + 5 = 1+ 9 = 1+ 3 = 4
Checking your solution in this type of problem is especially important. The act of
squaring to remove the radical causes problems. Solutions to the equation that we
created by squaring may not be solutions to the equation we had before squaring
(extraneous solution)
Example:
w −1 + w = 3
(
w −1 = 3 − w
w −1
)
2
= (3 − w )
2
w −1 = 9 − 6w + w 2 .
w 2 − 7w + 10 = 0
(w − 2)(w − 5) = 0
(w − 2) = 0 ⇒ w = 2
(w − 5) = 0 ⇒ w = 5
Check w = 2
2 −1 + 2
Check w = 5
5 −1 + 5
1+2
1+ 2
3
TRUE
4 +5
2+5
7
NOT TRUE
Translating English in Algebra
The hardest part of solving word problems, is the translation of what was written
into the algebra needed to solve the problem. Once we get the algebra equation set up it
is usually easy to solve the algebra.
The first important issue in translating the problem is to read the problem
carefully enough to figure out what the author is asking for. This usually requires more
than one reading.
Once you understand what information is needed then you need to look at the
problem to find a relationship between what is known and the unknown. Many times
there will be keywords in the problem to help you find that relationship. Listed below is
a partial list of the normal keywords (this is by no means an exhaustive list).
Addition
increased by
more than
combined, together
total of
sum
added to
Subtraction
decreased by
minus, less
difference between/of
less than, fewer than
Multiplication
of
times, multiplied by
product of
increased/decreased by
a
factor of
Division
per, a
out of
ratio of, quotient of
percent (divide by 100)
Equals
is, are, was, were, will
be
gives, yields
sold for
After knowing the relationship you can set up a word equation that represents the
problem. My age + her age*3 = his age . And then you can change the words into
variables and solve.
If the problem allows for it, one of the best ways to set up a relationship for a
problem could be to draw a picture of the problem (this is not appropriate for all
problems but it is good for some types)
Number Problems
There is no more common or more useless type of word problem than the
standard number problem. It is important to learn how to solve them for no other reason
than to learn to manipulate the algebra and put the words into algebra.
Ex.
Find two consecutive integers whose sum is 99
Let the smaller number = x
The next number is x + 1 (consecutive integers are 1 unit apart)
Sum implies addition
x + (x +1) = 99
2x + 1 = 99
2x = 98
x = 49
x + 1 =50
Ex.
Find the number such that 3 times the number + ½ of the number is
14
Let x = the number
3x + ½ x = 14
3.5x = 14
x = 4
Age Problems
Problems involving the age of various people are always interesting. They occur in all
algebra books and have family relationships.
Ex.
Jerry is 10 times older than his daughter Carrie, but in 5 years he will be 5
times as old as Carrie. How old are they today?
Since Carrie is the youngest, let’s allow x to be her age now
Then Jerry’s age now is 10x
In five years, Carries age is x + 5 and Jerry’s age is 10x + 5
We know in 5 years Jerry’s age (10x + 5) is 5 times Carrie’s age (x+5) so
we write
10x + 5 = 5(x + 5) and solve
10x + 5 = 5x + 25
subtract 5 from both sides
10x = 5x + 20
subtract 5x from both side
5x = 20
divide by 5
x=4
today Carrie is 4 years old and Jerry is 40
years old (10x)
Money Problems
Money problems are always interesting, since we all like to work with money.
They generally involve a certain number of coins that total to specific amount.
You have to set up two relationships in this type of problem (one for the number
of coins and one for the value of the coins.
Ex.
Joe has a handful of coins (nickels, dimes, and quarters). He has a total of
$1.20. He has 4 times as many dimes as he does nickels and 3 times as many quarters as
he does nickels. How many of each type of coin does Joe have?
Since everything is based on the number of nickels, allow x = number of
nickels
Coin
How Many
Value
Nickels
x
.05x
Dimes
4x
.10(4x) = .4x
Quarters
3x
.25(3x) = .75x
We know that Joe has $1.20 in total, so we set up a total value equation
Nickels + Dimes + Quarters = $1.20
.05x + .4x + .75x = $1.20
1.20x = 1.20
x=1
so we have 1 nickel , 4(1) = 4 dimes and 3(1) = 3 quarters
Work Problems
Probably, one of the more confusing types of word problems is the problems
involving two people doing one job. In this type of problem, you generally know how
long it takes each of the people to complete the job alone and are asked to determine the
time it will take to complete the job together.
In this type of problem, it is helpful to look at how much of the job each can do in
one hour(minute) and determine the number of hours(minutes) x it takes together.
Ex.
It takes Tom 3 hours to plow a mall parking lot and it take Joe 4 hours to
plow the same lot. How long will it take them to plow the lot together(assuming they
don’t stop for coffee and to talk)?
Let x = the number of hours it takes to complete the job
Tom completes the job in 3 hours so he completes 1/3 in 1 hour
Tom completes x/3 of the job in x hours
Joe completes the job in 4 hours so he completes ¼ in 1 hour
Joe completes x/4 of the job in x hours
Together they perform 1 job (together = addition)
x x
+ =1
multiply both sides by 12 to eliminate the fractions
3 4
4x + 3x = 12
7x = 12
x = 12/7 hours
remember to include the units for the answer
Mixture Problems
Mixture problems are a real world type of problem that can be useful. Generally
you are given two type of items that need to be mixed together to form a third item.
Ex.
You need to have 10 liters of 25% acid for an experiment in Chemistry
Class, but all you have is bottles of 10% acid and 30% acid. How much
each type do you need to combine to make the type you need?
Allow x = the amount of 10% acid needed
Since you need a total of 10 liters 10 – x = amount of 30% acid
Amount of acid in 10% = .10x
Amount of acid in 30% = .30(10 – x) = 3 - .3x
Amount of acid in final = .25 (10) = 2.5
By Jerry’s Law of the Conservation of Acid, there was the same amount of
acid in the mixture before they were added as there was after the added
10% acid + 30% acid = total Acid
.10x
+ (3 - .3x) = 2.5
-.2x + 3 = 2.5
-.2x = -.5
x = .25 liters of 10% acid
10 – x = 9.75 liters of 30% acid
Distance Problems
Problems involving distances traveled will use one from of the formula relating
distance, rate and time
(d = rt, t = d/r, r =d/t) . Generally there will something in the problem that is equal in the
problem (they traveled the same distance, time, or at the same rate)
Ex.
Joe and Tom each went to a party. . Joe got the in 4 hrs and Tom got the in
5 hrs. Joe averaged 15 mph more than Tom. What was the average speed
of each? How far away was the party?
Since Tom was the slowest let his rate be x so Joe’s rate is x + 15
Tom
d = rt
d = x(5)
Joe
d = rt
d = (x + 15)4 = 4x + 60
Since they each traveled the same distance we can set their distances
equal.
5x = 4x + 60 solve
x = 60
Tom’s speed
x + 15 = 75 Joe’s speed
Distance to the party, using Tom’s eauation = 60(5) = 300 miles