**Unformatted text preview: **Math 0980
Explanations & Examples
(updated Summer 2018) Table of Contents
Module 1 – Linear Equations with One Variable
Module 1 Study Skills (Where Am I?)……………………………………………………………………………….….1
Module 1 Reality Check (Analyzing the Formula: D=RT)……………………………………………………..2
Module 1 Vocabulary………………………………………………………………………………………………………….3
1.1 Solving One-Step Linear Equations………………………………………………………………………………4
1.2 Solving Two-Step Linear Equations………………………………………………………………………………6
1.3 Solving Linear Equations – Like Terms & The Distributive Property……………………………..8
1.4 Solving Linear Equations with Variables on Both Sides………………………………………………10
1.5 Solving Linear Equations Containing Fractions…………………………………………………………..11
1.6 Solving Linear Equations Containing Decimals…………………………………………………………..14
1.7 Applications Using Linear Equations………………………………………………………………………….16
1.8 Applications – Variation and Percents……………………………………………………………………….21
1.9 Linear Equations – Formulas……………………………………………………………………………………..24
Module 2 – Linear Inequalities and Absolute Value
Module 2 Study Skills (Help Me!)……………………………………………………………………….…………………………29
Module 2 Reality Check (An Unfair Coin)…………………………………………………………………………..30
Module 2 Vocabulary………………………………………………………………………………………………………..31
2.1 Solve and Graph Linear Inequalities…………………………………………………………………………..32
2.2 Applications Using Linear Inequalities……………………………………………………………………….34
2.3 Sets and Compound Inequalities (Disjunction)…………………………………………………………..36
2.4 Compound Inequalities (Conjunction)……………………………………………………………………….39
2.5 Solve Absolute Value Equations and Inequalities..……………………………………………………..42
2.6 Introduction To Functions………………………………………………………………………………………… 48 Module 3 – Linear Equations With Two Variables
Module 3 Study Skills (Time Tracker)…………………………………………………………………………………51
Module 3 Reality Check (Analyzing the Average Rate of Change)………………………………………54
Module 3 Vocabulary………………………………………………………………………………………………………..55
3.1 Describing Data………………………………………………………………………………………………………….57
3.2 Rectangular Coordinate System and Ordered Pairs……..…………………………………………….61
3.3 Graph Linear Equations………………………………………………………………………………………………66
3.4 Graph Linear Inequalities in Two Variables..…………………..………………………………………….71
3.5 Slope of a Line.……………….………………………………………………..………………………………………..76
3.6 Rate of Change….………………………………………………………………..……………………………………..82
3.7 Write a Linear Equation..………………..……………………………………..…………………………………..84
3.8 Approximate Linear Relationships………………………………………………………………………………89
Module 4 – Systems of Linear Equations
Module 4 Study Skills (Organization)…………………………………………………………………………………94
Module 4 Reality Check (Bringing the NBA Into Math 0980)……………………………………………..95
Module 4 Vocabulary………………………………………………………………………………………………………..96
4.1 Solve Systems of Equations by Graphing……………………………………………………………………97
4.2 Solve Systems of Equations by the Substitution Method……………………………………….…102
4.3 Solve Systems of Equations by the Addition/Elimination Method…………………………….107
4.4 Applications Using Systems of Linear Equations…………………………………………………….…114
Module 5 – Exponents, Radicals, and Polynomials
Module 5 Study Skills (Yes, I Can!)………………………………………………….………………………………..120
Module 5 Reality Check (Tsunamis)………………………………………………………………………………….121
Module 5 Vocabulary…………………………………………………………………………………………………….…122
5.1 Exponent Properties………………………………………………………………………………………………...123
5.2 Exponent Properties Used Together…………………………………………………………….……..……128 5.3 Scientific Notation ……………………………………………………………………..……………………………131
5.4 Introduction to Radicals………………………………………………………………..………………………….135
5.5 Simplify Radicals…..…………………………………………………………………………………………….…….139
5.6 Applications – The Pythagorean Theorem…………………………………………………………..……142
5.7 Introduction to Polynomials and Multiplication………………………………………………………..145
5.8 Introduction to Factoring Polynomials………………………………………………………………………149
Module 6 – Exponential and Logarithmic Functions
Module 6 Study Skills (It Ain’t Over)………………………………………………………………….……………..154
Module 6 Reality Check (Retirement Calculations)…………………………………………….…………….157
Module 6 Vocabulary……………………………………………………………………………………………………….158
6.1 Exponential Functions………………………………………………………………………………………………159
6.2 Applications Using Exponential Functions………………………………………………………………..163
6.3 Logarithmic Functions………………………………………………………………………………………………167
6.4 Common Logarithms………………………………………………………………………………………………..171
6.5 Base e and Natural Logarithms………………………………………………………………………………...173
6.6 Solve Exponential Equations…………………………………………………………………………………….177
6.7 Solve Logarithmic Equations…………………………………………………………………………………….179
6.8 Applications Using Logarithmic Functions……………………………………………………………..182 Name _________________________ Where Am I?
Course Information:
Course Title _____________________________________
Section Number __________________________________
My Professor/Instructor:
Name __________________________________________
Office Hours ____________________________________
____________________________________
Office Location __________________________________
Phone Number ___________________________________
Email Address _________________________________________________
Supplies:
Textbook _____________________________________________________
Other Supplies I Need ___________________________________________
_____________________________________________________________ Spend 5 minutes talking to a fellow classmate. Complete the following:
Classmate’s Name ____________________________________
Something interesting that my classmate told me about himself/herself
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________ Analyzing the Formula D = RT: An Application of Linear Equations Name _______________________________________
One application of linear equations is the concept of the relationship between the distance traveled, the rate of travel, and
the time traveled. If you assume that the rate is a constant, the relationship becomes the formula: = ∙ 1. To estimate your average speed (rate) from home to either school or work, do the following:
A. Estimate the distance in miles you travel from home to either school or work:
D=__________miles
B. Next, estimate the time, in hours, you travel from home to either school or work. If your estimate is in minutes,
convert the minutes to hours: T=__________hr
C. Using the distance formula given above and your estimates in A and B, calculate your estimated average speed.
Round your answer to the nearest tenth, if rounding is necessary.
Write the resulting equation and solve for R in the space provided here: R=__________mph.
2. If you needed to get there in 0.8 hours (48 minutes), how fast would you have to go? Using your estimated distance from
A, calculate the average speed you would need to go. Round your answer to the nearest tenth, if rounding is necessary.
Write the resulting equation and solve for R in the space provided here: R=___________mph.
3. If you were traveling 33 mph, how long would it take? Using your estimated distance from A, calculate your time in hours.
Round your answer to the nearest hundredth, if rounding is necessary.
Write the resulting equation and solve for T in the space provided here: T=__________hr. Module 1 Vocabulary
Area – The size of the surface of a shape measured in square units.
Area of a Rectangle – The measurement of the surface of a rectangle and is found
by multiplying the length L and width W.
Area = L × W
Coefficient – A number used to multiply a variable.
Consecutive integers – Integers that follow each other in order.
x, x+1, x+2, x+3, …
Consecutive even integers – Even integers that follow each other in order.
x, x+2, x+4, …
Consecutive odd integers – Odd integers that follow each other in order.
x, x+2, x+4, …
Distributive property – Allows one to multiply each number or term of a sum by a
specific number/term.
a(b + c) = ab + ac
Equation – A statement that two expressions are the same.
Evaluate – Substituting a given value for a variable(s) and simplifying the
expression.
Inequality – A statement that compares the quantities of two expressions.
Like terms – Terms having the same variable(s) raised to the same exponent.
Perimeter – The distance around an object or shape, found by adding the
measurement of sides of the object or shape.
Proportion – A fraction equal to a fraction.
multiplication.
Variable – An unknown value. Commonly solved using cross Solving One-Step Linear Equations
Solving linear equations is an important and fundamental skill in algebra. Solving an equation
means that we are finding the value of the variable, the unknown number. While the variable
can be any letter of the alphabet, x is commonly used.
To solve linear equations, the general rule is to do the opposite. The opposite of addition is
subtraction. The opposite of subtraction is addition. The opposite of multiplication is division.
The opposite of division is multiplication. When we do the opposite to one side of the equation,
we must do the same to the other side of the equation to maintain equality. Example 1: x+7= -5
- 7 - 7 Subtract 7 from both sides
x
= -12 Example 2: x–5= 4
+5 +5
x
= 9 Example 3: Example 4: Example 5: Add 5 to both sides -6+x=-2
+6
+6
x= 4 4x = 20
4
4
x= 5 - 5x = 30
-5
-5
x=-6 Add 6 to both sides Divide both sides by 4 Divide both sides by -5 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
4 Example 6: 5 =-3 (5) 5 = - 3 (5) Multiply both sides by 5 x = - 15 Example 7: −7 =-2 (- 7) −7 = - 2 (- 7) Multiply both sides by -7 x = 14 Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons
Attribution 3.0 Unported License. Modified by Carla Kulinsky.
5 Solving Two-Step Linear Equations
As we solve two-step equations, the important thing to remember is that we do the opposite to both
sides of the equation. With each step, we get closer to getting the variable by itself, basically
working backwards. This means we will add or subtract first, then multiply or divide second. Example 1: 4x – 20 = - 8
+ 20 = +20
4x
= 12
4
4
x
= 3 Add 20 to both sides
Divide both sides by 4 To check our answer: 4(3) – 20 = - 8
12 – 20 = - 8
- 8=-8 Example 2: 5x + 7 = 7
-7 -7
5x
=0
5
5
x
=0 Subtract 7 from both sides
Divide both sides by 5 To check our answer: 5(0) + 7 = 7
0+7=7
7=7 A common error students make with two-step equations is with negative signs. The sign always
stays with the number, sometimes resulting in a negative number. Example 3: 4 – 2x = 10
-4
-4
- 2x = 6
-2 -2
x=-3 Subtract 4 from both sides
Divide both sides by -2 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
6 Example 4: 8–x= 2
-8
-8
-x=-6
- 1x = - 6
-1 -1
x= 6 Subtract 8 from both sides
No number in front of the variable means 1
Divide both sides by -1 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
7 Solving Linear Equations by Combining Like Terms
And/Or Using The Distributive Property
Often as we are solving linear equations, we will need to do some work to set them up into a form
we are familiar with solving. This section will focus on manipulating an equation we are asked to
solve in such a way that we can use our pattern for solving two-step equations to ultimately arrive
at the solution.
One such issue that needs to be addressed is combining like terms. Example 1: Example 2: Example 3: -x – 4 = -8 + 3
-x – 4 = -5
+4 +4
-x
= -1
-1
-1
x
= 1 6x – 6 – 9x
-3x – 6
+6
-3x
-3
x = 9
= 9
+6
= 15
-3
= -5 Combine -8 + 3
Add 4 to both sides
Divide both sides by -1 Combine 6x – 9x
Add 6 to both sides
Divide both sides by -3 3x + 1 + 2x + 3 = 6 + 3
5x + 4 = 9
-4 -4
5x
=5
5
5
x
=1 Combine 3x+2x and 1+3
Subtract 4 from both sides
Divide both sides by 5 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
8 Another issue to be addressed is parentheses. The first step is to get rid of the parentheses by
using the Distributive Property. Example 4: Example 5: 4(2x – 6) = 16
8x – 24 = 16
+ 24 + 24
8x
= 40
8
8
x
= 5 3(2x – 4) + 9 = 15
6x – 12 + 9 = 15
6x – 3 = 15
+3 +3
6x
= 18
6
6
x
= 3 Distribute 4 through parentheses
Add 24 to both sides
Divide both sides by 8 Distribute 3 through parentheses
Combine -12 + 9
Add 3 to both sides
Divide both sides by 6 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
9 Solving Linear Equations with Variables on Both Sides
In this section, the equations have variable terms on both sides of the equation. In addition, some
examples will require the Distributive Property and some will require combining like terms. To
get the variable terms together, we add or subtract one of the variable terms to the other side. To
avoid a negative coefficient, move the smaller term. Example 1: 4x – 6 = 2x + 10
- 2x
- 2x
2x – 6 = 10
+6 +6
2x
= 16
2
2
x
=8 To check our answer: Subtract 2x from both sides
Add 6 to both sides
Divide both sides by 2 4(8) – 6 = 2(8) + 10
32 – 6 = 16 + 10
26 = 26 Example 2: x + 18 = 2(x – 5) + 3x
x + 18 = 2x – 10 + 3x
x + 18 = 5x – 10
-x
-x
18 = 4x – 10
+ 10
+ 10
28 = 4x
4
4
7=x Distribute the 2 through parentheses
Combine 2x + 3x
Subtract x from both sides
Add 10 to both sides
Divide both sides by 4 Example 3: 4(2x – 6) + 9 = 3(x – 7) + 8x Distribute 4 and 3
8x – 24 + 9 = 3x – 21 + 8x Combine -24+9 and 3x+8x
8x – 15 = 11x – 21
- 8x
- 8x
Subtract 8x from both sides
- 15 = 3x - 21
+ 21
+ 21
Add 21 to both sides
6 = 3x
3 3
Divide both sides by 3
2=x
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
10 Solving Linear Equations Containing Fractions
Some linear equations contain fractions. First, we find the least common denominator (LCD) and
clear the fractions by multiplying each term by the LCD. 3 −
4 Example 1:
12 3 7 5 =
2 The LCD = 12 6 12 7 5 (1 )4 − (1 )2 = 12 (1) 6 9x – 42 = 10
+ 42 + 42
9x
= 52
9
9 Example 2:
6 2 2 Divide both sides by 9 9 3 1 +
2 6 Add 42 to both sides 52 = −2=
3 12 Multiply each term by 12 ( 1 ) The LCD = 6 6 6 3 6 1 6 (1) 3 − (1) 2 = (1) 2 + (1) 6 Multiply each term by 6 (1) 4x – 12 = 9x + 1
- 4x
- 4x
- 12 = 5x + 1
-1
-1
-13 = 5x
5
5
− 13
5 To check our answer: Subtract 4x from both sides
Subtract 1 from both sides
Divide both sides by 5 =x 2
3 (− 13 3 ) − 2 = 2 (− 5
26 13
5 39 − 15 − 2 = − 10 +
26 − 15 − 30 117 56 30
112 56 30
56 = −
15 − 15 = − 1 )+6 + 1
6
5
30 − 15 = − 15 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
11 If the equation contains parentheses, distribute the coefficient in front of the parentheses first, and
then clear the fractions.
3 5 4 2 9 27 ( + Example 3: 5
6 + 18 5 )=
2 9
18 2 = 32 3 Distribute 2 through parentheses 9
32
9
18 32 (1 )6 + (1 )9 = (1 )
15x + 4 = 64
-4 -4
15x
= 60
15
15
x
= 4 9 The LCD = 18
18 Multiply each term by 18 ( 1 )
Subtract 4 from both sides
Divide both sides by 15 When a fraction is equal to a fraction, the equation is often referred to as a proportion.
3 Example 4: 4
3 15 = 2x
= 15 The LCD = 4 ∙ 2x = 8x 4 2x
8x 3 15 8x ( 1 ) 4 = 2x ( 1 ) Multiply each term by 8x 6x = 60
x = 10 Example 5: Divide both sides by 6 x 2 x 2 x 2 5(x+10) =
x + 10 5
=
x + 10 5 5(x+10) ( 1 ) x + 10 = 5 ( The LCD = 5(x + 10)
1 ) 5x = 2(x + 10)
5x = 2x + 20
–2x –2x
3x =
20
3
3
x= Multiply each term by 5(x + 10)
Distribute 2
Subtract 2x from both sides
Divide both sides by 3 20
3 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
12 Notice that a shortcut would be to use a technique called cross multiplication or cross products.
Given the proportion:
cross multiply a c =
b d a c b = d which results in ad = bc
where a and d are called the extremes, and b and c are called the means Example 6: 3
2
3
2 =
= 24
n
24 Cross multiply: 3 ∙ n and 2 ∙ 24 n 3n = 48
3
3
n = 16 Example 7: x+8
4
x+8
4 =
= Divide both sides by 3 x−3
5
x−3
5 5(x + 8) = 4(x – 3)
5x + 40 = 4x – 12
-4x
-4x
x + 40 =
– 12
-40
-40
x
=
–52 Cross multiply: 5 ∙ (x + 8) and 4 ∙ (x – 3)
Distribute 5 and distribute 4
Subtract 4x from both sides
Subtract 40 from both sides Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License. Modified by Carla Kulinsky.
13 Solving Linear Equations Containing Decimals
Recall that a fraction can be converted into a decimal. Thus, the process for clearing decimals is
similar to the process for clearing fractions. Rather than multiplying each term by the LCD, we
multiply each term by the appropriate power of ten – 10, 100, 100, and so on – that will convert
all of the decimals to a whole number.
Multiply by
10
100
100
1000 Move decimal
1 place to the right
2 places to the right
2 places to the right
3 places to the right Example
3.4 × 10 = 34
5.21 × 100 = 521
7.2 × 100 = 720
6.09 ×1000 = 6090 **Some students feel comfortable working with decimals. If you do not mind working with
decimals, you do not need to clear them** Example 1: 2.5x + 3.4 = 6
(10) 2.5x + (10) 3.4 = 6 (10)
25x + 34 = 60
- 34 - 34
x
= 26
25
25
x
= 1.04 Multiply each term by 10
Subtract 34 from both sides
Dvide both sides by 25 Example 2:
2x + 18.2 = 2.8x + 1.48
(100) 2x + (100) 18.2 = (100) 2.8x + (100) 1.48 Multiply by 100
200x + 1820 = 280x + 148
- 200x
- 200x
Subtract 200x from both sides
1820 = 80x + 148
- 148
- 148
Subtract 148 from both sides
1672 = 80x
80
80
Divide both sides by 80
20.9 = x Carla Kulinsky, Salt Lake Community College, 2014 14 Example 3: 0.5(x + 3) = 6.83
Distribute 0.5 through parentheses
0.5x + 1.5 = 6.83
(100) 0.5x + (100) 1.5 = (100) 6.83
Multiply by 100
50x + 150 = 683
- 150 -150
Subtract 150 from both sides
50x
= 533
50
50
Divide both sides by 50
x
= 10.66 To check our answer: 0.5(10.66 + 3) = 6.83
0.5(13.66) = 6.83
6.83 = 6.83 Carla Kulinsky, Salt Lake Community College, 2014 15 Applications Using Linear Equations
This section will discuss several types of applications. No matter how simple or complex the
application is, the following steps will help to translate and solve the problem.
1.
2.
3.
4.
5. Read through the entire problem
Organize the information
Write the equation
Solve the equation
Check the answer Direct Translation
With this type of application, you will translate word to symbol. Look for key words and phrases.
The following table gives some common translations. Word or Phrase
is
the same as
sum
more than
difference
less than
product
of Symbol
=
=
+
+
×
× Example 1:
The sum of four and a number is ten.
The sum of four and a number is ten.
4 + x = 10
-4
-4
x= 6 Subtract 4 from both sides Check: The sum of 4 and 6 is 10.
Or 4+6=10. Modified from Prealgebra Textbook, by the Department of Mathematics, College of the Redwoods, CCBY 2009. Licensed under a Creative Commons Attribution 3.0 Unported License.
Modified by Carla Kulinsky.
16 Example 2:
If 28 less than five times a number is 232. What is the number?
If 28 less than five times a number is 232.
5x – 28 = 232
+ 28 + 28 Add 28 to both sides
5x
= 260
5
5 Divide both sides ...

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- Elementary algebra, Negative and non-negative numbers, Carla Kulinsky