Equations and Inequalities in One Variable

1
Equations and Inequalities in One Variable

• Chapter 2

2
Chapter Sections
2.1 Linear Equations The Addition and
Multiplication Properties of Equality 2.2
Linear Equations Using the Properties
Together 2.3 Solving Linear Equations Involving
Fractions and Decimals Classifying Equations 2.4
Evaluating Formulas and Solving Formulas for a
Variable 2.5 Introduction to Problem Solving
Direct Translation Problems 2.6 Problem
Solving Direct Translation Problems Involving
Percent 2.7 Problem Solving Geometry and
Uniform Motion Problems 2.8 Solving
Inequalities in One Variable
3
Linear Equations The Addition and
Multiplication Properties of Equality

• Section 2.1

4
Linear Equations in One Variable
A linear equation in one variable is an equation
that can be written in the form ax b c where
a b and c are real numbers and a 0.
3x 5 25
The expressions are called the sides of the
equation.
5
Solutions
The solution of a linear equation is the value or
values of the variable that make the equation a
true statement. The set of all solutions of an
equation is called the solution set. The
solution satisfies the equation.
Example Determine if x 1 is a solution to
the equation.
3(x 3) 4x 3 5x
3( 1) 3 4( 1) 3 5( 1)
3( 4) 4 3 5
12 12
True. x 1 is a solution
6
Solving Equations
Linear equations are solved by writing a series
of steps that result in the equation x a
number One method for solving equations is to
write a series of equivalent equations.
Two or more equations that have precisely the
same solutions are called equivalent equations.
3 5 8
1 7 2 6
7
Addition Property
The Addition Property of Equality states that for
real numbers a b and c if a b then a c
b c.
6 y 11
6 y (6) 11 (6)
y 5
8
Using the Addition Property
Example Solve the linear equation x 9 22.
Step 1 Isolate the variable x on the left side
of the equation.
x 9 9 22 9
Add 9 to both sides of the equation.
Step 2 Simplify the left and right sides of the
equation.
x 31
Apply the Additive Inverse Property.
Step 3 Check to verify the solution.
x 9 22
31 9 22
22 22

9
Multiplication Property
The Multiplication Property of Equality states
that for real numbers a b and c where c
0 if a b then ac bc.
x 28
10
Using the Multiplication Property
Example Solve the linear equation 3x 81
Step 1 Get the coefficient of the variable x to
be 1.
Step 2 Simplify the left and right sides of the
equation.
x 27
Apply the Multiplicative Inverse Property.
Step 3 Check to verify the solution.
3x 81
3(27) 81
81 81

11
Linear Equations Using the Properties Together

• Section 2.2

12
Solving Linear Equations
Example Solve the linear equation 3x 9 24.
3x 9 24
Add 9 to both sides of the equation.
3x 9 9 24 9
3x 15
Divide both sides of the equation by 3
x 5
Check your answer in the original equation.
3( 5) 9 24
15 9 24

24 24
13
Solving by Combining Like Terms
Example Solve the equation 3x 8 2x 15
3x 8 2x 15
3x 2x 7
Add 8 to both sides.
x 7
Subtract 2x from both sides.
3( 7) 8 2( 7) 15
Check your answer in the original equation.
21 8 14 15
29 29

14
Using the Distributive Property
Example Solve the equation 12 2x 3(x 2)
4x 12 x.
12 2x 3(x 2) 4x 12 x
12 2x 3x 6 4x 12 x
Use the Distributive Property.
6 5x 3x 12
Combine like terms.
6 8x 12
Add 5x to both sides.
6 8x
Subtract 12 from both sides.
Divide both sides by 8 and simplify.
Be sure to check your answer.
15
Summary Solving a Linear Equation
Summary Steps for Solving an Equation in
One Variable Step 1 Remove any parentheses
using the Distributive Property. Step 2 Combine
like terms on each side of the equation. Step 3
Use the Addition Property of Equality to get all
variables on one side and all constants on the
other side. Step 4 Use the Multiplication
Property of Equality to get the coefficient of
the variable to equal 1. Step 5 Check the
solution to verify that it satisfies the original
equation.
16
Solving Problems
Example A Chunky Cheeseburger contains 12 more
grams of fat than a Happy Hamburger. Find the
number of grams of fat in each sandwich if there
are a total of 98 grams of fat in the two
sandwiches by solving the equation x (x 12)
98 where x represents the number of fat grams
in a Happy Hamburger and x 12 represents the
number of fat grams in a Chunky Cheeseburger.
x (x 12) 98
2x 12 98
Simplify.
2x 86
Subtract 12 from both sides.
x 43
Divide both sides by 2.
There are 43 fat grams in a Happy Hamburger and
43 12 55 fat grams in a Chunky Cheeseburger.
17
Solving Linear Equations Involving Fractions and
Decimals Classifying Equations

• Section 2.3

18
Solving Equations Containing Fractions
Example Solve the equation
Fractions can be removed by multiplying both
sides of the equation by the LCD of all the
fractions.
The LCD is 10.
Remove all parentheses.
Divide out common factors.
Simplify.
Simplify.
Continued.
19
Solving Equations Containing Fractions
Example continued
Add 4 to both sides of the equation.
Divide both sides by 10
Check your answer in the original equation.

20
Solving Equations Containing Decimals
Example Solve the equation
Decimals can be eliminated by multiplying both
sides of the equation by the LCD of all the
decimals.
Distribute before eliminating the decimals.
Combine like terms.
Subtract 3.6x from both sides.
Subtract 5.4 from both sides.
Continued.
21
Solving Equations Containing Decimals
Example continued
Multiply both sides of the equation by the LCD 10
to eliminate the decimal.
Simplify.
Divide both sides by 38.
Check your answer in the original equation.

22
Conditional Equations Contradictions
A conditional equation is an equation that is
true for some values of the variable and false
for other values of the variable.
A contradiction is an equation that is false for
every replacement value of the variable.
Example Solve the equation 4x 8 4x 1.
4x 8 4x 1
8 1
Subtract 4x from both sides.
This is a false statement so the equation is a
contradiction. The solution set is the empty
set written or .
23
Identities
An identity is an equation that is satisfied for
all values of the variable for which both sides
of the equation are defined.
Example Solve the equation 5x 3 2x 3(x
1).
5x 3 2x 3(x 1).
Distribute to remove parentheses.
5x 3 2x 3x 3
Combine like terms.
5x 3 5x 3
Subtract 5x from both sides.
3 3
This is a true statement for all real numbers x.
The solution set is all real numbers.
24
Evaluating Formulas and Solving Formulas for a
Variable

• Section 2.4

25
Simple Interest
A mathematical formula is an equation that
describes how two or more variables are related.
Interest is money paid for the use of money.
The total amount borrowed is called the principal.
The rate of interest expressed as a percent is
the amount charged for the use of the principal
for a given period of time usually on a yearly
basis.
Simple Interest Formula If an amount of money P
called the principal is invested for a period of
t years at an annual interest rate r expressed
as a decimal the interest I earned is I
Prt This interest earned is called simple
interest.
26
Simple Interest
Example Jordan received a bonus check for 1000.
He invested it in a mutual fund that earned
6.75 simple interest. Find the amount of
interest Jordan will earn in two years.
The principal P is 1000.
The interest rate r is 6.75
The time t is 2.
Use the simple interest formula to solve the
problem
I Prt
(1000)(0.0675)(2)
135
Jordan will earn 135 interest at the end of two
years.
27
Geometric Formulas
Definitions The perimeter is the sum of the
lengths of all the sides of a figure. The area is
the amount of space enclosed by a two-dimensional
figure measured in square units. The surface area
of a solid is the sum of the areas of the
surfaces of a three-dimensional figure. The
volume is the amount of space occupied by a
figure measured in cubic units. The radius r of a
circle is the line segment that extends from the
center of the circle to any point on the
circle. The diameter of a circle is any line
segment that extends from one point on the circle
through the center to a second point on the
circle. The diameter is two times the length of
the radius d 2r. In circles we use the term
circumference to mean the perimeter.
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Plane Figure Formulas
Continued.
29
Plane Figure Formulas
b
c
a
h
B
30
Solid Formulas
s
s
s
r
Continued.
31
Solid Formulas
32
Solving a Formula for a Variable
The formula for the area of a trapezoid is
Example Solve the formula for b.
Multiply both sides by 2.
Divide both sides by h.
Subtract a from both sides.
33
Evaluating a Formula
The formula for the area of a triangle is
Example If the area of a triangle is 66 inches
and the base is 8 inches find the height of the
triangle.
Substitute in the values.
Multiply both sides by 2.
Divide both sides by 8.
The height of the triangle is 16.5 inches.
34
Evaluating a Formula
Example Carla is making a planter out of an
empty can for her mothers birthday. She has 157
cubic inches of soil to use. Find the radius of
the can if it has a height of 8 inches.
The volume of a circular cylinder is V r2h.
Substitute the known values into the formula
V r2h
157 r2(8)
Simplify. (8 25.13)
157 25.13r2
6.25 r2
Divide both sides by 25.13.
Continued.
35
Evaluating a Formula
Example continued
6.25 r2
2.5 r
Take the square root of both sides.
The radius of the can is 2.5 inches.
Check Since r represents the radius of the
cylinder
V r2h
(2.5)2(8)
157

36
Introduction to Problem Solving Direct
Translation Problems

• Section 2.5

37
Translating English Expressions
38
Translating Sentences into Expressions
Example Translate each of the following into a
mathematical statement.
a.) Twelve more than a number is 25.
x 12

25
b.) One-third of the sum of a number and four
yields 6.
39
Problem Solving
Problem solving is the ability to use
information tools and our own skills to achieve
a goal.
The process of taking a verbal description of the
problem and developing it into an equation that
can be used to solve the problem is mathematical
modeling.
The equation that is developed is the
mathematical model.
40
Categories of Problems

• Five Categories of Problems
• Direct Translation problems that must be
  translated from English into mathematics using
  key words in the verbal description
• Mixtures problems where two or more quantities
  are combined in some fashion
• Geometry problems where the unknown quantities
  are related through geometrical formulas
• Uniform Motion problems where an object travels
  at a constant speed
• Work problems problems where two or more
  entities join forces to complete a job

41
Steps for Solving Problems
Solving Problems with Mathematical Models Step 1
Identify What You Are Looking For Read the
problem carefully. Identify the type of problem
and the information we wish to learn. Typically
the last sentence in the problem indicates what
it is we wish to solve for. Step 2 Give Names
to the Unknowns Assign variables to the unknown
quantities. Choose a variable that is
representative of the unknown quantity it
represents. For example use t for time. Step 3
Translate into the Language of Mathematics
Determine if each sentence can be
translated into a mathematical statement. If
necessary combine the statements into an
equation that can be solved.
Continued.
42
Steps for Solving Problems
Solving Problems with Mathematical Models Step 4
Solve the Equation(s) Found in Step 3 Solve
the equation for the variable and then answer the
question posed by the original problem. Step 5
Check the Reasonableness of Your Answer Check
your answer to be sure that it makes sense. If
it does not go back and try again. Step 6
Answer the Question Write your
answer in a complete sentence.
43
Direct Translation Problem
Example In a baseball game the Yankees scored 4
more runs than the White Sox. A total of 12 runs
were scored. How many runs were scored by each
team
Step 1 Identify. This is a direct translation
problem. We are looking for the number of runs
scored by each team.
Step 2 Name. Let x represent the number of runs
scored by the White Sox. The number of runs
scored by the Yankees is equal to x 4.
Continued.
44
Direct Translation Problem
Example continued
Step 3 Translate. Since we know that the total
number of runs is 12 we have
Yankees runs
White Sox runs
x
x 4

12

Step 4 Solve.
x x 4 12
Combine like terms.
2x 4 12
2x 8
Subtract 4 from both sides.
x 4
Divide both sides by 2.
Continued.
45
Direct Translation Problem
Example continued
Step 5 Check. Since x represents the number of
runs scored by the White Sox the White Sox
scored 4 runs. The Yankees scored x 4 4 4
8 runs.
4 8 12

Step 6 Answer the Question.
The Yankees scored 8 runs and the White Sox
scored 4 runs.
46
Problem Solving Direct Translation Problems
Involving Percent

• Section 2.6

47
Solving an Equation Involving Percent
Example A number is 9 of 65. Find the number.
Step 1 Identify We want to know the unknown
number.
Step 2 Name Let n represent the number.
Step 3 Translate
A number is 9 of 65
Continued.
48
Solving an Equation Involving Percent
Example continued
n 0.09 65
Step 4 Solve Solve the equation.
n 0.09 65
5.85
Step 5 Check Check the multiplication.
0.09 65 5.85
Step 6 Answer the Question
5.85 is 9 of 65.
49
Solving an Equation Involving Percent
Example 36 is 6 of what number
Step 1 Identify We want to know a number.
Step 2 Name Let x represent the number.
Step 3 Translate
36 is 6 of what number
Continued.
50
Solving an Equation Involving Percent
Example continued
36 0.06x
Step 4 Solve Solve the equation.
36 0.06x
600
Divide both side by 0.06.
Step 5 Check Is 6 of 600 equal to 36
0.06 600 36
36 36

Step 6 Answer the Question
36 is 6 of 600.
51
Solving a Business Problem
One type of percent problem involves discounts or
mark-ups that businesses use in determining their
prices.
Original Price Discount Sale Price Wholesale
Price Markup Selling Price
Example Julie bought a leather sofa that was
on sale for 35 off the original price. If she
paid 780 what was the original price of the
sofa
Step 1 Identify This is a direct translation
problem. We are looking for the original price
of the sofa.
Step 2 Name Let p represent the original price.
Continued.
52
Solving a Business Problem
Example continued
Step 3 Translate The original price minus the
amount of the discount will equal the sale price.
p discount 780
The discount is 35 of the original price.
p 0.35p 780
Step 4 Solve Solve the equation.
Combine like terms.
0.65p 780
Divide both sides by 0.65.
p 1200
Continued.
53
Solving a Business Problem
Example continued
Step 5 Check If the original price of the couch
was 1200 then the discount would be 0.35(1200)
420. Subtracting 420 from the original price
of 1200 results in a sale price of 780.
Step 6 Answer the Question
The original price of the couch was 1200.
54
Problem Solving Geometry and Uniform Motion

• Section 2.7

55
Angles
Two angles whose sum is 90 are called
complementary angles. Each angle is called the
complement of the other.
Two angles whose sum is 180 are called
supplementary angles. Each angle is called the
supplement of the other.
56
Solving Angle Problems

• Example
• Angle A and angle B are complementary angles and
  angle A is 21º more than twice angle B. Find the
  measure of both angles.

Step 1 Identify This is complementary problem.
We are looking for the measure of the two angles
whose sum is 90.
Step 2 Name Let a represent the measure of
angle A.
Continued.
57
Solving Angle Problems

• Example continued

Step 3 Translate Angle A is 21 more than twice
the measure of angle B.
a 21 2 m B
a 21 2 (90 a)
Step 4 Solve Solve the equation.
a 21 2 (90 a)
Distribute.
a 21 180 2a
Combine like terms.
a 201 2a
Continued.
58
Solving Angle Problems

• Example continued

a 201 2a
Add 2a to both sides.
3a 201
Divide both sides by 3.
a 67
Step 5 Check The measure of A is 67. The
measure of B is 90 a 90 67 23.
67 23 90

Step 6 Answer the Question
The two complementary angles measure 67 and 23.
59
Solving Triangle Problems

• Example
• Find the measure of Angle C.

Remember that the sum of the measures of the
interior angles of a triangle is 180.
Step 1 Identify This is an angles of the
triangle problem.
Step 2 Name Let c represent the measure of
angle C.
Continued.
60
Solving Triangle Problems

• Example continued

Step 3 Translate The three angles add up to
180.
84 42 c 180
Step 4 Solve Solve the equation.
84 42 c 180
Combine like terms.
126 c 180
Subtract 126 from both sides.
c 54
Continued.
61
Solving Triangle Problems

• Example continued

Step 5 Check Is the sum of the three angles
equal to 180
84 42 54 180
180 180

Step 6 Answer the Question The measure of Angle
C is 54.
62
Solving Geometry Problems

• Example
• Julie is making cone-shaped candles. The mold
  for the candles is 4 in. in diameter and 7 in.
  high. How many cubic inches of wax does Julie
  need to buy if she wants to make 50 candles

Step 1 Identify This is a geometry volume
problem. We want to find the volume of the
cone-shaped candle to determine the amount of wax
needed for 50 candles.
Step 2 Name Let V represent the volume of one
cone.
Continued.
63
Solving Geometry Problems

• Example continued

Step 3 Translate We need to use the formula for
the volume of a cone.
3.14 r 2 h 7
Step 4 Solve Solve the equation.
Continued.
64
Solving Geometry Problems

• Example continued

This is the amount needed for one candle.
This is the amount needed for 50 candles.
29.3 in.3 50 1465 in.3
Step 5 Check
29.3 in.3 50 1465 in.3

Step 6 Answer the Question Julie needs to buy
approximately 1465 in.3 of wax to make 50 candles.
65
Uniform Motion
Objects that move at a constant velocity (speed)
are said to be in uniform motion.
Uniform Motion Formula If an object moves at an
average speed r the distance d covered in time t
is given by the formula d rt.
The following table is helpful in solving motion
problems.
66
Uniform Motion Problem
Example Steve jogs at an average rate of 8
kilometers per hour. How long would it take him
to jog 14 kilometers
Step 1 Identify This is a uniform motion
problem. We are looking for the length of time
it would take Steve to jog 14 kilometers.
Step 2 Name Let t represent the length of time
it would take Steve to jog 14 kilometers.
Continued.
67
Uniform Motion Problem
Example continued
Step 3 Translate Organize the information in a
table.
d rt
14 8t
Step 4 Solve
Divide both sides by 8.
Simplify.
Continued.
68
Uniform Motion Problem
Example continued
Step 5 Check t 1.75 represents the length of
time.
d rt
14 (8)(1.75)
14 14

Step 6 Answer the Question
It takes Mark 1.75 hours (or 1 hour and 45
minutes) to run 14 kilometers.
69
Uniform Motion Problem
Example Nina drove her car to Cleveland while
Paula drove her car to Columbus. Nina drove 360
kilometers while Paula drove 280 kilometers.
Nina drove 20 kilometers per hour faster than
Paula on her trip. What was the average speed in
kilometers per hour for each driver
Step 1 Identify
Distance problems can be solved using the
formula distance rate time (d rt).
Step 2 Name
Let r the rate of Paulas car.
Let r 20 the rate of Ninas car.
The time t for each driver was the same.
Continued.
70
Uniform Motion Problem
Example continued
Step 3 Translate
Step 4 Solve
Since the time for each driver was the same we
can set the times equal to each other.
Continued.
71
Uniform Motion Problem
Example continued
Multiply both sides by r(r 20).
Simplify.
Distribute.
Subtract 280r from each side.
Divide each side by 80.
Paulas rate was 70 kilometers per hour. Ninas
rate was r 20 90 kilometers per hour.
Continued.
72
Uniform Motion Problem
Example continued
Step 5 Check

Step 6 Answer the Question
Paulas rate was 70 kilometers per hour. Ninas
rate was 90 kilometers per hour.
73
Solving Linear Inequalities in One Variable

• Section 2.8

74
Linear Inequalities
A linear inequality in one variable is an
inequality that can be written in the form ax
b lt c or ax b c or ax b gt c or ax b
c where a b and c are real numbers and a 0.
Inequalities that contain one inequality symbol
are called simple inequalities.
means the set of all real numbers x greater
than four
x gt 4
75
Set-Builder Notation
Set-builder notation is used to express the
inequality in written form.
Representing an inequality on a number line is
called graphing the inequality and the picture
is called the graph of the inequality.
x gt 4
76
Graphing Inequalities
Example Graph each interval.
77
Interval Notation
Interval notation is also used to represent
inequalities.
(4 )
x gt 4
Graph
Interval Notation
Set-Builder Notation
x x a
The interval a)
x x gt a
The interval (a )
x x a
The interval ( a
x x lt a
The interval ( a)
x x is a real number
The interval ()
78
Interval Notation
Example Write each inequality using interval
notation.
( 2 0
( 3)
( 1.5 3)
79
Inequalities and Interval Notation

• SUMMARY Graphing Inequalities and Interval
  Notation
• If the inequality contains the symbol less than
  or the symbol greater than. lt or gt use
  parentheses ( or ) on the number line.
• If the inequality contains the symbol less than
  or equal to or the symbol greater than or equal
  to or use brackets or on the
  number line.
• The symbols or always use parentheses.

80
Addition Property of Inequality
Addition Property of Inequality For real numbers
a b and c If a lt b then a c lt b c
If a gt b then a c gt b c
Example Solve the linear inequality and state
the solution set using set-builder notation and
interval notation. Graph the solution set.
x 24 gt 19
x gt 5
Subtract 24 from both sides.
Set-builder notation x x gt 5
Interval notation ( 5 )
81
Multiplication Property of Inequality
Multiplication Properties of Inequality Let a b
and c be real numbers. If a lt b and if c gt 0
then ac lt bc If a gt b and if c gt 0 then
ac gt bc If a lt b and if c lt 0 then ac gt bc
If a gt b and if c lt 0 then ac lt bc
Example Solve the linear inequality and state
the solution set using set-builder notation and
interval notation.
2y lt 16
y gt 8
Divide both sides by 2.
Set-builder notation y y gt 8
Interval notation ( 8 )
82
Solving Linear Inequalities
Steps for Solving Linear Inequalities Step 1
Remove parentheses. Step 2 Combine like terms
on each side of the inequality. Step 3 Get the
variable expressions on the left side of the
inequality and the constants on the right
side. Step 4 Get the coefficient of the
variable term to be one.
83
Solving Linear Inequalities
Example Solve the linear inequality and state
the solution set using set-builder notation and
interval notation. Graph the solution set.
Multiply by the LCD 24.
Simplify.
Distribute.
Simplify.
Continued.
84
Solving Linear Inequalities
Example continued
Subtract 8x from both sides.
Divide both sides by 5.
Set-builder notation x x 8
Interval notation ( 8
85
Linear Inequality Problems
86
Linear Inequality Problems
Example Federation Express will not deliver a
package if its height plus girth (circumference
around the widest part) is more than 130 inches.
If you are preparing a package that is 33 inches
wide and 8 inches long how high is the package
permitted to be
Step 1 Identify. This is a geometry problem. We
are looking for the height of the package.
Step 2 Name. Let h represent the height.
Continued.
87
Linear Inequality Problems
Example continued
Step 3 Translate. The height plus the girth
cannot be more than 130 inches.
height circumference 130
h (2)(33) (2)(8) 130
Step 4 Solve.
h (2)(33) (2)(8) 130
Simplify.
h 66 16 130
h 82 130
Simplify.
Subtract 82 from both sides.
h 48
Continued.
88
Linear Inequality Problems
Example continued
Step 5 Check. Since h represents the height of
the package check to make sure that the height
girth is not more than 130 inches.
h circumference 130
48 (2)(33) (2)(8) 130
48 66 16 130
130 130

Step 6 Answer the Question.
The height of the package cannot be more than 48
inches.