Download CHAPTER 1 Two-Step Equations and Properties of Equality

www.ck12.org
C HAPTER
Chapter 1. Two-Step Equations and Properties of Equality
1
Two-Step Equations and
Properties of Equality
Here you’ll learn how to use the Addition Property and Multiplication Property to solve equations that you have
written in two steps.
Suppose you are saving up to buy a leather jacket that costs $250. You currently have $90 put away, and you’re
saving at a rate of $40 per month. Can you write an equation to represent this situation and then solve it in two steps
to find the number of months needed to reach your goal? In this Concept, you’ll learn how to solve problems just
like this.
Guidance
Suppose Shaun weighs 146 pounds and wants to lose enough weight to wrestle in the 130-pound class. His
nutritionist designed a diet for Shaun so he will lose about 2 pounds per week. How many weeks will it take
Shaun to weigh enough to wrestle in his class?
This is an example that can be solved by working backward. In fact, you may have already found the answer by
using this method. The solution is 8 weeks.
By translating this situation into an algebraic sentence, we can begin the process of solving equations. To solve an
equation means to “undo” all the operations of the sentence, leaving a value for the variable.
Translate Shaun’s situation into an equation.
−2w + 146 = 130
This sentence has two operations: addition and multiplication. To find the value of the variable, we must use both
properties of equality: the Addition Property of Equality and the Multiplication Property of Equality.
Procedure to Solve Equations of the Form ax + b =some number:
1. Use the Addition Property of Equality to get the variable term ax alone on one side of the equation:
ax = some number
2. Use the Multiplication Property of Equality to get the variable x alone on one side of the equation:
x = some number
Example A
Solve Shaun’s problem.
Solution: −2w + 146 = 130
Apply the Addition Property of Equality: −2w + 146 − 146 = 130 − 146.
Simplify: −2w = −16.
1
www.ck12.org
Apply the Multiplication Property of Equality: −2w ÷ −2 = −16 ÷ −2.
The solution is w = 8.
It will take 8 weeks for Shaun to weigh 130 pounds.
Solving Equations by Combining Like Terms
Michigan has a 6% sales tax. Suppose you made a purchase and paid $95.12, including tax. How much was the
purchase before tax?
Begin by determining the noun that is unknown and choose a letter as its representation.
The purchase price is unknown so this is our variable. Call it p. Now translate the sentence into an algebraic
equation.
price + (0.06)price = total amount
p + 0.06p = 95.12
To solve this equation, you must know how to combine like terms.
Like terms are expressions that have identical variable parts.
According to this definition, you can only combine like terms if they are identical. Combining like terms only
applies to addition and subtraction! This is not a true statement when referring to multiplication and division.
The numerical part of an algebraic term is called the coefficient. To combine like terms, you add (or subtract) the
coefficients of the identical variable parts.
Example B
Identify the like terms, and then combine.
10b + 7bc + 4c + (−8b)
Solution: Like terms have identical variable parts. The only terms having identical variable parts are 10b and −8b.
To combine these like terms, add them together.
10b + 7bc + 4c + −8b = 2b + 7bc + 4c
You will now apply this concept to the Michigan sales tax situation.
Example C
What was the purchase amount from this section’s opening scenario?
Solution: p + 0.06p = 95.12
Combine the like terms: p + 0.06p = 1.06p, since p = 1p.
Simplify: 1.06p = 95.12.
Apply the Multiplication Property of Equality: 1.06p ÷ 1.06 = 95.12 ÷ 1.06.
Simplify: p = 89.74.
2
www.ck12.org
Chapter 1. Two-Step Equations and Properties of Equality
The price before tax was $89.74.
The next several examples show how algebraic equations can be created to represent real-world situations.
Vocabulary
Like terms: Like terms are expressions that have identical variable parts. To solve equations, you often need to
combine like terms.
Coefficient: The numerical part of an algebraic term is called the coefficient. To combine like terms, you add (or
subtract) the coefficients of the identical variable parts.
Solving equations: To solve an equation means to “undo” all the operations of the sentence, leaving a value for the
variable.
Guided Practice
1. An emergency plumber charges $65 as a call-out fee plus an additional $75 per hour. He arrives at a house at
9:30 and works to repair a water tank. If the total repair bill is $196.25, at what time was the repair completed?
2. To determine the temperature in Fahrenheit, multiply the Celsius temperature by 1.8 and then add 32. Determine
the Celsius temperature if it is 89◦ F.
Solutions:
1. Translate the sentence into an equation. The number of hours it took to complete the job is unknown, so call it h.
Write the equation: 65 + 75(h) = 196.25.
Apply the Addition Property and simplify.
65 + 75(h) − 65 = 196.25 − 65
75(h) = 131.25
Apply the Multiplication Property of Equality: 75(h) ÷ 75 = 131.25 ÷ 75.
Simplify: h = 1.75.
The plumber worked for 1.75 hours, or 1 hour, 45 minutes. Since he started at 9:30, the repair was completed at
11:15.
2. Translate the sentence into an equation. The temperature in Celsius is unknown; call it C.
Write the equation: 1.8C + 32 = 89.
3
www.ck12.org
Apply the Addition Property and simplify.
1.8C + 32 − 32 = 89 − 32
1.8C = 57
Apply the Multiplication Property of Equality: 1.8C ÷ 1.8 = 57 ÷ 1.8.
Simplify: C = 31.67.
If the temperature is 89◦ F, then it is 31.67◦C.
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note
that there is not always a match between the number of the practice exercise in the video and the number of the
practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Ba
sic Algebra:Two-Step Equations (13:50)
MEDIA
Click image to the left for more content.
1. Define like terms. Give an example of a pair of like terms and a pair of unlike terms.
2. Define coefficient.
In 3 – 7, combine the like terms.
3.
4.
5.
6.
7.
8.
−7x + 39x
3x2 + 21x + 5x + 10x2
6xy + 7y + 5x + 9xy
10ab + 9 − 2ab
−7mn − 2mn2 − 2mn + 8
Explain the procedure used to solve −5y − 9 = 74.
Solve and check your solution.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
4
1.3x − 0.7x = 12
6x − 1.3 = 3.2
5x − (3x + 2) = 1
4(x + 3) = 1
5q − 7 = 23
3
5
2
5x+ 2 = 3
3s
5
s− 8 = 6
0.1y + 11 = 0
5q−7
2
12 = 3
5(q−7)
2
12 = 3
www.ck12.org
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
Chapter 1. Two-Step Equations and Properties of Equality
33t − 99 = 0
5p − 2 = 32
14x + 9x = 161
3m − 1 + 4m = 5
8x + 3 = 11
24 = 2x + 6
66 = 23 k
5
1
8 = 2 (a + 2)
16 = −3d − 5
Jayden purchased a new pair of shoes. Including a 7% sales tax, he paid $84.68. How much did his shoes cost
before sales tax?
A mechanic charges $98 for parts and $60 per hour for labor. Your bill totals $498.00, including parts and
labor. How many hours did the mechanic work?
An electric guitar and amp set costs $1195.00. You are going to pay $250 as a down payment and pay the rest
in 5 equal installments. How much should you pay each month?
Jade is stranded downtown with only $10 to get home. Taxis cost $0.75 per mile, but there is an additional
$2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money.
Determine how many miles she can ride.
Jasmin’s dad is planning a surprise birthday party for her. He will hire a bouncy castle and provide party food
for all the guests. The bouncy castle costs $150 dollars for the afternoon, and the food will cost $3.00 per
person. Andrew, Jasmin’s dad, has a budget of $300. Write an equation to help him determine the maximum
number of guests he can invite.
Mixed Review
33. Trish showed her work solving the following equation. What did she do incorrectly?
−2c = 36
c = 18
34. Write an expression for the following situation: Yoshi had d dollars, spent $65, and earned $12. He had $96
left.
35. Find the domain of the following graph.
36. Does the graph represent a function? Explain your answer.
37. Find the difference:
1
2
− 15
9.
5
www.ck12.org
38. What is the additive identity?
39. Find the opposite of –4.1398.
6
www.ck12.org
C HAPTER
Chapter 2. Distributive Property for Multi-Step Equations
2
Distributive Property for
Multi-Step Equations
Here you’ll learn how to use the Distributive Property to solve problems with multiple steps.
Suppose you were a contestant on a game show where they gave you the answer and you had to think of the
corresponding question. If the answer were, "M(N + K) = MN + MK or M(N − K) = MN − MK," what would you
give for the question? How about, "What is the Distributive Property?" In this Concept, you will learn all about the
Distributive Property and how to use it to solve equations with multiple steps.
Guidance
Solving Multi-Step Equations by Using the Distributive Property
When faced with an equation such as 2(5x + 9) = 78, the first step is to remove the parentheses. There are two
options to remove the parentheses. You can apply the Distributive Property or you can apply the Multiplication
Property of Equality. This Concept will show you how to use the Distributive Property to solve multi-step equations.
Example A
Solve for x: 2(5x + 9) = 78.
Solution: Apply the Distributive Property: 10x + 18 = 78.
Apply the Addition Property of Equality: 10x + 18 − 18 = 78 − 18.
Simplify: 10x = 60.
Apply the Multiplication Property of Equality: 10x ÷ 10 = 60 ÷ 10.
The solution is x = 6.
Check: Does 10(6) + 18 = 78? Yes, so the answer is correct.
Example B
Solve for n when 2(n + 9) = 6n.
Solution:
2(n + 9) = 5n
2 · n + 2 · 9 = 5n
2n + 18 = 5n
− 2n + 2n + 18 = −2n + 5n
18 = 3n
1
1
· 18 = · 3n
3
3
6=n
Checking the answer:
7
www.ck12.org
2(6 + 9) = 5(6)
2(15) = 30
30 = 30
Example C
Solve for d when 3(d + 15) − 18d = 0.
Solution:
3(d + 15) − 18d = 0
3 · d + 3 · 15 − 18d = 0
3d + 45 − 18d = 0
− 15d + 45 = 0
− 15d + 45 − 45 = 0 − 45
− 15d = −45
1
1
− · −15d = − · −45
15
15
d=3
Checking the answer:
3(3 + 15) − 18(3) = 0
3(3 + 15) − 18(3) = 0
3(18) − 18(3) = 0
54 − 54 = 0
0=0
Vocabulary
Distributive Property: For any real numbers M, N, and K:
M(N + K) = MN + MK
M(N − K) = MN − MK
The Addition Property of Equality: For all real numbers a, b, and c:
If a = b, then a + c = b + c.
The Multiplication Property of Equality: For all real numbers a, b, and c:
If a = b, then a(c) = b(c).
8
www.ck12.org
Chapter 2. Distributive Property for Multi-Step Equations
Guided Practice
Solve for x when 3(2x + 5) + 2x = 7.
Solution:
Step 1: Apply the Distributive Property.
3(2x + 5) + 2x = 7
3 · 2x + 3 · 5 + 2x = 7
6x + 15 + 2x = 7
Step 2: Combine like terms.
6x + 15 + 2x = 7
8x + 15 = 7
Step 3: Isolate the variable and its coefficient by using the Addition Property.
8x + 15 = 7
8x + 15 − 15 = 7 − 15
8x = −8
Step 4: Isolate the variable by applying the Multiplication Property.
8x = −8
1
1
· 8x = −8 ·
8
8
1
1
· 8x = −8 ·
8
8
x = −1
Step 5: Check your answer.
Substitute x = −1 into 3(2x + 5) = 7.
3(2(−1)x + 5) + 2(−1) = 3(−2 + 5) − 2 = 3(3) − 2 = 9 − 2 = 7.
Therefore, x = −1.
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note
that there is not always a match between the number of the practice exercise in the video and the number of the
9
www.ck12.org
practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Ba
sic Algebra:Multi-Step Equations (15:01)
MEDIA
Click image to the left for more content.
In 1 – 22, solve the equation.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
10
3(x − 1) − 2(x + 3) = 0
7(w + 20) − w = 5
9(x
� − 2)1=
� 3x 2+ 3
2 �5a − �3 = 7
2
2
2
9 �i + 3 � = 5
4 v + 14 = 35
2
22 = 2(p + 2)
−(m + 4) = −5
48� = 4(n� + 4)
6
3
6
5 v − 5 = 25
−10(b − 3) = −100
6v + 6(4v + 1) = −6
−46 = −4(3s + 4) − 6
8(1 + 7m) + 6 = 14
0 = −7(6 + 3k)
35 = −7(2 − x)
−3(3a
� + 1)
� − 7a = −35
−2 n + 37 = − 14
3 �
�
59
− 60
= 16 − 43 r − 5
4y+3
7 =9
(c + 3) − 2c − (1 − 3c) = 2
5m − 3[7 − (1 − 2m)] = 0
www.ck12.org
C HAPTER
Chapter 3. Inequality Expressions
3
Inequality Expressions
Here you’ll learn how to express inequalities in various forms, as well as how to graph inequalities on a number line.
Suppose you’re having a party, and you know that the number of people attending will be greater than or equal to
25. How would you write this inequality? If you had to graph the solutions to this inequality on a number line,
could you do it? After completing this Concept, you’ll not only be able to express inequalities such as this one with
a graph, but you’ll also be able to look at a graph and determine what inequality it represents.
Guidance
Verbs that translate into inequalities are:
> “greater than”
≥ “greater than or equal to”
< “less than”
≤ “less than or equal to”
�= “not equal to”
Definition: An algebraic inequality is a mathematical sentence connecting an expression to a value, a variable, or
another expression with an inequality sign.
Solutions to one-variable inequalities can be graphed on a number line or in a coordinate plane.
Example A
Graph the solutions to t > 3 on a number line.
Solution: The inequality is asking for all real numbers larger than 3.
You can also write inequalities given a number line of solutions.
Example B
Write the inequality pictured below.
Solution: The value of four is colored in, meaning that four is a solution to the inequality. The red arrow indicates
values less than four. Therefore, the inequality is:
x≤4
11
www.ck12.org
Inequalities that “include” the value are shown as ≤ or ≥. The line underneath the inequality stands for “or equal to.”
We show this relationship by coloring in the circle above this value on the number line, as in the previous example.
For inequalities without the “or equal to,” the circle above the value on the number line remains unfilled.
Four Ways to Express Solutions to Inequalities
1. Inequality notation: The answer is expressed as an algebraic inequality, such as d ≤ 12 .
�
�
2. Set notation: The inequality is rewritten using set notation brackets . For example, d|d ≤ 12 is read, “The set
of all values of d, such that d is a real number less than or equal to one-half.”
3. Interval notation: This notation uses brackets to denote the range of values in an inequality.
1. Square or “closed” brackets [ ] indicate that the number is included in the solution
2. Round or “open” brackets ( ) indicate that the number is not included in the solution.
Interval notation also uses the concept of infinity ∞ and negative �infinity�−∞. For example, for all values of d that
are less than or equal to 12 , you could use set notation as follows: −∞, 12 .
4. As a graphed sentence on a number line.
Example C
Describe the set of numbers contained by the given set notation for the following:
a) (8, 24)
b) [3, 12)
Solution:
(8, 24) states that the solution is all numbers between 8 and 24 but does not include the numbers 8 and 24.
[3, 12) states that the solution is all numbers between 3 and 12, including 3 but not including 12.
Vocabulary
Algebraic inequality: An algebraic inequality is a mathematical sentence connecting an expression to a value, a
variable, or another expression with an inequality sign.
Guided Practice
Describe and graph the solution set expressed by (∞, 3.25).
Solution:
The solution set contains all numbers less than 3.25, not including 3.25.
The graph on the number line is:
Practice
1. What are the four methods of writing the solution to an inequality?
12
www.ck12.org
Chapter 3. Inequality Expressions
Graph the solutions to the following inequalities using a number line.
2.
3.
4.
5.
6.
7.
8.
9.
x < −3
x≥6
x>0
x≤8
x < −35
x > −17
x ≥ 20
x≤3
Write the inequality that is represented by each graph.
10.
11.
12.
13.
14.
15.
16.
17.
13
www.ck12.org
C HAPTER
4
Inequalities with Addition
and Subtraction
Here you’ll learn how to use addition and subtraction to find the solutions to one-step inequalities.
Suppose your favorite baseball team had w wins last year, and this year it vows to win 10 more games. It also
promises that the number of its wins this year will be greater than or equal 85. Could you determine how many wins
the team had last year? In this Concept, you’ll learn to solve inequalities such as the one representing this scenario
by using addition and subtraction.
Guidance
Inequalities Using Addition or Subtraction
To solve inequalities, you need some properties.
Addition Property of Inequality: For all real numbers a, b, and c:
If x < a, then x + b < a + b.
If x < a, then x − c < a − c.
The two properties above are also true for ≤ or ≥.
Because subtraction can also be thought of as “add the opposite,” these properties also work for subtraction
situations.
Just like one-step equations, the goal is to isolate the variable, meaning to get the variable alone on one side of the
inequality symbol. To do this, you will cancel the operations using inverses.
Example A
Solve for x : x − 3 < 10.
Solution: To isolate the variable x, you must cancel “subtract 3” using its inverse operation, addition.
x − 3 + 3 < 10 + 3
x < 13
Now, check your answer. Choose a number less than 13 and substitute it into your original inequality. If you choose
0, and substitute it you get:
0 − 3 < 10 = −3 < 10
What happens at 13? What happens with numbers greater than 13?
Example B
Solve for x : x + 4 > 13
14
www.ck12.org
Chapter 4. Inequalities with Addition and Subtraction
Solution:
x + 4 > 13
Subtract 4 from both sides of the inequality.
Simplify.
x + 4 − 4 > 13 − 4
x>9
The solution is shown below in a graph:
Example C
Solve for x: x + 23 ≥ − 13 .
Solution:
2
1
≥−
3
3
2 2
1 2
x+ − ≥ − −
3 3
3 3
x ≥ −1
x+
Subtract
2
from both sides of the inequality.
3
Simplify.
Vocabulary
Addition Property of Inequality: For all real numbers a, b, and c:
If x < a, then x + b < a + b.
If x < a, then x − c < a − c.
Guided Practice
Solve for y: 5.6 > y − 3.4.
Solution:
Add 3.3 to both sides of the inequality.
Simplify.
5.6 > y − 3.3
5.6 + 3.3 > y − 3.3 + 3.3
8.9 > y
15
www.ck12.org
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note
that there is not always a match between the number of the practice exercise in the video and the number of the
practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Ba
sic Algebra:Inequalities Using AdditionandSubtraction (7:48)
MEDIA
Click image to the left for more content.
Solve each inequality and graph the solution on a number line.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
x − 1 > −10
x − 1 ≤ −5
−20 + a ≥ 14
x+2 < 7
x + 8 ≤ −7
5 + t ≥ 34
x − 5 < 35
15 + g ≥ −60
x−2 ≤ 1
x − 8 > −20
11 + q > 13
x + 65 < 100
x − 32 ≤ 0
x + 68 ≥ 75
16 + y ≤ 0
Mixed Review
16.
17.
18.
19.
20.
16
Write an equation containing (3, –6) and (–2, –2).
Simplify: |2 − 11 × 3|+1.
Graph y = −5 on a coordinate plane.
y varies directly as x. When x = −1, y = 54 . Find y when x =
Rewrite in slope-intercept form: −2x + 7y = 63.
16
3.
www.ck12.org
C HAPTER
Chapter 5. Inequalities with Multiplication and Division
5
Inequalities with
Multiplication and Division
Here you’ll learn how to use multiplication and division to find the solutions to one-step inequalities.
What if the distance in miles of a bowling alley from your house is 15 of the distance of a roller-skating rink from
your house? The distance of the roller-skating rink is represented by r, and you know that the bowling alley is less
than or equal to 6 miles from your house. How far is the roller-skating rink from your house? After completing
this Concept, you’ll be able to solve inequalities like the one representing this scenario by using multiplication and
division.
Watch This
Multimedia Link: For help with solving inequalities involving multiplication and division, visit Khan Academy’s
website: http://khanexercises.appspot.com/video?v=PNXozoJWsWc.
Guidance
Equations are mathematical sentences in which the two sides have the same “weight.” By adding, subtracting,
multiplying, or dividing the same value to both sides of the equation, the balance stays in check. However,
inequalities begin off-balance. When you perform inverse operations, the inequality will remain off-balance. This is
true with inequalities involving both multiplication and division.
Before we can begin to solve inequalities involving multiplication or division, you need to know two properties: the
Multiplication Property of Inequality and the Division Property of Inequality.
Multiplication Property of Inequality: For all real positive numbers a, b, and c:
If x < a, then x(c) < a(c).
If x > a, then x(c) > a(c).
Division Property of Inequality: For all real positive numbers a, b, and c:
If x < a, then x ÷ (c) < a ÷ (c).
If x > a, then x ÷ (c) > a ÷ (c).
Example A
Consider the inequality 2x ≥ 12. To find the solutions to this inequality, we must isolate the variable x by using the
inverse operation of “multiply by 2,” which is dividing by 2.
2x ≥ 12
2x 12
≥
2
2
x≥6
This solution can be expressed in four ways. One way is already written: x ≥ 6. Below are the three remaining ways
to express this solution:
17
www.ck12.org
• {x|x ≥ 6}
• [6, ∞)
• Using a number line:
Example B
Solve for y :
y
5
≤ 3. Express the solution using all four methods.
Solution: The inequality above is read, “y divided by 5 is less than or equal to 3.” To isolate the variable y, you must
cancel division using its inverse operation, multiplication.
y 5
5
· ≤ 3·
5 1
1
y ≤ 15
One method of writing the solution is y ≤ 15.
The other three are:
• (−∞, 15]
• {y|y ≤ 15}
•
Multiplying and Dividing an Inequality by a Negative Number
Notice that the two properties in this Concept focused on c being only positive. This is because those particular
properties of multiplication and division do not apply when the number being multiplied (or divided) is negative.
Think of it this way. When you multiply a value by –1, the number you get is the negative of the original.
6(−1) = −6
Multiplying each side of a sentence by –1 results in the opposite of both values.
5x(−1) = 4(−1)
−5x = −4
When multiplying by a negative, you are doing the “opposite” of everything in the sentence, including the verb.
x>4
x(−1) > 4(−1)
−x < −4
This concept is summarized below.
18
www.ck12.org
Chapter 5. Inequalities with Multiplication and Division
Multiplication/Division Rule of Inequality: For any real number a, and any negative number c,
If x < a, then x · c > a · c
If x < a, then
x
c
>
a
c
As with the other properties of inequalities, these also hold true for ≤or ≥.
Example C
Solve for r : −3r < 9.
Solution: To isolate the variable r, we must cancel “multiply by –3” using its inverse operation, dividing by –3.
−3r
9
<
−3
−3
Since you are dividing by –3, everything becomes opposite, including the inequality sign.
r > −3
Example D
Solve for p : 12p < −30.
Solution: To isolate the variable p, we must cancel “multiply by 12” using its inverse operation, dividing by 12.
12p −30
<
12
12
Because 12 is not negative, you do not switch the inequality sign.
p<
−5
2
.
�
�
In set notation, the solution would be: −∞, −5
2 .
Vocabulary
Multiplication Property of Inequality: For all real positive numbers a, b, and c:
If x < a, then x(c) < a(c).
If x > a, then x(c) > a(c).
Division Property of Inequality: For all real positive numbers a, b, and c:
If x < a, then x ÷ (c) < a ÷ (c).
If x > a, then x ÷ (c) > a ÷ (c).
19
www.ck12.org
Guided Practice
Solve for m : − m3 < 2.4.
Solution:
To isolate the variable m, we must cancel “divide by 3” using its inverse operation, multiplying by 3. We must also
cancel out the negative, so we would multiply by -1. Multiplying by 3 and -1 means multiplying by -3.
−3 · −
m
< −3 · 2.4
3
Because -3 is negative, you need to switch the inequality sign.
m > −7.2
.
In set notation, the solution would be: (−7.2, ∞).
This means that m must be greater than -7.2, but not equal to it.
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note
that there is not always a match between the number of the practice exercise in the video and the number of the
practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Ba
sic Algebra:Inequalities Using Multiplication and Division (10:27)
MEDIA
Click image to the left for more content.
1. In which cases do you change the inequality sign?
Solve each inequality. Give the solution using inequality notation and with a solution graph.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
20
3x ≤ 6
x
3
5 > − 10
−10x > 250
x
−7 ≥ −5
9x > − 34
x
−15 ≤ 5
620x > 2400
x
7
20 ≥ − 40
−0.5x ≤ 7.5
75x ≥ 125
x
10
−3 > − 9
www.ck12.org
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
Chapter 5. Inequalities with Multiplication and Division
k
−14 ≤ 1
x
−15 < 8
x
2 > 40
x
−3 ≤ −12
x
3
25 < 2
x
−7 ≥ 9
4x < 24
238 < 14d
−19m ≤ −285
−9x ≥ − 35
−5x ≤ 21
Mixed Review
24. After 3 dozen cookies, Anna has fewer than 24 to make.
1. Write an inequality to represent this situation. Let c =the number of cookies Anna had to make.
2. Write the solutions in set notation.
25. Tracey’s checking account balance is $31.85. He needs to deposit enough money to pay his satellite T.V. bill,
which is $97.12.
1. Write an inequality to represent this situation.
2. Write the solutions as an algebraic sentence.
26. Solve for v : v = −|2 − (−19) + 6|.
27. A piggy bank is filled with dimes and quarters. The total amount of money is $26.00.
1. Graph all the combinations that make this statement true.
2. If $13.50 is in quarters, how many dimes must be in the piggy bank?
21