2-5
Solving Inequalities with
Variables on Both Sides
Going Deeper
Essential question: How can you use properties to justify solutions to inequalities
with variables on both sides?
Standards for
Mathematical Content
Extra Example
Solve 2x - 3 ≤ 3(4 - x). Justify each step and graph
your solution.
A-REI.2.3 Solve linear inequalities…in one
variable…
2x - 3 ≤ 12 - 3x [Distributive Property]
2x + 3x - 3 ≤ 12 - 3x + 3x [Add. Prop. of Inequality]
5x - 3 ≤ 12 [Combine like terms; Simplify]
5x - 3 + 3 ≤ 12 + 3 [Addition Property of Inequality]
5x ≤ 15 [Combine like terms; Simplify]
Prerequisites
Solving Inequalities by Adding and Subtracting
Solving Inequalities by Multiplying and Dividing
Solving Two-Step and Multi-Step Inequalities
___
​ 5x
  ​ ≤ ___
​ 15
  ​ [Division Property of Inequality]
5
5
x ≤ 3 [Simplify.]
{x | x ≤ 3}; the graph has a closed circle at 3 and the
line to the left of 3 is shaded.
Math Background
Students have solved a variety of inequalities, but
the variables have been on one side only. This
lesson extends to inequalities with variables on
both sides of the inequality symbol. Students will
use many of the techniques they have used before,
but these inequalities are more complex and often
require more steps and justifications. These more
complex inequalities provide opportunities to vary
the order of steps and justification in the solutions.
CLOSE
Essential Question
How can you use properties to justify solutions to
inequalities with variables on both sides?
For each step, cite the property you use, or state
that you are simplifying or combining like terms.
INTR O D U C E
Review the first goal of solving any equation or
inequality with students: getting the variable term
alone on one side of the equal/inequality sign
and getting the constant alone on the other side.
Ask them to verbalize the goal in their own words.
Discuss whether the Properties of Inequality would
apply to variable terms also.
Highlighting
the Standards
TEAC H
1
The exercises provide opportunities to
address Mathematical Practices Standards
2 (Reason abstractly and quantitatively), 3
(Construct viable arguments and critique
the reasoning of others), and 6 (Attend to
precision).
example
Questioning Strategies
• Does it matter if you isolate the variable term first
or the constant term first? Explain. No; the order
of the steps would be different, but the solution
would be the same.
PRACTICE
• What property allows you to isolate the x-term on
the left? Subtraction Property of Inequality
Where skills are
taught
1 EXAMPLE
Chapter 2 95
Where skills are
practiced
EXS. 1–3
Lesson 5
© Houghton Mifflin Harcourt Publishing Company
Summarize
Have students write step-by-step methods for
solving inequalities with variables on both sides.
One method should involve the distributive
property.
Name
Class
Notes
2-5
Date
Solving Inequalities with Variables
on Both Sides
Going Deeper
Essential question: How can you use properties to justify solutions of inequalities with
variables on both sides?
You can use the Properties of Inequality and the Distributive Property to justify the
steps in a solution when solving inequalities that have variables on both sides.
A-REI.2.3
1
EXAMPLE
Using Properties to Justify Solutions
Find the solution set. Justify each step and graph your solution.
3(2x - 3) ≤ x + 1
Distributive
6x - 9 ≤ x + 1
5x - 9 ≤
1
Simplify.
Addition Property of Inequality
1+9
5x - 9 + 9 ≤
5x ≤ 10
5x
__
10
≤ __
x
≤ 2
5
© Houghton Mifflin Harcourt Publishing Company
Simplify.
Division Property of Inequality
5
Simplify.
{x x ≤ 2}
Solution set:
Property
Subtraction Property of Inequality
6x - x - 9 ≤ x - x + 1
–4 –3 –2 –1
0
1
2
3
4
5
6
7
8
9
REFLECT
1a. Why is the Distributive Property applied first in the solution?
The Distributive Property is used to eliminate the parentheses.
1b. Could the properties have been applied in a different order than shown above? If
so, would this make finding the solution easier or more difficult? Explain.
Sample answer: Yes, you could apply the Division Property of Inequality first, but
that would give you fractions on the right side of the inequality, which would
make it more difficult to solve.
1c. How would the solution change if the simplified coefficient of x were negative?
If the simplified coefficient were negative, you would have to divide by a negative
number. This means you would have to reverse the inequality symbol at that point.
Chapter 2
Lesson 5
95
PRACTICE
Find the solution set. Justify each step and graph your solution.
1. 21x + 28 < 10 - 3x
24x + 28 < 10
24x + 28 - 28 < 10 - 28
24x < -18
18
24x < -__
___
24
24
3
x < -__
Solution set:
4
⎧
3⎫⎬
⎨x x < -__
4⎭
⎩
Addition Property of Inequality
Simplify.
Subtraction Property of Inequality
Simplify.
Division Property of Inequality
Simplify.
–4 –3 –2 –1
0
1
2
3
4
5
6
7
8
9
1 (x + 2) ≥ 7x + 3
2. -__
3
1 (x + 2) ≤ -3(7x + 3)
-3 -__
3
( )
x + 2 ≤ -21x - 9
x + 21x + 2 ≤ -21x + 21x - 9
22x + 2 ≤ -9
22x + 2 - 2 ≤ -9 - 2
22x ≤ -11
22x ÷ 22 ≤ -11 ÷ 22
1
x ≤ -__
2
⎧
1⎫⎬
⎨x x ≤ -__
2⎭
Solution set:
⎩
Multiplication Property of Inequality
Simplify.
Addition Property of Inequality
Simplify.
Subtraction Property of Inequality
Simplify.
© Houghton Mifflin Harcourt Publishing Company
© Houghton Mifflin Harcourt Publishing Company
21x + 3x + 28 < 10 - 3x + 3x
Division Property of Inequality
Simplify.
–4 –3 –2 –1
0
1
2
3
4
5
6
7
8
9
3. 2(4 - 3x) ≤ 4x - 2
8 - 6x ≤ 4x - 2
8 - 6x - 4x ≤ 4x - 4x - 2
8 - 10x ≤ - 2
8 - 8 - 10x ≤ -2 - 8
-10x ≤ -10
-10x ≤ ____
-10
_____
-10
-10
x≥1
Solution set:
Chapter 2
Chapter 2
{x x ≥ 1}
Distributive Property
Subtraction Property of Inequality
Simplify.
Subtraction Property of Inequality
Simplify.
Division Property of Inequality
Reverse inequality symbol; simplify.
-4 -3 -2 -1
96
0
1
2
3
4
5
6
7
8
9
Lesson 5
96
Lesson 5
Problem Solving
ADD I T I O NA L P R AC TI C E
AND PRO BL E M S O LV I N G
1. 5r > 50 + 2r; r > 16
2. 25,000 + 1000y > 19,000 + 1500y; y < 12
Assign these pages to help your students practice
and apply important lesson concepts. For
additional exercises, see the Student Edition.
3. 75m > 510 + 60m; m > 34
4. 15 + 5s > 8s; s < 5
Answers
5. B
6. F
Additional Practice
1. x ≤ 6
2. k > 3
3. b ≤ 3
4. n > -14
5. s < -1
6. x ≥ 25
© Houghton Mifflin Harcourt Publishing Company
7. z < 2
2
8. p ≤ _
3
9. all real numbers
10. no solutions
11. no solution
12. 9.95 m < 4.95 m + 49.95; m < 9.99;
for 0 to 9 months
13. 7(x + 2) > 7 + (x + 2) + 7 + (x + 2); x > 0.8
Chapter 2
97
Lesson 5
Name
Class
Notes
2-5
Date
Additional Practice
© Houghton Mifflin Harcourt Publishing Company
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