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1.6: Solve Linear
Inequalities
I can:
Solve a linear and/or compound inequality and graph
the solution set.
Manipulate a formula to solve for a particular variable.

Key Vocabulary
• Linear Inequality
• Compound Inequality
• Equivalent Inequalities

Linear Inequality
• Can be written in one of the following forms where a and b are
real numbers and a does not equal 0.

Algebra Inequality presentation power point.pptx

Compound Inequality
• Consists of two simple inequalities joined by “and” or “or”

GUIDED PRACTICE for Examples 1 and 2
Graph the inequality.
1. x > – 5
The solutions are all real numbers greater than 5.
An open dot is used in the graph to indicate – 5 is
not a solution.

2. x ≤ 3
The solutions are all real numbers less than or
equal to 3.
A closed dot is used in the graph to indicate 2 is a
solution.

3. – 3 ≤ x < 1
The solutions are all real numbers that are greater
than or equalt to – 3 and less than 1.

4. x < 1 or x ≥ 2
The solutions are all real numbers that are less than 1
or greater than or equal to 2.

Why?
• Near the end of the semester or school year, I have students
ask me what scores they must get on a final in order to earn
a particular grade. You can use linear inequalities to
answer this question.
• You budgeted x amount of dollars for movies. You have
already spend $50 of the amount. You can use inequalities
to determine staying under budget or reaching your
budgeted target.

Solve the inequality. Then graph the solution.
5. 4x + 9 < 25
4x + 9 < 25 Write original inequality.
4x < 16 Subtract 9 from each side.
x < 4 Divide each side by 4.
6. 1 – 3x ≥ – 14
1 – 3x ≥ – 14 Write original inequality.
– 3x ≥ – 15 Subtract –1 from each side.
x ≤ 5

7. 5x – 7 ≤ 6x
8. 3 – x > x – 9
5x – 7 ≤ 6x Write original inequality.
x > – 7 Subtract 5x from each side.
3 – x > x – 9 Write original inequality.
3 – 2x > – 9
– 2x > – 12
x < 6
Subtract x from each side.
Subtract 3 from each side.
Divide each side by –2 and reverse.

EXAMPLE 6 Solve an “or” compound inequality
Solve 3x + 5 ≤
11 or 5x – 7 ≥ 23
. Then graph the solution.
SOLUTION
A solution of this compound inequality is a solution of
either of its parts.
First Inequality Second Inequality
3x + 5 ≤
11 3x ≤
6
x ≤
2
Write first inequality.
Subtract 5 from each side.
Divide each side by 3.
5x – 7 ≥
23 5x ≥
30
x ≥ 6
Write second inequality.
Add 7 to each side.

GUIDED PRACTICE for Examples 5,6, and 7
9. – 1 < 2x + 7 < 19
– 1 < 2x + 7 < 19 Write original inequality.
Subtract 7 from each expression.
Simplify.
Divide each expression by 2.
– 1– 7 < 2x + 7 – 7 < 19 – 7
– 8 < 2x < 12
– 4 < x < 6
ANSWER
The solutions are all real numbers greater than – 4
and less than 6.

GUIDED PRACTICE for Examples 5,6 and 7
10. – 8 ≤
– x – 5 ≤
6
– 8 ≤– x – 5 ≤
6
– 8+5 ≤– x – 5 + 5
≤ 6 + 5
– 3 ≤– x ≤ 11
– 11 ≤ x ≤
3
Write original inequality.
Add 5 to each expression.
Simplify.
The solutions are all real numbers greater than and
equal to – 11 and less than and equal to 3.
ANSWER

11. x + 4 ≤
9 or x – 3 ≥ 7
SOLUTION
x ≤
5
Subtract 4 from each side. x ≥
10
Add 3 to each side.
x + 4 ≤
9
x – 3 ≥ 7

ANSWER
The graph is shown below. The solutions are all real
numbers.
less than or equal to 5 or greater than or equal to 10.

12. 3x – 1 < – 1 or 2x + 5 ≥
11
SOLUTION
3x ≤
0
x ≤
0
Add 1 each side .
2x + 5 ≥
11 2x ≥ 6
x ≥ 3
Subtract 5 from each side
3x – 1< – 1

less than 0 or greater than or equal to 3.
ANSWER
The graph is shown below. The solutions are all real
numbers.

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Algebra Inequality presentation power point.pptx

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