Page 31 - Math Course 3 (Book 1)

More Absolute Value Inequalities

Mo. 1

Lesson 6 Solve |x + 6| < –8.

Since |x + 6| cannot be negative, |x + 6| cannot be
less than –8. So, the solution is the empty set Ø.
KEY CONCEPTS:
1. Solving absolute value inequalities.
Answer Ø

MO. 1 - L6a Solve an Absolute Value Inequality (>)

Solving Absolute Value Examples
Inequalities

Solve |3y – 3| > 9. Then graph the solution set.
Concept Summary Case 1 3y – 3 is positive

Absolute Value Equations and Inequalities
3y – 3 > 9 Original inequality
If | x | = n, then x = –n or x = n.

If | x | = n, then x < n and x > –n. 3y – 3 + 3 > 9 + 3 Add 3 to each side.
If | x | = n, then x > n or x < –n.
3y > 12 Simplify.
These properties are also true when > or < is 3y 12
replaced with > or < > Divide each side by 3.
3 3
y > 4 Simplify.
Let’s Begin Case 2 3y – 3 is negative.

3y – 3 < –9 Original inequality
Solve an Absolute Value Inequality (<) 3y – 3 + 3 < –9 + 3 Add 3 to each side.

Examples 3y < –6 Simplify.

3y < –6
3 3 Divide each side by 3.
Solve |s – 3| ≤ 12. Then graph the solution set.
y < –2 Simplify.
Write |s – 3| ≤ 12 as s – 3 ≤ 12 and s – 3 ≥ –12.
The solution set is
Case 1 Case 1 Answer {y | y < –2 or y > 4}.

s – 3 ≤ 12 Original s –3 ≥ –12
inequality.
–3 –2 –1 0 1 2 3 4 5
s – 3 + 3 ≤ 12 + 3 Add 3 to s – 3 + 3 ≥ –12 + 3
each side.
Solve |2x + 7| ≥ –11.
s ≤ 15 Simplify. s ≥ –9
Since |2x + 7| is always greater
The solution set is
Answer Answer than or equal to 0, the solution
{s | –9 ≤ s ≤ 15}. set is {x | x is a real number}.

–15 –10 –5 0 5 10 15
–5 –4 –3 –2 –1 0 1 2 3 4 5

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