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March 31, 2015

March 31, 2015

Posted by **Nick K.** on Sunday, June 7, 2009 at 9:49pm.

A. {4/3, -4}

B. {x|x > 4 or x < -4/5}

C. No Solution

D. {x| -11 < x < -7}

E. All Real Numbers

Solution set

ABCDE|x + 9| < 2

ABCDE|5x - 8| > 12

ABCDE|3k + 4| = 8

ABCDE|x - 6| = -10

ABCDE|6x + 8| > -4

- Algebra -
**MathMate**, Sunday, June 7, 2009 at 11:53pmIt is easier to solve for the unknown and then look for the question.

Working backwards:

|x + 9| < 2

Solution set: -7<x<-11

check: |-7 + 9| =2

|-11 + 9| = -2

|5x - 8| > 12

x>4 or x<-4/5

|3k + 4| = 8

k=4/3 or k=-4

|x - 6| = -10

no solution, ∀ y ∈ ℝ: |y| ≥ 0

ABCDE|6x + 8| > -4

x∈ℝ

Can you take it from here?

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