# Mat 116 Algebra 1A

7[3m-(m+2)]>-6(m+3)

Use set-builder notation to describe the complete solution

{m|m > ? }

(type an equality symbol; then type an integer or a simplified fraction.)

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1. 7[3m-(m+2)] > -6(m+3).
Since the absolute value of a positive
number is = the absolute value of a negative number of the same magnitude,
we have 2 solutions.

+-7(3m-(m+2)) > -6(m+3),

+7(3m-m-2) > -6m-18,
7(2m-2) > -6m-18,
14m-14 > -6m-18,
14m + 6m > -18 + 14,
20m > -4,
m > -1/5.

-7(3m-(m+2)) > -6(m+3),
-7(3m-m-2) > -6m-18,
-7(2m-2) > -6m-18,
-14m + 14 > -6m-18,
-14m + 6m > -18-14,
-8m > -32,
m < 4.

Solution set: m > -1/5, and m < 4.

OR
-1/5 < m < 4.

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