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May 25, 2015

May 25, 2015

Posted by **Martin** on Thursday, June 16, 2011 at 9:21pm.

Use set-builder notation to describe the complete solution

{m|m > ? }

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

- Mat 116 Algebra 1A -
**Henry**, Friday, June 17, 2011 at 1:52pm7[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.