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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