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8 January, 20:06

How do you solve, 3r+2 (12r+7) less then or equal to 5r-8

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  1. 8 January, 20:29
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    r ≤ - 1

    Step-by-step explanation:

    Given in the question

    3r+2 (12r+7) ≤ 5r-8It can be solve by Simplifying and Reordering the terms

    3r + 2 (12r) + 2 (7) ≤ 5r - 8

    3r + 24r + 14 ≤ 5r - 8

    27r + 14 ≤ 5r - 8

    27r - 5r + 14 + 8 ≤ 0

    22r + 22 ≤ 0

    22r ≤ - 22

    Move term containing r to the left, all term to the right.

    r ≤ - 22/22

    simplifying

    r ≤ - 1

    all negative integers which are smaller than or equal to - 1
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