How do you solve $\frac{- 2 (3 - 4 m)}{5} = \frac{2 m + 1}{2}$ $(-2(3-4m))/5=( 2m+1)/2$ ?

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How do you solve $\frac{- 2 (3 - 4 m)}{5} = \frac{2 m + 1}{2}$ $(-2(3-4m))/5=( 2m+1)/2$ ?

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Answer 1 · 2016-10-09T09:21:18

$m = 2 \frac{5}{6}$ $m=2(5)/6$

Explanation:

$\frac{- 2 (3 - 4 m)}{5} = \frac{2 m + 1}{2}$ $(-2(3-4m))/5 = (2m+1)/2$
$\frac{- 6 + 8 m}{5} = \frac{2 m + 1}{2}$ $(-6+8m)/5 = (2m+1)/2$
$2 (8 m - 6) = 5 (2 m + 1)$ $2(8m-6) = 5(2m+1)$
$16 m - 12 = 10 m + 5$ $16m-12 = 10m+5$
$16 m - 10 m = 5 + 12$ $16m-10m = 5+12$
$6 m = 17$ $6m = 17$
$m = \frac{17}{6}$ $m=17/6$
$m = 2 \frac{5}{6}$ $m=2(5)/6$

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How do you solve $\frac{- 2 (3 - 4 m)}{5} = \frac{2 m + 1}{2}$ $(-2(3-4m))/5=( 2m+1)/2$ ?

1 Answer

Explanation:

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How do you solve −2(3−4m)5=2m+12(-2(3-4m))/5=( 2m+1)/2?

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How do you solve $\frac{- 2 (3 - 4 m)}{5} = \frac{2 m + 1}{2}$ $(-2(3-4m))/5=( 2m+1)/2$ ?