How do you solve $m - 3 = \frac{4}{6} m - 2$ $m- 3= \frac { 4} { 6} m - 2$ ?

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Answer 1 · 2018-06-04T11:11:34

$m = 3$ $m = 3$

Firstly, multiply both sides by 6 as the $\frac{4}{6} m$ $4/6m$ looks ugly and we don't want it

$6 m - 18 = 4 m - 12$ $6m - 18 = 4m-12$

from there, take the 4m over so it ends up being $6 m - 4 m - 18 = - 12$ $6m-4m-18=-12$ simplify $2 m - 18 = - 12$ $2m-18 = -12$

next, move the constants to one side, i'll move the -18 over making + 18, so that $2 m = 6$ $2m=6$ and thus $m = 3$ $m=3$

Answer 2 · 2018-06-04T12:50:32

$m = 3$ $m=3$

$m - 3 = \frac{4}{6} m - 2$ $m-3=4/6m-2$

multiply both sides by $6$ $6$

$:.6m-18=4m-12$

$:.6m-4m-18+12=0$

$:.2m-6=0$

$:.2m=6$

$:.m=6/2$

$:.m=3$

How do you solve m- 3= \frac { 4} { 6} m - 2m−3=46m−2m- 3= \frac { 4} { 6} m - 2?