# How do you solve 12(6m+2)=-6(3m+67) ?

Mar 16, 2018

$m = - \frac{71}{15} = - 4 \frac{11}{15}$

#### Explanation:

Solve:

$12 \left(6 m + 2\right) = - 6 \left(3 m + 67\right)$

Divide both sides by $- 6$.

$\frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}}}^{2} \left(6 m + 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 6}}}} ^ 1 = \frac{- {\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}^{1} \left(3 m + 67\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 6}}}} ^ 1$

Simplify.

$- 2 \left(6 m + 2\right) = 3 m + 67$

Expand.

$- 12 m - 4 = 3 m + 67$

Add $4$ to both sides.

$- 12 m - 4 + 4 = 3 m + 67 + 4$

Simplify.

$- 12 m - 0 = 3 m + 71$

$- 12 m = 3 m + 71$

Subtract $3 m$ from both sides.

$- 12 m - 3 m = 3 m - 3 m + 71$

Simplify.

$- 15 m = 0 + 71$

$- 15 m = 71$

Divide both sides by $- 15$.

$\frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}}}^{1} m}{{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}}}^{1}} = \frac{71}{- 15}$

Simplify.

$m = - \frac{71}{15}$

We can convert this improper fraction to a mixed number $\textcolor{red}{a} \frac{\textcolor{b l u e}{b}}{\textcolor{g r e e n}{c}}$. Divide the numerator by the denominator by long division. The whole number quotient is the whole number $\left(\textcolor{red}{a}\right)$ of the mixed number, the remainder is the numerator $\left(\textcolor{b l u e}{b}\right)$, and the denominator (divisor) stays the same $\left(\textcolor{g r e e n}{c}\right)$.

$- 71 \div 15 = - \text{4 remainder 11}$

m=-71/15=-color(red)(4) color(blue)(11)/color(green)(15