# How do you solve \frac{m}{m + 5} = \frac{6}{8}?

Sep 16, 2016

Explained using first principles showing every step.
Much faster to use the shortcuts.

$m = 15$

#### Explanation:

$\textcolor{b l u e}{\text{Principle for add, subtract, multiply and divide:}}$

To move to the other side of the = reverse the process. So for multiply you divide, for add you subtract and so on.

$\textcolor{b r o w n}{\text{What you do to one side you do to the other.}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

Given:$\text{ } \frac{m}{m + 5} = \frac{6}{8}$

Multiply both sides by $\left(m + 5\right) \leftarrow \text{ gets rid of the LHS } \frac{1}{m + 5}$

$\frac{m + 5}{m + 5} \times m = \frac{6}{8} \times \left(m + 5\right)$

But $\frac{m + 5}{m + 5} = 1$

$1 \times m = \frac{6}{8} \times \left(m + 5\right)$

$m = \frac{6}{8} \left(m + 5\right)$

Multiply both sides by 8 $\leftarrow \text{ gets rid of the RHS } \frac{1}{8}$

$8 m = \frac{8}{8} \times 6 \left(m + 5\right)$

But $\frac{8}{8} = 1$

$8 m = 6 \left(m + 5\right)$

Multiply out the bracket

$8 m = 6 m + 30$

Subtract 6m from both sides $\leftarrow \text{ gets rid of the RHS } 6 m$

$2 m = 30$

Divide both sides by 2 $\leftarrow \text{ gets rid of the RHS 2 from } 2 m$

$m = 15$

Sep 16, 2016

$8 m = 6 \left(m + 5\right)$
$8 m = 6 m + 30$
$8 m - 6 m = 30$
$2 m = 30$
$m = \frac{30}{2}$
$m = 15$