# Question #a71e9

Sep 19, 2016

$x = \textcolor{g r e e n}{- \frac{10}{3}}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \frac{4}{x + 2} + \frac{1}{x + 3} = \frac{8}{x + 2}$

Multiplying both sides by $\left(x + 2\right) \left(x + 3\right)$
$\textcolor{w h i t e}{\text{XXX}} 4 \left(x + 3\right) + 1 \left(x + 2\right) = 8 \left(x + 3\right)$

Simplifying
$\textcolor{w h i t e}{\text{XXX}} 5 x + 14 = 8 x + 24$

$\textcolor{w h i t e}{\text{XXX}} - 3 x = 10$

$\textcolor{w h i t e}{\text{XXX}} x = - \frac{10}{3} \left(= 3 \frac{1}{3}\right)$

Sep 19, 2016

$x = - \frac{10}{3}$

#### Explanation:

Basically we need to isolate $x$ so I will first multiply both sides by $x + 2$

$4 + \frac{x + 2}{x + 3} = 8$
now move $4$ over
$\frac{x + 2}{x + 3} = 4$
now multiply both sides by $x + 3$
$x + 2 = 4 x + 12$
Move like terms to each side
$- 3 x = 10$
Solve for $x$
$x = - \frac{10}{3}$

Sep 19, 2016

$x = - \frac{10}{3}$
This solution is not the most efficient on purpose: The objective is to demonstrates a basic principle that may be used else ware.

#### Explanation:

Given:$\text{ } \frac{4}{x + 2} + \frac{1}{x + 3} = \frac{8}{x + 2}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\left[\frac{4}{x + 2} \textcolor{red}{\times 1}\right] + \left[\frac{1}{x + 3} \textcolor{b l u e}{\times 1}\right] \text{ "=" } \left[\frac{8}{x + 2} \textcolor{g r e e n}{\times 1}\right]$

$\textcolor{w h i t e}{.}$

$\left[\frac{4}{x + 2} \textcolor{red}{\times \frac{x + 3}{x + 3}}\right] + \left[\frac{1}{x + 3} \textcolor{b l u e}{\times \frac{x + 2}{x + 2}}\right] = \left[\frac{8}{x + 2} \textcolor{g r e e n}{\times \frac{x + 3}{x + 3}}\right]$

$\textcolor{w h i t e}{.}$

$\left[\frac{4 \left(x + 3\right)}{\left(x + 2\right) \left(x + 3\right)}\right] + \left[\frac{x + 2}{\left(x + 2\right) \left(x + 3\right)}\right] = \left[\frac{8 \left(x + 3\right)}{\left(x + 2\right) \left(x + 3\right)}\right]$
$\textcolor{w h i t e}{.}$

Multiply both sides by $\left(x + 2\right) \left(x + 3\right)$ giving:

$4 \left(x + 3\right) \text{ "+" "x+2" "=" } 8 \left(x + 3\right)$

$\textcolor{w h i t e}{.} 4 x + 12 \text{ "+" "x+2" "=" } 8 x + 24$

$\text{ "5x+14" "=" } 8 x + 24$

$\text{ "-10" "=" } 3 x$

$\text{ } x = - \frac{10}{3}$