# How do you solve x/(x-5) - 2/(x+5) = 50/(x^2-25)?

Apr 9, 2016

Start by multiplying both sides by the Least Common Denominator.

#### Explanation:

$\frac{x}{x - 5} - \frac{2}{x + 5} = \frac{50}{{x}^{2} - 25}$

The Least Common Denominator is $\left(x + 5\right) \left(x - 5\right)$ since that is the simplest expression that all denominators will divide into. So all three terms get multiplied by $\left(x + 5\right) \left(x - 5\right)$

$x \left(x + 5\right) - 2 \left(x - 5\right) = 50$
Above we have multiplied all numerators by the LCD and canceled out terms with the denominator where needed.

Now distribute and add like terms.
${x}^{2} + 5 x - 2 x + 10 = 50$
${x}^{2} + 3 x - 40 = 0$
Note that we need to solve for 0 in order to solve by factoring.

Factor the left side: $\left(x + 8\right) \left(x - 5\right) = 0$
Solve each factor for 0: $x = \left\{- 8 , 5\right\}$

HOWEVER the second answer must be eliminated since it would result in a zero denominator (extraneous solution).
So the only valid answer is $x = - 8$

Hope this helps!

Apr 9, 2016

Multiplying both sides by ${x}^{2} - 25$ we get

$\frac{x \left({x}^{2} - 25\right)}{x - 5} - \frac{2 \left({x}^{2} - 25\right)}{x + 5} = \frac{50 \left({x}^{2} - 25\right)}{{x}^{2} - 25}$
$\implies x \left(x + 5\right) - 2 \left(x - 5\right) = 50$
$\implies {x}^{2} + 5 x - 2 x + 10 - 50 = 0$
$\implies {x}^{2} + 8 x - 5 x - 40 = 0$
$\implies x \left(x + 8\right) - 5 \left(x + 8\right)$
$\implies \left(x + 8\right) \left(x - 5\right)$
$\therefore x = - 8 \mathmr{and} x = 5$