# How do you solve 10/(x^2-25) - 1/10 = 1/(x - 5)?

Apr 15, 2018

${x}_{1} = 5$
${x}_{2} = - 15$

#### Explanation:

${x}^{2} - 25 = \left(x - 5\right) \cdot \left(x + 5\right)$ so we have

$\frac{10}{\left(x + 5\right) \cdot \left(x - 5\right)} - \frac{1}{10} = \frac{1}{x - 5}$

multiply both sides by $10 \left(x + 5\right) \left(x - 5\right)$

$10 \left(x + 5\right) \left(x - 5\right) \frac{10}{\left(x + 5\right) \left(x - 5\right)} - 10 \left(x + 5\right) \left(x - 5\right) \frac{1}{10} = 10 \left(x + 5\right) \left(x - 5\right) \frac{1}{x - 5}$

$10 \textcolor{red}{\cancel{x + 5}} \textcolor{red}{\cancel{x - 5}} \frac{10}{\textcolor{red}{\cancel{x + 5}} \textcolor{red}{\cancel{x - 5}}} - \textcolor{b l u e}{\cancel{10}} \left(x + 5\right) \left(x - 5\right) \frac{1}{\textcolor{b l u e}{\cancel{10}}} = 10 \left(x + 5\right) \textcolor{g r e e n}{\cancel{x - 5}} \frac{1}{\textcolor{g r e e n}{\cancel{x - 5}}}$

$100 - \left({x}^{2} - 25\right) = 10 x + 50$

$100 - {x}^{2} + 25 - 10 x - 50 = 0$

$- {x}^{2} - 10 x + 75 = 0$

${x}^{2} + 10 x - 75 = 0$

now use the formula ${x}_{1 / 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = 1$
$b = 10$
$c = - 75$

${x}_{1 / 2} = \frac{- 10 \pm \sqrt{100 - 4 \left(- 75\right)}}{2}$
${x}_{1 / 2} = \frac{- 10 \pm \sqrt{400}}{2}$
${x}_{1 / 2} = \frac{- 10 \pm 20}{2}$
${x}_{1} = \frac{- 10 + 20}{2} = \frac{10}{2} = 5$
${x}_{2} = \frac{- 10 - 20}{2} = - \frac{30}{2} = - 15$