# How do you evaluate \frac { 2x } { x ^ { 2} - 25} - \frac { 1} { x - 5} = \frac { 1} { 6}?

Jan 5, 2017

$x = 1$ is the only solution
$x = 5$ is an extraneous solution.

#### Explanation:

$\frac{2 x}{{x}^{2} - 25} - \frac{1}{x - 5} = \frac{1}{6}$

multiply both sides by$\left({x}^{2} - 25\right)$

Note that $\textcolor{red}{{x}^{2} - 25 = \left(x + 5\right) \left(x - 5\right)}$

$2 x - \frac{\textcolor{red}{\left(x + 5\right) \cancel{\left(x - 5\right)}}}{\cancel{\left(x - 5\right)}} = \frac{{x}^{2} - 25}{6}$

$2 x - \left(x + 5\right) = \frac{{x}^{2} - 25}{6}$

multiply both sides by 6

$12 x - 6 x - 30 = {x}^{2} - 25$

$6 x - 30 = {x}^{2} - 25$

$6 x - {x}^{2} + 25 - 30 = 0 \text{ }$ (make a quadratic equal to 0)

$- {x}^{2} + 6 x - 5 = 0$

${x}^{2} - 6 x + 5 = 0 \text{ }$ make $+ {x}^{2}$

$\left(x - 5\right) \left(x - 1\right) = 0 \text{ }$ find factors

$x = 5 \mathmr{and} x = 1$

Check:
$x = 5$ will give : $\frac{2 x}{{x}^{2} - 25} \rightarrow \frac{10}{0}$

Division by 0 is not defined, so $x \ne 5$

$x = 1$

$\frac{2 \left(1\right)}{{1}^{2} - 25} - \frac{1}{1 - 5} = \frac{1}{6}$

$\frac{2 \left(1\right)}{1 - 25} - \frac{1}{1 - 5} = \frac{1}{6}$

${\cancel{2}}^{1} / {\cancel{- 24}}^{-} 12 - \frac{1}{- 4} = \frac{1}{6}$

$\frac{1}{- 12} - \frac{1}{- 4} = \frac{1}{6}$

$- \frac{1}{12} + \frac{1}{4} = \frac{1}{6}$

$\frac{- 1 + 3}{12} = \frac{1}{6}$

${\cancel{2}}^{1} / {\cancel{12}}^{6} = \frac{1}{6}$

$\frac{1}{6} = \frac{1}{6}$