# How do you solve x/(x-2)+1/(x-4)=2/(x^2-6x+8) and check for extraneous solutions?

Aug 26, 2016

$x = - 1$

#### Explanation:

We have an equation with fractions. This means we can get rid of the denominators.

Factor the right side denominator first.
${x}^{2} - 6 x + 8 = \left(x - 4\right) \left(x - 2\right)$

$\frac{x}{\left(x - 2\right)} + \frac{1}{\left(x - 4\right)} = \frac{2}{\left(x - 2\right) \left(x - 4\right)}$

$x \ne 4 , x \ne 2$

To make it fit on one line, let color(red)((x-4) = a) and color(blue)((x-2) = b)

The LCM = $\textcolor{red}{\left(x - 4\right) \textcolor{b l u e}{\left(x - 2\right)} = \textcolor{red}{a}} \times \textcolor{b l u e}{b}$

Multiplying all the terms by this LCM .

$\frac{\textcolor{red}{a} \times \textcolor{b l u e}{\cancel{b}} \times x}{\cancel{x - 2}} + \frac{\textcolor{red}{\cancel{a}} \times \textcolor{b l u e}{b} \times 1}{\cancel{x - 4}} = \frac{\textcolor{red}{\cancel{a}} \times \textcolor{b l u e}{\cancel{b}} \times 2}{\cancel{x - 2} \cancel{x - 4}}$

Much better! $x \left(x - 4\right) + x - 2 = 2$

${x}^{2} - 4 x + x - 2 = 2 \text{ make} = 0$

${x}^{2} - 3 x - 4 = 0 \text{ find factors}$

$\left(x - 4\right) \left(x + 1\right) = 0$

$x = 4 , \mathmr{and} x = - 1$

However, $x = 4$ is not permissible.

$x = - 1$