# How do you solve 7/(x+1)=6/(x-5) and check for extraneous solutions?

Aug 26, 2016

$x = 41$

#### Explanation:

Before we even start with solving, we can immediately exclude two values of x as solutions.

The denominator of a fraction may not be equal to 0 so any value of x which causes this may not be a solution.

If $x + 1 = 0 \Rightarrow x = - 1 \text{ " and " } \mathmr{if} x - 5 = 0 \Rightarrow x = 5$

Therefore $x \ne - 1 \mathmr{and} x \ne 5$

We have an equation with one term on each side.
Cross-multiply.

$\frac{7}{x + 1} = \frac{6}{x - 5}$

$7 \left(x - 5\right) = 6 \left(x + 1\right)$

$7 x - 35 = 6 x + 6$

$x = 41$