# How do you solve 7/(x+5)+3/(x-5)=30/(x^2-25) and check for extraneous solutions?

Feb 17, 2017

I got that it has no real solutions and only one extraneous.

#### Explanation:

Have a look:

Feb 17, 2017

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

#### Explanation:

First we assert that $x \ne - 5 \mathmr{and} x \ne + 5$

After that, we we multiply the first two terms by disguised $1$'s:

$\frac{7}{x + 5} \times \frac{x - 5}{x - 5} = \frac{7 \left(x - 5\right)}{{x}^{2} - 25}$ and:

$\frac{3}{x - 5} \times \frac{x + 5}{x + 5} = \frac{3 \left(x + 5\right)}{{x}^{2} - 25}$

$\frac{7 \left(x - 5\right)}{{x}^{2} - 25} + \frac{3 \left(x + 5\right)}{{x}^{2} - 25} = \frac{30}{{x}^{2} - 25} \to$
$\frac{7 x - 35 + 3 x + 15}{{x}^{2} - 25} = \frac{30}{{x}^{2} - 25} \to$
$10 x - 20 = 30 \to 10 x = 50 \to x = + 5$