# How do you solve for x in  7/(x-2) + 3/(x+2) = -x/(x^2-4 )?

Jul 26, 2018

Here is your answer $x = - \frac{8}{11}$

#### Explanation:

$\frac{7 \left(x + 2\right) + 3 \left(x - 2\right)}{\left(x - 2\right) \left(x + 2\right)} = - \frac{x}{{x}^{2} - 4}$

$\frac{7 x + 14 + 3 x - 6}{{x}^{2} + 2 x - 2 x - 4} = - \frac{x}{{x}^{2} - 4}$

$\frac{10 x + 8}{\cancel{{x}^{2} - 4}} = - \frac{x}{\cancel{{x}^{2} - 4}}$

$10 x + 8 = - x$

$11 x = - 8$

$x = - \frac{8}{11}$

$x = - \frac{8}{11}$

#### Explanation:

Given equation:

$\setminus \frac{7}{x - 2} + \frac{3}{x + 2} = - \frac{x}{{x}^{2} - 4}$

$\setminus \frac{7 \left(x + 2\right) + 3 \left(x - 2\right)}{\left(x - 2\right) \left(x + 2\right)} = - \frac{x}{{x}^{2} - 4}$

$\setminus \frac{10 x + 8}{{x}^{2} - 4} = - \frac{x}{{x}^{2} - 4}$

$\setminus \frac{10 x + 8}{{x}^{2} - 4} + \frac{x}{{x}^{2} - 4} = 0$

$\setminus \frac{10 x + 8 + x}{{x}^{2} - 4} = 0$

$\setminus \frac{11 x + 8}{{x}^{2} - 4} = 0$

$11 x + 8 = 0 \setminus \setminus \quad \left(\setminus \forall \setminus \setminus x \setminus \ne \setminus \pm 2\right)$

$11 x = - 8$

$x = - \frac{8}{11}$

Jul 26, 2018

$x = - \frac{8}{11}$

#### Explanation:

It must be $x \setminus \ne 2 , x \setminus \ne - 2$
multimplying by $\left(x - 2\right) \left(x + 2\right)$ we get

$7 \left(x + 2\right) + 3 \left(x - 2\right) = - x$ expanding we get

$7 x + 14 + 3 x - 6 = - x$
combining like Terms

$10 x + 8 = - x$
adding $- 8$ and $x$
We get

$11 x = - 8$

so $x = - \frac{8}{11}$

Jul 26, 2018

$x = - \frac{8}{11}$

#### Explanation:

$\frac{7}{x - 2} + \frac{3}{x + 2} = - \frac{x}{\textcolor{b l u e}{\left({x}^{2} - 4\right)}} \text{ } \leftarrow$ factorise

$\frac{7}{x - 2} + \frac{3}{x + 2} = - \frac{x}{\textcolor{b l u e}{\left(x + 2\right) \left(x - 2\right)}}$

Multiply each tern in the equation by the LCM of the denominators which is $\textcolor{m a \ge n t a}{\left(x + 2\right) \left(x - 2\right)}$ and cancel where possible:

$\frac{\textcolor{m a \ge n t a}{\left(x + 2\right) \cancel{\left(x - 2\right)}} \times 7}{\cancel{\left(x - 2\right)}} + \frac{3 \times \textcolor{m a \ge n t a}{\cancel{\left(x + 2\right)} \left(x - 2\right)}}{\cancel{\left(x + 2\right)}} = \frac{- x \times \textcolor{m a \ge n t a}{\cancel{\left(x + 2\right) \left(x - 2\right)}}}{\cancel{\left(x + 2\right) \left(x - 2\right)}}$

This leaves:

$7 \left(x + 2\right) + 3 \left(x - 2\right) = - x$

$7 x + 14 + 3 x - 6 = - x$

$10 x + x = - 8$

$11 x = - 8$

$x = - \frac{8}{11}$