# How do you solve (4-8x)/(x+1)=8/(x+1) and find any extraneous solutions?

May 21, 2016

Solve using the rule $\frac{a}{b} = \frac{m}{n} \to a \times n = b \times m$

#### Explanation:

$8 \left(x + 1\right) = \left(4 - 8 x\right) \left(x + 1\right)$

$8 x + 8 = 4 x - 8 {x}^{2} - 8 x + 4$

$8 {x}^{2} + 12 x + 4 = 0$

$4 \left(2 {x}^{2} + 3 x + 1\right) = 0$

$4 \left(2 {x}^{2} + 2 x + x + 1\right) = 0$

$4 \left(2 x \left(x + 1\right) + 1 \left(x + 1\right)\right) = 0$

$4 \left(2 x + 1\right) \left(x + 1\right) = 0$

$x = - \frac{1}{2} \mathmr{and} - 1$

Since $x = - 1$ in the original équation makes the denominator equal to 0, and division by 0 in mathematics is undefined, this is an extraneous solution.

Therefore, our solution set is $\left\{x = - \frac{1}{2}\right\}$.

Hopefully this helps!