# How do you solve the rational equation (4-8x)/(1-x+4)=8/(x+1)?

Jun 4, 2018

$\frac{4 - 8 x}{1 - x + 4} = \frac{8}{x + 1}$

$\frac{4 - 8 x}{5 - x} = \frac{8}{x + 1}$

Cross multiply to remove the fractions

$\left(4 - 8 x\right) \left(x + 1\right) = 8 \left(5 - x\right)$

Expand the brackets

$4 x + 4 - 8 {x}^{2} - 8 x = 40 - 8 x$

$4 - 4 x - 8 {x}^{2} = 40 - 8 x$

Add $8 {x}^{2}$ to both sides

$4 - 4 x = 8 {x}^{2} - 8 x + 40$

add $4 x$ to both sides

4=8x^2-4x+40#

subtract 4 from both sides

$8 {x}^{2} - 4 x + 36 = 0$

Divide both sides by 4

$2 {x}^{2} - x + 9 = 0$

Put into the quadratic formula

$x = \frac{1 \setminus \pm \sqrt{1 - 4 \times 2 \times 9}}{2 \times 2}$

$x = \frac{1 \setminus \pm \sqrt{- 71}}{4}$

No solutions as you have a negative number in the square root sign