# How do you solve \frac { 4} { x - 3} = \frac { x + 5} { 5} ]?

Jan 11, 2018

${x}_{1} = 5$ and ${x}_{2} = - 7$

#### Explanation:

$\frac{4}{x - 3} = \frac{x + 5}{5}$

Perform this
$\frac{4}{x - 3}$χ$\frac{x + 5}{5}$

$\iff$ $4 \cdot 5 = \left(x - 3\right) \left(x + 5\right)$ $\iff$

$20 = {x}^{2} + 5 x - 3 x - 15 \iff$

${x}^{2} + 2 x - 35 = 0$

Discriminant will be Δ=b^2-4ac=4-4*1*(-35)=140+4=144

x_(1,2)=(-b+-sqrt(Δ))/(2α) so

${x}_{1} = \frac{- 2 + 12}{2} = 5$

${x}_{2} = \frac{- 2 - 12}{2} = - \frac{14}{2} = - 7$

As a result the roots are ${x}_{1} = 5$ and ${x}_{2} = - 7$