# How do you solve (3x+2)/(2x-1)=(5x+6)/(x+4)?

Sep 25, 2016

$x = 2$ or $x = - 1$

#### Explanation:

Multiply both sides of the equation by $\left(2 x - 1\right) \left(x + 4\right)$ to get:

$\left(3 x + 2\right) \left(x + 4\right) = \left(5 x + 6\right) \left(2 x - 1\right)$

Multiply out both sides to get:

$3 {x}^{2} + 14 x + 8 = 10 {x}^{2} + 7 x - 6$

Subtract the left hand side from both sides to get:

$0 = 7 {x}^{2} - 7 x - 14$

$\textcolor{w h i t e}{0} = 7 \left({x}^{2} - x - 2\right)$

$\textcolor{w h i t e}{0} = 7 \left(x - 2\right) \left(x + 1\right)$

Hence $x = 2$ or $x = - 1$

It remains to check that neither of these values causes a denominator to become $0$. The only values that do are $x = \frac{1}{2}$ and $x = - 4$ so we're good.