# How do you solve (x-2)/(3x+5)=6/(x-3) and check for extraneous solutions?

Nov 9, 2016

$x = 24 \text{ or } x = - 1$

#### Explanation:

Identify the restrictions on $x$ before you even solve the equation.
The denominators may not be equal to 0.

$3 x + 5 \ne 0 \text{ and } x - 3 \ne 0$
$3 x \ne - 5 \textcolor{w h i t e}{\times \times \times \times \times x} x \ne 3$
$x \ne - \frac{5}{3}$

As there is one fraction on each side of the equal sign you can cross -multuiply.

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

${x}^{2} - 3 x - 2 x + 6 = 18 x + 30 \text{ } \leftarrow$ a quadratic, so set =0

${x}^{2} - 5 x - 18 x + 6 - 30 = 0$

${x}^{2} - 23 x - 24 = 0 \text{ } \leftarrow$ find factors of 24 which differ by 23

$\left(x - 24\right) \left(x + 1\right) = 0$

$x = 24 \text{ or } x = - 1$

Both solutions are valid.