# How do you solve 2/3+5/(y-4)=(y+6)/(3y-12) and find any extraneous solutions?

Nov 24, 2017

$y = - 1$ and the solution is not extraneous.

#### Explanation:

First, find the LCM between $3$, $y - 4$, and $3 y - 12$. In this case, it is $3 y - 12$. Then change the fractions so that they share a common denominator. This yields

$\left(\frac{2}{3}\right) \times \left(\frac{y - 4}{y - 4}\right) + \left(\frac{5}{y - 4}\right) \times \left(\frac{3}{3}\right) = \frac{y + 6}{3 y - 12}$

After simplifying, you get

$\frac{2 y - 8}{3 y - 12} + \frac{15}{3 y - 12} = \frac{y + 6}{3 y - 12}$

Multiply both sides by $3 y - 12$ to get

$2 y - 8 + 15 = y + 6$

Simplify

$2 y + 7 = y + 6$

Subtract both sides by $7$, then subtract both sides by $y$ to get

$y = - 1$

Check whether this is extraneous by plugging in $- 1$ to the original equation. If you do, you get

$- \frac{1}{3} = - \frac{1}{3}$

Therefore, our solution is not extraneous.