Question

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

Answer 1

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

Explanation:

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

`(2/3)xx((y-4)/(y-4))+(5/(y-4))xx(3/3)= (y+6)/(3y-12)`

After simplifying, you get

`(2y-8)/(3y-12) + 15/(3y-12) = (y+6)/(3y-12)`

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

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

Simplify

`2y+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

`-1/3=-1/3`

Therefore, our solution is not extraneous.