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

1 Answer
Nov 24, 2017

Answer:

#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.