How do you solve #(y^2-4)/(y+3) = 2- (y-2)/(y+3)#?

Question

1563 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-26T20:20:17

First multiply all the terms by #(y+3)# to get:

#y^2-4 = 2(y+3) - (y-2)#

#=2y+6-y+2#

#=y+8#

Subtract #(y+8)# from both sides to get:

#y^2 - y - 12 = 0#

This factors as #(y-4)(y+3) = 0#

So #y^2 - y - 12 = 0# has solutions #y=4# and #y=-3#

Of these, #y = -3# is not a solution of the original problem, because it causes the denominators of two of the terms to be zero.

So the unique solution of the original problem is #y = 4#.