SOLUTION: 3y/3y-6 + 8/y^2-4 = 2y/2y+4 I got thi problem wrong in my test please help me

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Question 162854This question is from textbook beginning and intermidate algebra
: 3y/3y-6 + 8/y^2-4 = 2y/2y+4 I got thi problem wrong in my test please help meThis question is from textbook beginning and intermidate algebra

Answer by ankor@dixie-net.com(15660) About Me  (Show Source):
You can put this solution on YOUR website!
%283y%29%2F%28%283y-6%29%29 + 8%2F%28%28y%5E2-4%29%29 = %282y%29%2F%28%282y%2B4%29%29
:
First, recognize that you can factor out 3 and 2 in the denominators
Note that y^2-4 is the "difference of squares"
%283y%29%2F%283%28y-2%29%29 + 8%2F%28%28y-2%29%28y%2B2%29%29 = %282y%29%2F%282%28y%2B2%29%29
:
Note that the 3's and 2's will cancel leaving:
y%2F%28%28y-2%29%29 + 8%2F%28%28y-2%29%28y%2B2%29%29 = y%2F%28%28y%2B2%29%29
:
Multiply equation by (x-2)(x+2), results:
y(y+2) + 8 = y(y-2)
:
y^2 + 2y + 8 = y^2 - 2y
:
arrange the variables on the left, and numbers on the right
y^2 - y^2 + 2y + 2y = -8
:
4y = -8
y = %28-8%29%2F4
y = -2
This should be the solution, but when we substitute -2 for y in the original
equation, we end up with 2 denominators of 0, which we can't have
therefore we have to say, " No solution"