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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 Click here to see ALL problems on test 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)   (Show Source): You can put this solution on YOUR website! + = : First, recognize that you can factor out 3 and 2 in the denominators Note that y^2-4 is the "difference of squares" + = : Note that the 3's and 2's will cancel leaving: + = : 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 = 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"