SOLUTION: I am working on solving linear and quadratic equations. I need help with this problem: {{{(3y+1)(over)3y(squared)-4y-4 + (9)(over)9y(squared)-4 = (2y-2)(over)3y(squared)-8y+

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I am working on solving linear and quadratic equations. I need help with this problem: {{{(3y+1)(over)3y(squared)-4y-4 + (9)(over)9y(squared)-4 = (2y-2)(over)3y(squared)-8y+      Log On


   

Question 124607: I am working on solving linear and quadratic equations. I need help with this problem:

I believe the CD:(y-2)(3y-2)(3y+2)
Please help solve. Thanks.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Assume you mean:
%283y%2B1%29%2F%283y%5E2-4y-4%29 + 9%2F%289y%5E2-4%29 = %282y-2%29%2F%283y%5E2-8y%2B4%29
Factor the denominators:
%283y%2B1%29%2F%28%283y%2B2%29%28y-2%29%29 + 9%2F%28%283y-2%29%283y%2B2%29%29 = %282y-2%29%2F%28%283y-2%29%28y-2%29+%29
:
You're right about the common denominator, multiply each term by that:
(y-2)(3y-2)(3y+2)*%283y%2B1%29%2F%28%283y%2B2%29%28y-2%29%29 + (y-2)(3y-2)(3y+2)*9%2F%28%283y-2%29%283y%2B2%29%29 = (y-2)(3y-2)(3y+2)*%282y-2%29%2F%28%283y-2%29%28y-2%29+%29
:
Cancel out the denominators, leaving:
(3y-2)(3y+1) + 9(y-2) = (3y+2)(2y-2)
:
FOIL
(9y^2 - 3y - 2) + (9y-18) = 6y^2 - 2y - 4
:
Combine like terms on the left
9y^2 - 6y^2 - 3y + 9y + 2y - 2 - 18 + 4 = 0
:
3y^2 + 8y - 16 = 0
:
Factors to:
(3y - 4)(y + 4) = 0
:
3y = +4
y = 4%2F3
and
y = -4
:
:
I checked the y=-4 solution in the original equation.
:
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