Question 124607
Assume you mean:
{{{(3y+1)/(3y^2-4y-4)}}}  +  {{{9/(9y^2-4)}}}  =  {{{(2y-2)/(3y^2-8y+4)}}}
Factor the denominators:
{{{(3y+1)/((3y+2)(y-2))}}}  +  {{{9/((3y-2)(3y+2))}}}  =  {{{(2y-2)/((3y-2)(y-2) )}}}
:
You're right about the common denominator, multiply each term by that:
(y-2)(3y-2)(3y+2)*{{{(3y+1)/((3y+2)(y-2))}}}  +  (y-2)(3y-2)(3y+2)*{{{9/((3y-2)(3y+2))}}}  =  (y-2)(3y-2)(3y+2)*{{{(2y-2)/((3y-2)(y-2) )}}}
:
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 = {{{4/3}}}
and
y = -4
:
:
I checked the y=-4 solution in the original equation.
:
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