Question 33007
{{{ ((y+2)/(y^2-y-2))+((y+1)/(y^2-4))=((1)/(y+1)) }}}
LHS:
Take the LCM,
{{{ ((y+2)(y^2-4)+(y+1)(y^2-y-2))/((y^2-y-2)(y^2-4)) }}}
Factorizing,
{{{ ((y+2)^2(y-2)+(y+1)^2(y-2))/((y-2)^2(y+1)(y+2)) }}}
Taking out common factor (y-2)
{{{ ((y+2)^2+(y+1)^2)/((y-2)(y+1)(y+2)) }}}
This equals
{{{ ((1)/(y+1)) }}}
So we get,
{{{ ((y+2)^2+(y+1)^2)/((y-2)(y+1)(y+2)) = ((1)/(y+1)) }}}
Cancelling (y+1)
{{{ ((y+2)^2+(y+1)^2)/((y-2)(y+2)) = 1 }}}
{{{ ((y+2)^2+(y+1)^2) = ((y-2)(y+2)) }}}
Open the brackets
{{{ 2y^2+6y+5 = y^2-4 }}}
Take everything to one side
{{{ y^2+6y+9=0}}}
Which reduces to
{{{ (y+3)^2=0}}}
Hence,
<b>y=-3</b>
<P>
Hope this helps,
xC