<PRE><B>Question 189078</B>
please explain to me how to work out the following problem? 
[(3+y)/(y^2+9)] - [(y-3)/(9-y^2)]

I tried this and here is my answer: 
     1 / [(y-3)(y+3)]
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[(3+y)/(y^2+9)] - [(y-3)/(9-y^2)]

lcd = -(y^2+9)(y^2-9)

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Rewrite each fraction with the lcd as its denominator:

= [(y+3)(y^2-9)]/lcd - [(y-3)(y^2+9)]/lcd

Combine the numerators over the lcd:

= [[y^3+3y^2-9y-27] - (y^3-3y^2+9y-27)]]/lcd

= (6y^2-18y)/lcd

= [6y(y-3)]/[-(y^2+9)(y^2-9)]

= -(6y)/[(y^2+9)(y+3)]
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Cheers,
Stan H.
 </PRE>