Question 189078
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{{{(3+y)/(y^2+9)-(y-3)/(9-y^2)}}} Start with the given expression.



{{{(3+y)/(y^2+9)-(y-3)/((3-y)(3+y))}}} Factor {{{9-y^2}}} to get {{{(3-y)(3+y)}}} (use the difference of squares)



{{{(y+3)/(y^2+9)-(y-3)/(-(y-3)(y+3))}}} Factor a negative 1 from {{{3-y}}} to get {{{-(y-3)}}}. Also, rearrange the terms.



{{{(y+3)/(y^2+9)-cross((y-3))/(-cross((y-3))(y+3))}}} Cancel out the common terms.



{{{(y+3)/(y^2+9)-1/(-(y+3))}}} Simplify



{{{(y+3)/(y^2+9)+1/(y+3)}}} Reduce



{{{((y+3)(y+3))/((y^2+9)(y+3))+1/(y+3)}}} Multiply both the numerator and denominator of the first fraction by {{{y+3}}}



{{{(y^2+6y+9)/((y^2+9)(y+3))+1/(y+3)}}} FOIL



{{{(y^2+6y+9)/((y^2+9)(y+3))+(1(y^2+9))/((y^2+9)(y+3))}}} Multiply both the numerator and denominator of the second fraction by {{{y^2+9}}}



{{{(y^2+6y+9)/((y^2+9)(y+3))+(y^2+9)/((y^2+9)(y+3))}}} Distribute



{{{(y^2+6y+9+y^2+9)/((y^2+9)(y+3))}}} Add the fractions.



{{{(2y^2+6y+18)/((y^2+9)(y+3))}}} Combine like terms.



So {{{(3+y)/(y^2+9)-(y-3)/(9-y^2)=(2y^2+6y+18)/((y^2+9)(y+3))}}} where {{{y<>-3}}}