Question 195632

{{{((y^2-81)/(y^2))((y^2-9y)/(y^2+y-90))}}} Start with the given expression.



{{{(((y-9)(y+9))/(y^2))((y^2-9y)/(y^2+y-90))}}} Factor {{{y^2-81}}} to get {{{(y-9)(y+9)}}}.



{{{(((y-9)(y+9))/(y*y))((y^2-9y)/(y^2+y-90))}}} Factor {{{y^2}}} to get {{{y*y}}}.



{{{(((y-9)(y+9))/(y*y))((y(y-9))/(y^2+y-90))}}} Factor {{{y^2-9y}}} to get {{{y(y-9)}}}.



{{{(((y-9)(y+9))/(y*y))((y(y-9))/((y+10)(y-9)))}}} Factor {{{y^2+y-90}}} to get {{{(y+10)(y-9)}}}.



{{{(y(y-9)(y+9)(y-9))/(y*y(y+10)(y-9))}}} Combine the fractions. 



{{{(highlight(y)highlight((y-9))(y+9)(y-9))/(y*highlight(y)(y+10)highlight((y-9)))}}} Highlight the common terms. 



{{{(cross(y)cross((y-9))(y+9)(y-9))/(y*cross(y)(y+10)cross((y-9)))}}} Cancel out the common terms. 



{{{((y+9)(y-9))/(y(y+10))}}} Simplify. 



{{{(y^2-81)/(y(y+10))}}} FOIL



{{{(y^2-81)/(y^2+10y)}}} Distribute



So {{{((y^2-81)/(y^2))((y^2-9y)/(y^2+y-90))}}} simplifies to {{{(y^2-81)/(y^2+10y)}}}.



In other words, {{{((y^2-81)/(y^2))((y^2-9y)/(y^2+y-90))=(y^2-81)/(y^2+10y)}}} where {{{y<>-10}}}, {{{y<>0}}}, or {{{y<>9}}}.