Question 196844
Is the expression {{{((y^2-36)/(y^2-49))/((y+6)/(y+7))}}} ??? If it is, then...





{{{((y^2-36)/(y^2-49))/((y+6)/(y+7))}}} Start with the given expression.



{{{((y^2-36)/(y^2-49))((y+7)/(y+6))}}} Multiply the first fraction {{{(y^2-36)/(y^2-49)}}} by the reciprocal of the second fraction {{{(y+6)/(y+7)}}}.



{{{(((y-6)(y+6))/(y^2-49))((y+7)/(y+6))}}} Factor {{{y^2-36}}} to get {{{(y-6)(y+6)}}}.



{{{(((y-6)(y+6))/((y-7)(y+7)))((y+7)/(y+6))}}} Factor {{{y^2-49}}} to get {{{(y-7)(y+7)}}}.



{{{((y-6)(y+6)(y+7))/((y-7)(y+7)(y+6))}}} Combine the fractions. 



{{{((y-6)highlight((y+6))highlight((y+7)))/((y-7)highlight((y+7))highlight((y+6)))}}} Highlight the common terms. 



{{{((y-6)cross((y+6))cross((y+7)))/((y-7)cross((y+7))cross((y+6)))}}} Cancel out the common terms. 



{{{(y-6)/(y-7)}}} Simplify. 



So {{{((y^2-36)/(y^2-49))/((y+6)/(y+7))}}} simplifies to {{{(y-6)/(y-7)}}}.



In other words, {{{((y^2-36)/(y^2-49))/((y+6)/(y+7))=(y-6)/(y-7)}}} where {{{y<>-7}}}, {{{y<>-6}}}, or {{{y<>7}}}