Question 189465


{{{((y^2+5y+4)/(y^2-49))/((2y^2+5y-12)/(y^2+9y+14))}}} Start with the given expression.



{{{((y^2+5y+4)/(y^2-49))((y^2+9y+14)/(2y^2+5y-12))}}} Multiply the first fraction {{{(y^2+5y+4)/(y^2-49)}}} by the reciprocal of the second fraction {{{(2y^2+5y-12)/(y^2+9y+14)}}}.



{{{(((y+4)(y+1))/(y^2-49))((y^2+9y+14)/(2y^2+5y-12))}}} Factor {{{y^2+5y+4}}} to get {{{(y+4)(y+1)}}}.



{{{(((y+4)(y+1))/((y-7)(y+7)))((y^2+9y+14)/(2y^2+5y-12))}}} Factor {{{y^2-49}}} to get {{{(y-7)(y+7)}}}.



{{{(((y+4)(y+1))/((y-7)(y+7)))(((y+7)(y+2))/(2y^2+5y-12))}}} Factor {{{y^2+9y+14}}} to get {{{(y+7)(y+2)}}}.



{{{(((y+4)(y+1))/((y-7)(y+7)))(((y+7)(y+2))/((y+4)(2y-3)))}}} Factor {{{2y^2+5y-12}}} to get {{{(y+4)(2y-3)}}}.



{{{((y+4)(y+1)(y+7)(y+2))/((y-7)(y+7)(y+4)(2y-3))}}} Combine the fractions. 



{{{(highlight((y+4))(y+1)highlight((y+7))(y+2))/((y-7)highlight((y+7))highlight((y+4))(2y-3))}}} Highlight the common terms. 



{{{(cross((y+4))(y+1)cross((y+7))(y+2))/((y-7)cross((y+7))cross((y+4))(2y-3))}}} Cancel out the common terms. 



{{{((y+1)(y+2))/((y-7)(2y-3))}}} Simplify. 



{{{(y^2+3y+2)/(2y^2-17y+21)}}} FOIL



So {{{((y^2+5y+4)/(y^2-49))/((2y^2+5y-12)/(y^2+9y+14))}}} simplifies to {{{(y^2+3y+2)/(2y^2-17y+21)}}}.



In other words, {{{((y^2+5y+4)/(y^2-49))/((2y^2+5y-12)/(y^2+9y+14))=(y^2+3y+2)/(2y^2-17y+21)}}} where {{{y<>-7}}}, {{{y<>-4}}}, {{{y<>3/2}}}, or {{{y<>7}}} (these are the restrictions)