Question 173199


{{{((x^2+7x+10)/(x+5))/((x^2-4)/(5x-10))}}} Start with the given expression.



{{{((x^2+7x+10)/(x+5))((5x-10)/(x^2-4))}}} Multiply the first fraction {{{(x^2+7x+10)/(x+5)}}} by the reciprocal of the second fraction {{{(x^2-4)/(5x-10)}}}.



{{{(((x+5)(x+2))/(x+5))((5x-10)/(x^2-4))}}} Factor {{{x^2+7x+10}}} to get {{{(x+5)(x+2)}}}.



{{{(((x+5)(x+2))/(x+5))((5(x-2))/(x^2-4))}}} Factor {{{5x-10}}} to get {{{5(x-2)}}}.



{{{(((x+5)(x+2))/(x+5))((5(x-2))/((x-2)(x+2)))}}} Factor {{{x^2-4}}} to get {{{(x-2)(x+2)}}}.



{{{(5(x+5)(x+2)(x-2))/((x+5)(x-2)(x+2))}}} Combine the fractions. 



{{{(5*highlight((x+5))highlight((x+2))highlight((x-2)))/(highlight((x+5))highlight((x-2))highlight((x+2)))}}} Highlight the common terms. 



{{{(5*cross((x+5))cross((x+2))cross((x-2)))/(cross((x+5))cross((x-2))cross((x+2)))}}} Cancel out the common terms. 



{{{5}}} Simplify. 



So {{{((x^2+7x+10)/(x+5))/((x^2-4)/(5x-10))}}} simplifies to {{{5}}}.



In other words, {{{((x^2+7x+10)/(x+5))/((x^2-4)/(5x-10))=5}}} where {{{x<>-5}}}, {{{x<>-2}}} or {{{x<>2}}}