Question 188591

{{{((x^2+6x-7)/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))}}} Start with the given expression.



{{{(((x+7)(x-1))/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))}}} Factor {{{x^2+6x-7}}} to get {{{(x+7)(x-1)}}}.



{{{(((x+7)(x-1))/(x(x+5)))((3x^2+16x+5)/(2x^2+7x-9))}}} Factor {{{x^2+5x}}} to get {{{x(x+5)}}}.



{{{(((x+7)(x-1))/(x(x+5)))(((x+5)(3x+1))/(2x^2+7x-9))}}} Factor {{{3x^2+16x+5}}} to get {{{(x+5)(3x+1)}}}.



{{{(((x+7)(x-1))/(x(x+5)))(((x+5)(3x+1))/((2x+9)(x-1)))}}} Factor {{{2x^2+7x-9}}} to get {{{(2x+9)(x-1)}}}.



{{{((x+7)(x-1)(x+5)(3x+1))/(x(x+5)(2x+9)(x-1))}}} Combine the fractions. 



{{{((x+7)highlight((x-1))highlight((x+5))(3x+1))/(x*highlight((x+5))(2x+9)highlight((x-1)))}}} Highlight the common terms. 



{{{((x+7)cross((x-1))cross((x+5))(3x+1))/(x*cross((x+5))(2x+9)cross((x-1)))}}} Cancel out the common terms. 



{{{((x+7)(3x+1))/(x(2x+9))}}} Simplify. 



So {{{((x^2+6x-7)/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))}}} simplifies to {{{((x+7)(3x+1))/(x(2x+9))}}}.




In other words, {{{((x^2+6x-7)/(x^2+5x))((3x^2+16x+5)/(2x^2+7x-9))=((x+7)(3x+1))/(x(2x+9))}}} where {{{x<>-5}}}, {{{x<>-9/2}}}, {{{x<>0}}}, or {{{x<>1}}} (these are the restrictions to avoid dividing by zero).