Question 132135
{{{((x^2-9x+20)/(3x-15))/((x^2-16)/(9x+36))}}} Start with the given expression


{{{((x^2-9x+20)/(3x-15))*((9x+36)/(x^2-16))}}} Multiply the first fraction by the reciprocal of the second fraction



{{{(((x-5)(x-4))/(3x-15))((9x+36)/(x^2-16))}}}   Factor {{{x^2-9x+20}}} to get {{{(x-5)(x-4)}}} 


{{{(((x-5)(x-4))/(3(x-5)))((9x+36)/(x^2-16))}}}   Factor {{{3x-15}}} to get {{{3(x-5)}}} 


{{{(((x-5)(x-4))/(3(x-5)))((9(x+4))/(x^2-16))}}}   Factor {{{9x+36}}} to get {{{9(x+4)}}} 


{{{(((x-5)(x-4))/(3(x-5)))((9(x+4))/((x+4)(x-4)))}}}   Factor {{{x^2-16}}} to get {{{(x+4)(x-4)}}} 



{{{(x-5)(x-4)9(x+4)/3(x-5)(x+4)(x-4)}}} Combine the fractions



{{{cross(x-5)cross(x-4)9cross(x+4)/3cross(x-5)cross(x+4)cross(x-4)}}} Cancel like terms



{{{9/3}}} Simplify



{{{3}}} Reduce





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Answer:


So {{{((x^2-9x+20)/(3x-15))/((x^2-16)/(9x+36))}}} simplifies to {{{3}}}. In other words {{{((x^2-9x+20)/(3x-15))/((x^2-16)/(9x+36))=3}}} where {{{x<>-4}}}, {{{x<>4}}}, or {{{x<>5}}}