Question 74898
{{{((x^2 - 9x + 20) / (6x - 4))/((x^2 - 25)/(2x + 10))}}}Start with the given problem





{{{((x-5)(x-4) / (6x - 4))/((x-5)(x+5)/(2x + 10))}}}Factor the numerators of the 2 fractions. {{{x^2 - 9x + 20}}} factors to {{{(x-5)(x-4)}}} and {{{x^2 - 25}}} factors to {{{(x-5)(x+5)}}}





{{{((x-5)(x-4) / (2(3x - 2)))/((x-5)(x+5)/(2(x + 5)))}}}Factor out a 2 from the 2 denominators





{{{((x-5)(x-4) / (2(3x - 2)))*((2(x + 5))/(x-5)(x+5))}}}Multiply the 2 fractions by multiplying the 1st fraction by the reciprocal of the 2nd fraction





{{{(cross((x-5))(x-4) / (2(3x - 2)))*((2*cross((x + 5)))/(cross((x-5))*cross((x+5))))}}}Cancel like terms





{{{((x-4) / (2(3x - 2)))*((2)/1))}}}This is what your left with after the cancellations




{{{(2(x-4) / (2(3x - 2)))}}}Combine the 2 fractions by multiplying them






{{{(cross(2)(x-4) / (cross(2)(3x - 2)))}}}Notice the 2's cancel




{{{(x-4)/(3x-2)}}}This is the final simplified form.