Question 243596
{{{((2x^2-8x-42)/(6x^2))/((x^2-9)/(x^2-3x))}}} Start with the given expression.



{{{((2x^2-8x-42)/(6x^2))*((x^2-3x)/(x^2-9))}}} Multiply the first fraction by the reciprocal of the second fraction.



{{{((2*(x+3)*(x-7))/(6x^2))((x^2-3x)/(x^2-9))}}} Factor {{{2x^2-8x-42}}} to get {{{2*(x+3)*(x-7)}}}.



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



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



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



{{{(2*x*(x+3)*(x-7)(x-3))/(2*3*x*x(x-3)*(x+3))}}} Combine the fractions. 



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



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



{{{(x-7)/(3x)}}} Simplify. 



So {{{((2x^2-8x-42)/(6x^2))/((x^2-9)/(x^2-3x))}}} simplifies to {{{(x-7)/(3x)}}}.



In other words, {{{((2x^2-8x-42)/(6x^2))/((x^2-9)/(x^2-3x))=(x-7)/(3x)}}}