Question 200970
{{{((21x^2-49x-42)/(4x^2-28x+48))/((21x+14)/(8x^2-32x))}}} Start with the given expression.



{{{((21x^2-49x-42)/(4x^2-28x+48))((8x^2-32x)/(21x+14))}}} Multiply the first fraction {{{(21x^2-49x-42)/(4x^2-28x+48)}}} by the reciprocal of the second fraction {{{(21x+14)/(8x^2-32x)}}}.



{{{((7(3x+2)(x-3))/(4x^2-28x+48))((8x^2-32x)/(21x+14))}}} Factor {{{21x^2-49x-42}}} to get {{{7(3x+2)(x-3)}}}.



{{{((7(3x+2)(x-3))/(4(x-3)(x-4)))((8x^2-32x)/(21x+14))}}} Factor {{{4x^2-28x+48}}} to get {{{4(x-3)(x-4)}}}.



{{{((7(3x+2)(x-3))/(4(x-3)(x-4)))((4*2x(x-4))/(21x+14))}}} Factor {{{8x^2-32x}}} to get {{{4*2x(x-4)}}}.



{{{((7(3x+2)(x-3))/(4(x-3)(x-4)))((4*2x(x-4))/(7(3x+2)))}}} Factor {{{21x+14}}} to get {{{7(3x+2)}}}.



{{{(7*4*2x(3x+2)(x-3)(x-4))/(4*7(x-3)(x-4)(3x+2))}}} Combine the fractions. 



{{{(highlight(7)*highlight(4)*2x*highlight((3x+2))highlight((x-3))highlight((x-4)))/(highlight(4)*highlight(7)*highlight((x-3))highlight((x-4))highlight((3x+2)))}}} Highlight the common terms. 



{{{(cross(7)*cross(4)*2x*cross((3x+2))cross((x-3))cross((x-4)))/(cross(4)*cross(7)*cross((x-3))cross((x-4))cross((3x+2)))}}} Cancel out the common terms. 



{{{2x}}} Simplify. 




So {{{((21x^2-49x-42)/(4x^2-28x+48))/((21x+14)/(8x^2-32x))}}} simplifies to {{{2x}}}.



In other words, {{{((21x^2-49x-42)/(4x^2-28x+48))/((21x+14)/(8x^2-32x))=2x}}}