Question 370727
{{{((2x)/(x-4))/(8/(x^2-7x+12))}}}
As usual when dividing fractions, you change it to multiplying the reciprocal:
{{{((2x)/(x-4))*((x^2-7x+12)/8)}}}
When you first learned to multiply fractions you probably learned that common factors in the numerators and denominators can be canceled, even across from one fraction to another. You also probably learned that doing so made the multiplication easier. This is all still true. In fact the advantage of canceling before multiplying is even greater now than it was long ago.<br>
In order to cancel factors we need to know what the factors are. So we start by factoring the numerators and denominators:
{{{((2*x)/(1*(x-4)))*(((x-4)(x-3))/(2*4))}}}
Now we should be able to that several factors can be canceled:
{{{((cross(2)*x)/(1*cross((x-4))))*((cross((x-4))(x-3))/(cross(2)*4))}}}
which leaves:
{{{(x/1)*((x-3)/4)}}}
This is much easier to multiply that what we had before!
{{{(x*(x-3))/(1*4)}}}
which simplifies to:
{{{(x^2-3x)/4}}}