Question

How do you multiply `(8x-4x^2)/(xy-2y+3x-6) div (3x+6)/(y+3)`?

Answer 1

`-(4x)/(3(x+2)) =-(4x)/(3x+6)`

Explanation:

There is not the scope hear to demonstrate why you do this but the first part of the process is to invert (turn upside down) the divisor and then multiply. It is all to do with dividing counts (numerators) and applying a correction factor that is the equivalent of making the size indicators (denominators) the same. So we have:

`(8x-4x^2)/(xy-2y+3x-6)xx(y+3)/(3x+6)`

Lets go on the hunt for things we can cancel out.

Consider `8x-4x^2-> 4x(2-x) = =4x(x-2)`

I have done the above as I recognise something in the denominator. So now we have:

`(-4x(x-2))/(xy-2y+3x-6)xx(y+3)/(3x+6)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider `3x+6 ->3(x+2)` Not as I had hoped but pressing on:

`(-4x(x-2))/(xy-2y+3x-6)xx(y+3)/(3(x+2))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider: `xy-2y+3x-6 color(white)("d")->color(white)("d")y(x-2)+3(x-2)`giving:

`(x-2)(y+3)` Even better!

`(-4x(x-2))/((x-2)(y+3))xx(y+3)/(3(x+2))` So now we have

`(-4x)xx(x-2)/(x-2)xx(y+3)/(y+3)xx1/(3(x+2))`

`(-4x)xxcolor(white)("vd")1color(white)("vd")xxcolor(white)("dd")1color(white)("dd")xx1/(3(x+2)) color(white)("dd") = color(white)("dd") (-4x)/(3(x+2))`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer `->color(white)("d") -(4x)/(3(x+2))`