Question 74926
: x+4 over 3x + 4y times 9x squared - 16y squared over 2x squared + 5x - 12
:

{{{((x+4))/((3x+4y))}}} * {{{((9x^2 - 16y^2))/((2x^2 + 5x - 12))}}}
:
We can factor the difference of squares and the quadratic, so we have:
{{{((x+4))/((3x+4y))}}} * {{{((3x - 4y)(3x + 4y))/((2x - 3)(x + 4)))}}}
:
Cancel out the (3x+4y) and the (x+4), you end up with:
{{{((3x-4y))/((2x-3))}}}
:
Did this make sense to you? Any questions?