Question 66192
Divide the fraction:

6p-18     3p-9
-----  /  -----
  9p     p^2+2p

The only thing I understand of this problem is I need to flip the second eqaution. Other than this the information is slim. Thank you
:
You're right about that, invert and mult, you have;
{{{(6p-18)/(9p)}}} * {{{(p^2 + 2p)/(3p-9)}}}
:
Note all the factoring you can do
{{{(6(p-3))/(9p)}}} * {{{(p(p + 2))/(3(p-3))}}}
:
Cancel the 3 into the 6, and the p's, you have:
{{{(2(p-3))/(9)}}} * {{{((p + 2))/((p-3))}}}
:
Notice you can cancel the (p-3)'s also, resulting in:
{{{(2)/(9)}}} * (p + 2) = {{{(2(p+2))/9}}}
:
Has this fattened your information store somewhat?