Question 71268
Divide:
{{{((6p-18)/9p)/((3p-9)/(p^2+2p))}}}
Remember, to divide fractions, you "Copy the 1st fraction" "Flip the sign from divide to multiply" and "Flip (invert) the 2nd fraction" So we get:
{{{((6p-18)/9p)*((p^2+2p)/(3p-9))}}} Now you can factor most of the terms:
{{{(2(3p-9)/9p)*(p(p+2)/((3p-9)))}}} You can cancel (3p-9) and a p to get:
{{{2*(p+2)/9}}} or {{{(2p+4)/9}}}