Question 42311
From
[(6p - 18)/9p] / [(3p - 9)/(p^2 + 2p)] =
we factor and cancel giving us
[6(p - 3) / 9p] / [3(p - 3) / p(p+2)] =
now invert and multiply
[6(p - 3) / 9p] * [p(p+2) / 3(p - 3)] =
now cancel and reduce where we can...
(2 / 3p) * [p(p+2) / 3] =
2(p+2) / 9