Question 117935
{{{((6p-18)/(9p))/((3p-9)/(p^2+2p))}}} Start with the given expression


{{{((6p-18)/(9p))*(((p^2+2p))/((3p-9)))}}} Multiply the first fraction by the reciprocal of the second fraction



{{{((6(p-3))/(9p))((p^2+2p)/(3p-9))}}}   Factor {{{6p-18}}} to get {{{6(p-3)}}} 


{{{((6(p-3))/(9(p)))((p^2+2p)/(3p-9))}}}   Factor {{{9p}}} to get {{{9(p)}}} 


{{{((6(p-3))/(9(p)))((p(p+2))/(3p-9))}}}   Factor {{{p^2+2p}}} to get {{{p(p+2)}}} 


{{{((6(p-3))/(9(p)))((p(p+2))/(3(p-3)))}}}   Factor {{{3p-9}}} to get {{{3(p-3)}}} 


{{{(((2*3)(p-3))/(9(p)))((p(p+2))/(3(p-3)))}}}   Factor 6 into 2*3



{{{(2*3)(p-3)p(p+2)/9(p)3(p-3)}}} Combine the fractions



{{{(2*cross(3))cross((p-3))cross(p)(p+2)/9cross(p)cross(3)cross((p-3))}}} Cancel like terms



{{{2(p+2)/9}}} Simplify



So {{{((6p-18)/(9p))/((3p-9)/(p^2+2p))}}} simplifies to {{{2(p+2)/9}}}.



Note: This is not a typo even though this answer is identical to the previous one. The previous expression {{{((6p-18)/(3p-9))/((9p)/(p^2+2p))}}} is the same as {{{((6p-18)/(9p))/((3p-9)/(p^2+2p))}}}. Can you see why?


{{{((6p-18)/(3p-9))/((9p)/(p^2+2p))}}}  ---> {{{((6p-18)/(3p-9))*(((p^2+2p))/((9p)))}}} and 


{{{((6p-18)/(9p))/((3p-9)/(p^2+2p))}}} ---> {{{((6p-18)/(9p))*(((p^2+2p))/((3p-9)))}}} 



For each second expression, notice the denominators have swapped. Since they all get combined in one big denominator, the answer does not change.