Question 399998
{{{((m^2-9)/9m)(7m/(m+3))}}}
When multiplying fractions it is not only OK but a good idea to cancel factors that are common to the numerators and denominators before you multiply. One is even allowed to cancel factors from one fraction across to the other fraction.<br>
So the best place to start is to factor each numerator and denominator. We can factor the first numerator as a difference of squares, {{{a^2-b^2 = (a+b)(a-b)}}}. And we can factor a 1 out of the second denominator:
{{{(((m+3)(m-3))/(9*m))((7*m)/(1(m+3)))}}}
Now we can cancel the factors that are common to both the numerators and the denominators:
{{{((cross((m+3))(m-3))/(9*cross(m)))((7*cross(m))/(1*cross((m+3))))}}}
leaving:
{{{((m-3)/9)(7/1)}}}
These are much easier to multiply than the original two fractions. Multiplying we get:
{{{((m-3)7)/(9*1)}}}
Except for the m's that I canceled but you did not, this is what you had. This simplifies to:
{{{(7m-21)/9}}}