Question

Correct? http://prntscr.com/b7axg7

Answer 1

\[\Large \frac{(a-4)(a+3)}{a^2-4a}*\frac{2a^3}{(a+3)(a-1)}\]

\[\Large \frac{(a-4)(a+3)}{a(a-4)}*\frac{a*2a^2}{(a+3)(a-1)}\]

\[\Large \frac{(a-4)(a+3)}{\cancel{a}(a-4)}*\frac{\cancel{a}*2a^2}{(a+3)(a-1)}\]

\[\Large \frac{(a-4)(a+3)}{(a-4)}*\frac{2a^2}{(a+3)(a-1)}\]

\[\Large \frac{\cancel{(a-4)}(a+3)}{\cancel{(a-4)}}*\frac{2a^2}{(a+3)(a-1)}\]

\[\Large \frac{(a+3)}{1}*\frac{2a^2}{(a+3)(a-1)}\]

\[\Large \frac{\cancel{(a+3)}}{1}*\frac{2a^2}{\cancel{(a+3)}(a-1)}\]

\[\Large \frac{1}{1}*\frac{2a^2}{a-1}\]

\[\Large 1*\frac{2a^2}{a-1}\]

\[\Large \frac{2a^2}{a-1}\]

Answer 2

So
\[\Large \frac{(a-4)(a+3)}{a^2-4a}*\frac{2a^3}{(a+3)(a-1)}\]
simplifies to
\[\Large \frac{2a^2}{a-1}\]


You have the correct answer

Answer 3

Thanks ^0^

Answer 4

you're welcome