Algebraic Fractions. Oh no!

[(2a^2-9a-35)/(12x^3)]x[(9x^2)/(2a^2-3a-20)]

I got the answer of...
[3(a-7)]/[4(a-4)x]
But it says that it is wrong.. Helpp! Thanks :)

Question

Algebraic Fractions. Oh no!

[(2a^2-9a-35)/(12x^3)]x[(9x^2)/(2a^2-3a-20)]

I got the answer of...
[3(a-7)]/[4(a-4)x]
But it says that it is wrong.. Helpp! Thanks :)

jiteshmeghwal9 · Answer

$${2a^2-9a-35\over12x^3}\times{9x^2\over2a^2-3a-20}$$is this ur question???

anonymous · Answer

Yes it is! I'm not real sure how to make those..

jiteshmeghwal9 · Answer

$$\frac{2a^2-9a-35 \times \cancel {9x^2}3}{4x \cancel{12x^3} \times 2a^2-3a-20}$$

anonymous · Answer

And then the 2a^2 's cancel out & then you can simplify -9a & -3a into 3/1 and -35 & -20 ino 7/5. So would that give me 21/5 for a final answer?

anonymous · Answer

Oh wait. I forgot the other stuff. So would that give me 63/20x for a final answer?

anonymous · Answer

Whoops. Another mistake.. I meant it would turn it into 7/4 which would give me 63/16x, right?