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!
Thank you!

Answer 1

I don't know how to use the equation thing to make it look normal, so I hope you can understand! Sorry!

Answer 2

\[\large{\frac{2a^2-9a-35}{12x^3}\times x \times \frac{9x^2}{2a^2-3a-20}}\]

is this the question?

Answer 3

There is no x in the middle, it is just multiplying the two fractions. Sorry if that was confusing..

Answer 4

ok no problem :
\[\large{\frac{2a^2-9a-35}{12x^3} \times \frac{9x^2}{2a^2-3a-20}}\]
\[\large{\frac{2a^2+14a-5a-35}{12\cancel{x^3}^x}\times \frac{9\cancel{x^2}^1}{2a^2-3a-20}}\]

Answer 5

Wouldn't +14a-5a give you +9 ?a

Answer 6

**+9a

Answer 7

very right my mistake sorry :
\[\large{\frac{2a^2-14a+5a-35}{12\cancel{x^3}^x}\times \frac{9\cancel{x^2}^1}{2a^2-3a-20}}\]

Answer 8

Okay, just checking !

Answer 9

\[\large{\frac{2a^2-14a+5a-35}{12\cancel{x^3}^x}\times \frac{9\cancel{x^2}^1}{2a^2-3a-20}}\]
\[\large{\frac{2(a-7)+5(a-7)}{12x}\times \frac{9}{2a^2-8a+5a-20}}\]
\[\large{\frac{(2a+5)(a-7)}{12x}\times \frac{9}{(2a+5)(a-4)}}\]

Answer 10

\[\large{\frac{\cancel{2a+5}^1(a-7)}{12x}\times \frac{9}{\cancel{2a+5}^1(a-4)}}\]
\[\large{\frac{(a-7)}{12x}\times \frac{9}{a-4}}\]

Answer 11

Ohh. Okay. & One question.. Wouldn't it be 2(a^2-7) ?

Answer 12

In your second step on the left numerator

Answer 13

Well I guess technically, your third.

Answer 14

no i meant : it was : 2a(a-7)

Answer 15

Oh, right ! Oh, well thank you. You made it super easy to see when I originally made my mistake :) Thank you soo much ! (:

Answer 16

welcome @theequestrian and sorry for the mistakes made byy me

Answer 17

Oh no problem!!