Question

Help With Dividing Rational Equations!

Answer 1

\[\frac{ x ^{2}-81\ }{ 12x ^{2} }\div \frac{ 9-x }{ 4x}\]

Answer 2

just like with numbers, flip the second one and start with
\[\frac{x^2-81}{12x^2}\times \frac{4x}{9-x}\] then factor and cancel what you can

Answer 3

I tried to factor and got
(x-9)(x+9)
________
12x^2

Answer 4

can I cancel x-9 and 9-x? @satellite73

Answer 5

Is the answer 4x(x+9) over 12x^2?

Answer 6

Thats what I got at least

Answer 7

you cannot cancel x-9 and 9-x
but hey!
\(\Large x-9 = -(9-x)\)

so you can cancel two 9-x and leave behind a negative sign

Answer 8

so would it be a negative x?

Answer 9

I'm confused how I go from the factored equation to solving it

Answer 10

let me show you
\(\Large \dfrac{x-9}{9-x} = \dfrac{-\cancel{(9-x)}}{\cancel{9-x}} =-1 \)

Answer 11

okay, so that leaves me with the (x+9) and 4x on the numerators

Answer 12

\(\Large \frac{x^2-81}{12x^2}\times \frac{4x}{9-x} = \frac{(x+9)(x-9) \times 4 \times x}{4\times 3 \times x\times x\times (9-x)} \\ \Large = -\frac{(x+9)\cancel{(9-x)} \times \cancel 4 \times \cancel x}{\cancel 4\times 3 \times \cancel x\times x\times \cancel{(9-x)}}\)

the 'x' and the 4 are also common in the numerator and denominator, right ??
so they can be cancelled too

Answer 13

and don't forget that
\[\frac{4x}{12x^2}=\frac{1}{3x}\]

Answer 14

oh!

Answer 15

@hartnn nice use of \cancel

Answer 16

thanks! it helps understand better on whats happening :)

Answer 17

So the answer is x+9 over 3x?

Answer 18

and don't leave out that cute lil negative sign we got in the beginning
-(x+9)/3x

Answer 19

Oh thank you so much!

Answer 20

That helped tons

Answer 21

happy to help :)
welcome ^_^