Question

Stumped by rational expressions, please walk me through the steps if possible.

Answer 1

Question is: \[x^2-16/2x^2-9x+4\div2x^2+14+24/4x+4\]

Answer 2

Uhm, is it
\[\LARGE \frac{x^2-16}{2x^2-9x+4} \div \frac{2x^2+14x+24}{4x+2}\] ?

Answer 3

Yes, sorry I didn't know how to make fractions with the equation tool. I also didn't know how to put equations my original question.

Answer 4

Whoops, typo, 2, not 4, in the bottom second equation right?

But when you are dividing fractions, switch the sign ti multiplication and flip the second fraction.

Answer 5

I mean 4, not 2

Answer 6

Yeah, it's a 4, so it would be \[\frac{ x^2-16 }{ 2x^2-9x+4 }\times \frac{ 4x+4 }{2x^2 }\]?

Answer 7

... it got cut off?

Answer 8

I get what you mean, but yea, now, factor everything.

Answer 9

I'm a little slow at this it'll probably take me a minute or two

Answer 10

It's alright, take you time.

Answer 11

\[\frac{ (x-4)(x+4) }{ (2x-1)(x-4) }\]

Answer 12

That's the first half more coming, sorry I'm doing the work on paper, then trying to copy it over to this >_<

Answer 13

\[\frac{ 2(1+2x) }{ 2(19+x^2) }\]

Answer 14

I think you might need to re-check that. Here's what it should be:
\(\LARGE \frac{x^2-16}{2x^2-9x+4} \div \frac{2x^2+14x+24}{4x+4}==>\frac{(x+4)(x-4)}{(2x-1)(x-4)}*\frac{4(x+1)}{2(x+3)(x+4)}\)

Answer 15

\(\LARGE \frac{x^2-16}{2x^2-9x+4} \div \frac{2x^2+14x+24}{4x+4}=>\frac{(x+4)(x-4)}{(2x-1)(x-4)}*\frac{4(x+1)}{2(x+3)(x+4)}\)

Answer 16

So, I was right for the first half... Then no clue what happened to be honest

Answer 17

Yup, now all we do is just start canceling what we can, do you see anything we can cancel?

Answer 18

x-4,x+4...

Answer 19

Right! So those cancel out

So those cancel.. there is one more thing we can simplify, do you know what that is?