Question

how do you solve (x/x-4 + 6/x+4)/ 3/2x^2-32

Answer 1

it's \((x/x-4 + 6/x+4)/ (3/2x^2-32)\) right?

Answer 2

yes

Answer 3

if it's "solve", then there should be an equality sign, something like \((x/x−4+6/x+4)/(3/2x^2−32) = 0\) (or =1).

but even without that, you can do something aout this fraction.
1) work on the numerator: you must make one fraction (not a sum of two)
2) factorize all the expressions and make simplifications.

Answer 4

objective: to write your fraction as \(\frac{\frac{a}{b}}{\frac{c}{d}}\).
maybe \(a\) and \(c\) have common factors, maybe \(b\) and \(d\) have common factors.

Answer 5

It starts like: \((\frac{x}{x-4} + \frac{6}{x+4}) / (3/(2x^2-32)) = \frac{x(x+4)+6(x-4)}{(x-4)(x+4)} / (3/(2x^2-32)) = \frac{x(x+4)+6(x-4)}{(x-4)(x+4)} \frac{2x^2-32}{3}\).

when you factorize \(2x^2-32\) you'll see "something".

Answer 6

oops.
i wrote
\[
\frac{x(x+4)+6(x-4)}{(x-4)(x+4)} \frac{2x^2-32}{3}
\]on the right.

Answer 7

do you get something interesting?

Answer 8

no not really..

Answer 9

\(2x^2-32 = 2(x^2 - 16) = 2(x+4)(x-4)\)
(the last step used the formula for a difference of two squared numbers: \(a^2-b^2=(a-b)(a+b)\).)

here you got some factors that will cancel out other factors already present in the fraction. see that? :D

Answer 10

ya

Answer 11

write your final answer here, and can you tell why the question is "solve" ? im puzzled

Answer 12

its not it just needs to be awnsered

Answer 13

\(\frac{x^2+10x-24}{\text{cancels out}}\frac{2(\text{cancels out})}{3} = \frac23(x^2+10x-24)\). did you find that..?

Answer 14

ya

Answer 15

well done

Answer 16

is that the final awnser?

Answer 17

There's nothing more to do (except doing \(\frac23x^2 + \frac2310x - \frac23 ....\) but that's not interesting).
Yes it's the final answer.

Answer 18

ok thank you for the help!