Question

Simplify fully:?

`(9x^2-1)/(3x^2+2x-1)` `div (3x+1)/(x-2)`

Answer 1

`(x-2)/(x+1)` when `x!=+-1/3`and`x!=-1`

Explanation:

First, remember that:
`(a/b)/(c/d)=a/b*d/c`
Therefore, `((9x^2-1)/(3x^2+2x-1))/((3x+1)/(x-2))=(9x^2-1)/(3x^2+2x-1)*(x-2)/(3x+1)`
Let's factor the denominator and the numerator of `(9x^2-1)/(3x^2+2x-1)`

`9x^2-1=(3x+1)(3x-1)`

We use the quadratic formula `(-b+-sqrt (b^2-4(a)(c)))/(2(a))`
`(-b+-sqrt (b^2-4(a)(c)))/(2(a))=x`
`(-2+-sqrt (2^2-4(3)(-1)))/(2(3))=x`
`(-2+-sqrt 16)/6=x`
`(-2+-4)/6=x`
`-1=x=1/3`
`3x^2+2x-1=3(x+1)(x-1/3)`

So we now have: `((3x+1)(3x-1))/(3(x+1)(x-1/3))*(x-2)/(3x+1)`

Now, remember that: `(ab)/(cd)*(ed)/(fg)=(ab)/(c canceld)*(ecanceld)/(fg)`

Therefore, we now have:

`((3x-1)(x-2))/(3(x+1)(x-1/3))=>((3x-1)(x-2))/((x+1)(3x-1))`

We see that both the denominator and the numerator share `3x-1` in common.

`(cancel(3x-1)(x-2))/((x+1)cancel(3x-1))`
`(x-2)/(x+1)` This is our answer!

Remember, however, that our original expression is undefined when
`x` is `+-1/3` or `-1`

Answer 2

`(9x^2-1)/(3x^2+2x-1) -: (3x+1)/(x-2) = (x-2)/(x+1) = 1-3/(x+1)`

with exclusion `x != +-1/3`

Explanation:

`(9x^2-1)/(3x^2+2x-1) -: (3x+1)/(x-2)`

`=(9x^2-1)/(3x^2+2x-1) * (x-2)/(3x+1)`

`=(color(red)(cancel(color(black)((3x-1))))color(blue)(cancel(color(black)((3x+1)))))/(color(red)(cancel(color(black)((3x-1))))(x+1)) * (x-2)/color(blue)(cancel(color(black)((3x+1))))`

`=(x-2)/(x+1)`

`=(x+1-3)/(x+1)`

`=1-3/(x+1)`

with exclusions `x != +-1/3`