Question

How do you subtract `(-4)/(9-x^2)-(2x+1)/(x^2-3x)`?

Answer 1

Here's what I got.

Explanation:

The first thing to do here is make sure that the two fractions have the same denominator.

Notice that the first denominator can be written as

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

The second denominator can be written as

`x^2 - 3x = x * (x - 3)`

Now, you can rewrite this as

`x * (x-3) = - x * (3 - x)`

since

`-x * (3 -x ) = -3x - x * (-x) = x^2 - 3x`

This means that the second fraction can be rewritten s

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

The original expression becomes

`-4/((3-x) * (3 + x)) - [ - (2x+1)/(x (3-x))]`

`-4/((3-x)(3 + x)) + (2x + 1)/(x(3-x))`

Now, in order to find the common denominator, you must multiply the first fraction by `1 = x/x` and the second fraction by `1 = (3+x)/(3+x)`.

This will get you

`-4/((3-x)(3+x)) * x/x + (2x+1)/(x(3-x)) * (3+x)/(3+x)`

`-(4x)/(x(3-x)(3+x)) + ((2x+1)(3+x))/(x(3-x)(3+x))`

Now you're ready to focus on the numerators. You have

`-4x + (2x+1)(3+x)`

`-4x + (6x + 2x^2 + 3 + x)`

`2x^2 + 3x + 3`

Put this back into the resulting expression to get

`(2x^2 + 3x + 3)/(x(3-x)(3+x))`

Keep in mind that you need to have `x != {0, +- 3}`.