Question

How do you combine `(x^2)/(x^2 - 49) - x/(x-7)`?

Answer 1

`-(7x)/((x-7)(x+7))`

`=>-(7x)/(x^2-49)`

Explanation:

First, you have to realize that the denominator of the first fraction is a difference between two squares:

If you have
`(a^2-b^2)`

You can rewrite it as
`(a-b)(a+b)`

In this situation
`a=x`
`b=7`

So we can rewrite the first fraction as
`x^2/((x-7)(x+7))`

Now we have
`x^2/((x-7)(x+7))-x/(x-7)`

So in order to combine them, we need an LCD. In order to do so, we can multiply the second fraction by `(x+7)/(x+7)`. This works because it is the same thing as multiplying it by `1`, which wouldn't change it.

Now we have

`x^2/((x-7)(x+7))-((x+7)/(x+7))x/(x-7)`

`=>x^2/((x-7)(x+7))-(x(x+7))/((x-7)(x+7))`

Now distribute the `x` in the numerator of the second fraction

`x^2/((x-7)(x+7))-(x^2+7x)/((x-7)(x+7))`

Now subtract the fractions

`(x^2-x^2-7x)/((x-7)(x+7))`

`=>-(7x)/((x-7)(x+7))`

Answer

`-(7x)/((x-7)(x+7))`

or

`-(7x)/(x^2-49)`

Answer 2

`-(7x)/((x-7)(x+7))`
or
`-(7x)/(x^2-49)`

Explanation:

`(x^2)/(x^2-49)-(x)/(x-7)`
Notice that `x^2-49` is a difference of squares (`x^2-7^2`), so it can be factored to `(x+7)(x-7)` Inputting this,
`(x^2)/((x+7)(x-7))-(x)/(x-7)`
Now, multiply `(x)/(x-7)` by `(x+7)/(x+7)` to get common denominators:
`(x^2)/((x+7)(x-7))-((x)(x+7))/((x-7)(x+7))`
Since the denominators are now common, you can subtract the numerators and you get
`((x^2)-((x)(x+7)))/((x-7)(x+7))`. This simplifies to
`-(7x)/((x-7)(x+7))`
or
`-(7x)/(x^2-49)`