Question

How do you combine `4/(x^2+4x-5) - 3/(x^2-1)`?

Answer 1

We can factor both denominators in order to find the lowest common denominator among them.

Using Bhaskara for the first equation `x^2+4x-5`

`(-4+-sqrt(16-4(1)(-5)))/2`
`(-4+-6)/2`

`x_1=-5`, which is the same as the factor `x+5=0`
`x_2=1`, which is the same as the factor `x-1=0`

For the second equation, we can find the roots simply by equaling it to zero.

`x^2-1=0`
`x^2=1`
`x=sqrt(1)`

`x_1=1`, which is the same as the factor `x-1=0`
`x_1=-1`, which is the same as the factor `x+1=0`

Now, we can rewrite your subtraction as:

`4/((x+5)(x-1))-3/((x-1)(x+1))`

The lowest common denominator must comprise all terms in the denominators, which lead us to `(x+5)(x-1)(x+1)`, where all the elements of both denominators are included and, thus, we can proceed to subtract.

Now we know our denominator is `(x+5)(x-1)(x+1)`, let's complete the subtraction:

`(4(x+1)-3(x+5))/((x+5)(x-1)(x+1))`

If you want to redistribute, it'll end up like this:

`(x-11)/(x^3+4x^2-5)`

Answer 2

Factor `y = (x^2 + 4x - 5) = (x -1)(x + 5)`

`4/[(x - 1)(x + 5)] - 3/[(x - 1)(x + 1)]` =

Numerator: 4x + 4 - 3x - 15 = x - 11

`y = (x - 11)/[(x - 1)(x + 5)(x + 1)]`