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

May 19, 2015

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

Using Bhaskara for the first equation ${x}^{2} + 4 x - 5$

$\frac{- 4 \pm \sqrt{16 - 4 \left(1\right) \left(- 5\right)}}{2}$
$\frac{- 4 \pm 6}{2}$

${x}_{1} = - 5$, 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:

$\frac{4}{\left(x + 5\right) \left(x - 1\right)} - \frac{3}{\left(x - 1\right) \left(x + 1\right)}$

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

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

$\frac{4 \left(x + 1\right) - 3 \left(x + 5\right)}{\left(x + 5\right) \left(x - 1\right) \left(x + 1\right)}$

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

$\frac{x - 11}{{x}^{3} + 4 {x}^{2} - 5}$

May 19, 2015

Factor $y = \left({x}^{2} + 4 x - 5\right) = \left(x - 1\right) \left(x + 5\right)$

$\frac{4}{\left(x - 1\right) \left(x + 5\right)} - \frac{3}{\left(x - 1\right) \left(x + 1\right)}$ =

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

$y = \frac{x - 11}{\left(x - 1\right) \left(x + 5\right) \left(x + 1\right)}$