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

Oct 11, 2015

See the explanation below.

Explanation:

Just like with ratios of numbers, we need a common denominator to add these ratios.
In order to use the simplest (lowest degree) denominator, start by factoring the denominators.

$\frac{x - 1}{{x}^{2} + 4 x + 3} + \frac{x}{{x}^{2} - 9} = \frac{x - 1}{\textcolor{red}{\left(x + 1\right)} \left(x + 3\right)} + \frac{x}{\textcolor{b l u e}{\left(x - 3\right)} \left(x + 3\right)}$

So our common denominator will be $\textcolor{red}{\left(x + 1\right)} \textcolor{b l u e}{\left(x - 3\right)} \left(x + 3\right)$

$\frac{x - 1}{{x}^{2} + 4 x + 3} + \frac{x}{{x}^{2} - 9} = \frac{\textcolor{b l u e}{\left(x - 3\right)} \left(x - 1\right)}{\textcolor{red}{\left(x + 1\right)} \textcolor{b l u e}{\left(x - 3\right)} \left(x + 3\right)} + \frac{\textcolor{red}{\left(x + 1\right)} x}{\textcolor{red}{\left(x + 1\right)} \textcolor{b l u e}{\left(x - 3\right)} \left(x + 3\right)}$

$= \frac{{x}^{2} - 4 x + 3}{\left(x + 1\right) \left(x - 3\right) \left(x + 3\right)} + \frac{{x}^{2} + x}{\left(x + 1\right) \left(x - 3\right) \left(x + 3\right)}$

$= \frac{\left({x}^{2} - 4 x + 3\right) + \left({x}^{2} + x\right)}{\left(x + 1\right) \left(x - 3\right) \left(x + 3\right)}$

$= \frac{2 {x}^{2} - 3 x + 3}{\left(x + 1\right) \left(x - 3\right) \left(x + 3\right)}$

Check to see if we can reduce this. We can reduce it if we can factor the numerator using one of the factors in the denominator. We try, but it will not reduce, so we are finished.