# How do you add (x^2-x)/(x^2+9x-10) + 3/(x+10)?

Mar 5, 2016

$\frac{x + 3}{x + 10}$

#### Explanation:

The most importante thing is that you need to have all the parcels with the same denominator:

You easily can see that -10 is a zero of both ${x}^{2} + 9 x - 10$ and $x + 10$. So, in principle, we have to multiply $x + 10$ by something to obtain ${x}^{2} + 9 x - 10$. We know:

color(red)((x+a)(x+b)=x^2+(a+b)x+ ab

Knowing that a=10, we deduce that b=-1, because 10-1=9 and 10*(-1)=-10.

But -1 is not only a zero of ${x}^{2} + 9 x - 10$, but only of ${x}^{2} - x$. Instead of multiplying the second parcel by (x-1)/(x-1) (the standard procedure), we will cut the (x-1) factor in the first parcel:

Ultimately, you have:

$\frac{x \cancel{\left(x - 1\right)}}{\cancel{\left(x - 1\right)} \left(x + 10\right)} + \frac{3}{x + 10} = \frac{x + 3}{x + 10}$