Question

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

Answer 1

`(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+9x-10` and `x+10`. So, in principle, we have to multiply `x+10` by something to obtain `x^2+9x-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+9x-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:

`(xcancel((x-1)))/(cancel((x-1))(x+10))+3/(x+10)= (x+3)/(x+10)`