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

1 Answer
Mar 5, 2016

#(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)#