A(x+1)(x-1)+b(x-2)(x+1)+(x+1)^2=9x-x-10 What is a+b?

a(x+1)(x-1)+b(x-2)(x+1)+(x+1)^2=9x^2-x-10

(x+1)(a(x-1)+b(x-2)+(x+1))=(x+1)(9x-10)

a(x-1)+b(x-2)+x+1=9x-10

Now this equation is much simpler!

(a+b)x-a-2b+x+1=9x-10

(a+b)x-a-2b=8x-11

(a+b-8)x-a-2b-11=0

a+b-8=(a+2b-11)/x

I can't solve it 3 variables and 1 equations means infinite answers so yea.

For this case we have the following expression:
 y = (x-5) / (x ^ 2 - 6x +5)
 Rewriting the expression we have:
 y = (x-5) / ((x-5) (x-1))
 Therefore, discontinuity points are those that make the denominator zero:
 x = 5
 x = 1
 The point of removable discontinuity is:
 x = 5
 
 Discontinuity:
 x = 5
 x = 1
 Removable:
 x = 5