C/(x-2) + D/(x+1) = 6x/(x-2)(x+1)
Rewrite the terms on the left so that they have the same denominator as the term on the right.
C(x+1) + D(x-2)/[(x-2)(x+1)] =
6x/(x-2)(x+1)
Cx + C + Dx -2D = 6x
This can only be true if C+D= 6 and C-2D = 0
Solve those two equations to get C and D
This is called the method of partial fractions
I have the answer, but don't know the procedure.
The question is:
Find the values of C and D if
c/(x-2) + D/(x+1) = 6x/(x^2-x-2)
The answers are: C= 4, D=2
HERE IS MY WORK:
Well, I know that we can add the left side and the denominators become the same. So, the new equations becomes:
c(x+1)+ D(x-2) = 6x
but, how do you go from here to find the values of c and d?
1 answer