Step-by-step Answer:
This is a problem of partial fractions.
Step 1:
factor all denominators on the left-hand side (LHS).
LHS = a/[(x+1)(x-1)] + x/[(x-3)(x+1)]
Note the common factor (x+1) in both denominators, which makes the combined/common denominator [(x+1)(x-1)(x-3)]
Step 2:
multiply each term, top and bottom, by the factor "missing" from the common denominator.
The first term is missing (x-3), the second term is missing (x-1)
a(x-3)/[(x+1)(x-1)(x-3)] + x(x-1)/[(x+1)(x-1)(x-3)]
which can be simplified to:
[a(x-3)+x(x-1)]/[(x+1)(x-1)(x-3)]
Expand numerator:
[x^2 + ax-x -3a]/[(x+1)(x-1)(x-3)]
Step 3:
For the two expressions on each side of the equal sign (LHS and RHS) to be equivalent (for ALL values of x), the numerators and denominators must be identical when expanded, so
For denominator, we have factors [(x+1)(x-1)(x-3)], or b,c,d = -3, -1, 1 (in ascending order).
For the numerator, we need to have LHS = RHS
x^2 + (a-1)x -3a = x^2 + 2x -9
putting a=3 gives the
LHS = x^2 + (3-1)x -3(3) = x^2 + 2x - 9 which makes equality with the RHS.
So we have solved for all values required.