Find constants $A$ and $B$ such that \[\frac{x - 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]for all $x$ such that $x\neq -1$ and $x\neq 2$. Give your answer as the ordered pair $(A,B)$.
Note that x^2 - x - 2 can be factored as ( x -2) (x + 1)
Multiply through by this factorization and we have
x -7 = A(x + 1) + B( x -2) simplify
x - 7 = (A + B)x + ( A - 2B) equating terms we have this system
A + B = 1
A -2B = -7
subtract the second equation from the first and we have
3B = 8
B = 8/3
and
A + B = 1
A + 8/3 = 1
A = 1 - 8/3
A = - 5/3