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 $(A,B)$.
+128460 \(\[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]\)
\( $x\neq -1$\)
and
\($x\neq 2$\)
This is a partial fractions problem.....we want to solve this
x + 7 = A(x + 1) + B(x - 2) simplify
x + 7 = (A + B)x + (A - 2B) equating coefficients, we have the following system
A + B = 1 → B = 1 - A (1)
A - 2B = 7 (2)
Sub (1) into (2) and we have
A - 2(1 - A) = 7
3A - 2 = 7
3A = 9
A = 3
And B = 1 - 3 = -2
So (A, B) = (3 , -2)
