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).
I don't know how to have Only A and B on one side while the rest on the other side
+9673 You don't. That's not how you do it.
You finally get one side as a polynomial of x, with known coefficients, another side as a polynomial of x, with unknown coefficients.
You compare the coefficient of each term, then you can find the unknowns.
\(\dfrac{x+7}{x^2-x-2} = \dfrac{A}{x-2}+\dfrac B{x+1}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{A(x+1) + B(x-2)}{x^2-x-2}\\ \text{As the denominators are the same, so the numerators must be the same.}\\ x + 7 = A(x+1) + B(x-2)\\ x + 7 = Ax + A + Bx - 2B\\ x + 7 = (A+B)x + (A-2B)\\ \begin{cases} A + B = 1---(1)\\ A - 2B = 7---(2)\\ \end{cases}\\ (1)-(2):\\3B = -6\\ B = -2\\ A - 2 = 1 \implies A = 3\\ \therefore (A,B) = (3,-2)\)
