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 $(A,B)$.

By method of partial fractions, A - B = 1 and 4A + B = 1. Solving, we get (A,B) = (-3/5,2/5).

On the right-hand side make the denominator (x-2)(x+1)

Then the bottoms will be the same and you can just equate the coefficients on the top.

You want more hints ... ok

\(\frac{x - 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\\ \frac{x - 7}{x^2 - x - 2} = \frac{A(x+1)}{(x - 2)(x+1)} + \frac{B(x-2)}{(x + 1)(x-2)} \)

thank youuuu