Question

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 $x\\eq -1$ and $x\\eq 2$. Give your answer as the ordered pair $(A,B)$

Answer 1

If

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

then combining the fractions on the right gives

`(x-7)/(x^2-x-2) = (A(x+1) + B(x-2))/((x-2)(x+1))`

`(x-7)/(x^2-x-2) = ((A+B)x + A-2B)/(x^2 - x - 2)`

Solve for
`A` and
`B`.

`\begin{cases}A + B = 1 \\ A-2B = -7 \end{cases}`

Eliminating
`A`, we have

`(A + B) - (A-2B) = 1 - (-7) \implies 3B = 8 \implies B=\frac83`

which leaves us with

`A + \frac83 = 1 \implies A = -\frac53`

and so the solution is

`(A,B) = \boxed{\left(-\frac53,\frac83\right)}`