You can still show work using skeeter's method. This is what he was getting at:

\(\displaystyle 5-x = A(x+1) + B(2x-1)\)

choose two "strategic" values for x to help you find A and B

Choose a value of x such that the A is eliminated. So let x = -1:

\(\displaystyle \begin{aligned}

5 + 1 &= A(-1 + 1) + B(2(-1) - 1) \\

6 &= A(0) + B(-3) \\

6 &= -3B \\

B &= -2

\end{aligned}\)

See what happens? You'll just have one variable B, that you can solve for it easily.

Now choose a value of x so that B is eliminated.