help

If

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

then find A and B.

\(2x^2-x-1 = (2x+1)(x-1)\)

\(\begin{array}{|lrcll|} \hline & \dfrac{2x + 7}{2x^2 - x - 1} &=& \dfrac{A}{2x + 1} + \dfrac{B}{x - 1} \\\\ & \dfrac{2x + 7}{(2x+1)(x-1)} &=& \dfrac{A}{2x + 1} + \dfrac{B}{x - 1} \quad | \quad \times (2x+1)(x-1) \\\\ & \mathbf{2x + 7} &=& \mathbf{A(x - 1) + B(2x + 1)} \quad | \quad x\ne -\dfrac{1}{2}\ \text{and}\ x\ne 1 \\ \hline x = 1: & 2*1+7 &=& A(1-1) + B ( 2*1+1) \\ & 9 &=& A*0 + 3B \\ & 9 &=& 0+3B \\ & 3B &=& 9 \quad | \quad : 3 \\ & \mathbf{B} &=& \mathbf{3} \\ \hline x = -\dfrac{1}{2}: & 2\left(-\dfrac{1}{2}\right)+7 &=& A\left(-\dfrac{1}{2}-1\right) + B \left( 2\left(-\dfrac{1}{2}\right)+1\right) \\ & -1+7 &=& A\left(-\dfrac{3}{2}\right) + B*0 \\ & 6 &=& -\dfrac{3}{2}A \\ & A &=& -\dfrac{2}{3}*6 \\ & \mathbf{A} &=& \mathbf{-4} \\ \hline \end{array}\)