## Precalculus (6th Edition) Blitzer

The required value is $x=\frac{32}{7},y=\frac{-20}{7}$
We know that the system can be rewritten as: \begin{align} & 5x-6y=40 \\ & 3x+2y=8 \end{align} By adding three times the second equation to the first equation: \begin{align} & 5x-6y+3\left( 3x+2y \right)=40+3\left( 8 \right) \\ & 5x-6y+9x+6y=40+24 \\ & 14x=64 \\ & x=\frac{32}{7} \end{align} Substitute the value of $x=\frac{32}{7}$ in the first equation and solve for y: \begin{align} & 5\left( \frac{32}{7} \right)-6y=40 \\ & -6y=40-\frac{160}{7} \\ & \frac{-6y}{-6}=\frac{1}{-6}\left( \frac{120}{7} \right) \\ & y=\frac{-20}{7} \end{align} Thus, the value of $x=\frac{32}{7}$ and $y=\frac{-20}{7}$.