## Precalculus (6th Edition) Blitzer

The required value is $\left\{ \left( 1,4 \right) \right\}$
Multiply the second equation by $2$ and then add that to the first equation to obtain: \begin{align} & 2\left( 4x+y \right)+\left( 3x-2y \right)=2\left( 8 \right)+\left( -5 \right) \\ & 8x+2y+3x-2y=16-5 \\ & 11x=11 \\ & x=1 \end{align} Substitute the value of $x$ in the first equation and solve for y as follows: \begin{align} & 3-2y=-5 \\ & -2y=-8 \\ & y=4 \end{align} Thus, the solution set is $\left\{ \left( 1,4 \right) \right\}$.