## Precalculus (6th Edition) Blitzer

The required value is $x=4,y=1$
We add both equations to obtain, \begin{align} & \left( 3x+2y \right)+\left( 3x-2y \right)=14+10 \\ & 3x+3x+2y-2y=24 \\ & 6x=4 \\ & x=4 \end{align} Substitute the value of $x=4$ in the first equation to obtain, \begin{align} & 3\left( 4 \right)+2y=14 \\ & 12+2y=14 \\ & 2y=2 \\ & y=1 \end{align} Thus, the value of $x=4$ and $y=1$.