## Precalculus (6th Edition) Blitzer

The required value is $x=3,y=0$
We add both equations to obtain, \begin{align} & 2x+3y+\left( 2x-3y \right)=6+6 \\ & 4x=6+6 \\ & 4x=12 \\ & x=3 \end{align} Substitute the value of $x=3$ in the first equation and solve for y as follows: \begin{align} & 2\left( 3 \right)+3y=6 \\ & 6+3y=6 \\ & 3y=0 \\ & y=0 \end{align} Thus, the value of $x=3$ and .