## Precalculus (6th Edition) Blitzer

The required value is $x=-4,y=3$
We have to multiply the first equation with four and then add this to the second equation. \begin{align} & 4\left( x+2y \right)+\left( -4x+3y \right)=4\left( 2 \right)+25 \\ & 4x+8y-4x+3y=8+25 \\ & 11y=33 \\ & y=3 \end{align} Substitute the value of $y=3$ in the first equation and solve for x as follows: \begin{align} & x+2\left( 3 \right)=2 \\ & x+6=2 \\ & x=2-6 \\ & x=-4 \end{align} Thus, the value of $x=-4$ and $y=3$.