## Precalculus (6th Edition) Blitzer

The required value is $x=2,y=-1$
We have to multiply the first equation with 2 and then subtract this from the second equation: \begin{align} & 6x+5y-2\left( 3x-7y \right)=7-2\left( 13 \right) \\ & 6x+5y-6x+14y=7-26 \\ & 19y=-19 \\ & y=-1 \end{align} Substitute the value of $y=-1$ in the first equation to obtain, \begin{align} & 3x-7\left( -1 \right)=13 \\ & 3x+7=13 \\ & 3x=6 \\ & x=2 \end{align} Thus, the value of $x=2$ and $y=-1$.