## Precalculus (6th Edition) Blitzer

The required value is $\left\{ \left( 3,-2 \right) \right\}$
We have to multiply the second equation by $5$ and add this to the first equation: \begin{align} & 2x+5y+5\left( 3x-y \right)=-4+5\left( 11 \right) \\ & 2x+5y+15x-5y=-4+55 \\ & 17x=51 \\ & x=3 \end{align} Substitute the value of $x$ in the first equation to obtain, \begin{align} & 6+5y=-4 \\ & 5y=-10 \\ & y=-2 \end{align}. Thus, the solution set is $\left\{ \left( 3,-2 \right) \right\}$.