## Precalculus (6th Edition) Blitzer

The required value is $x=-2,y=-4$
We have to subtract 2 times the second equation from 5 times the first equation and obtain, \begin{align} & 5\left( 2x+3y \right)-2\left( 5x-10y \right)=5\left( -16 \right)-2\left( 30 \right) \\ & 10x+15y-10x+20y=-80-60 \\ & 35y=-140 \\ & y=-4 \end{align} Substitute the value of $y=-4$ in the first equation and solve for x: \begin{align} & 2x+3\left( -4 \right)=-16 \\ & 2x=-16+12 \\ & 2x=-4 \\ & x=-2 \end{align} Thus, the value of $x=-2$ and $y=-4$.