## Precalculus (6th Edition) Blitzer

The required value is $x=1,y=-2$
We have subtract three times the second equation from two times the first equation to obtain, Substitute the value of $y=-2$ in the first equation and solve for x as: \begin{align} & 3x-4\left( -2 \right)=11 \\ & 3x+8=11 \\ & 3x=3 \\ & x=1 \end{align}