## Precalculus (6th Edition) Blitzer

The solution set is $\left\{ \left( -1,-1 \right) \right\}$.
By rewriting the first equation, we get: \begin{align} & \frac{x}{6}-\frac{3y}{6}=\frac{2}{6} \\ & x-3y=2 \end{align} Therefore, the system can be rewritten as: \begin{align} & x-3y=2 \\ & x+2y=-3 \end{align} And subtract the second equation from the first equation: \begin{align} & x-3y-x-2y=2-\left( -3 \right) \\ & -5y=2+3 \\ & -5y=5 \\ & y=-1 \end{align} Substitute the value in the first equation and solve for x: \begin{align} & x-3\left( -1 \right)=2 \\ & x+3=2 \\ & x=-1 \end{align}. Thus, The solution set is $\left\{ \left( -1,-1 \right) \right\}$.