## Precalculus (6th Edition) Blitzer

The required solution is $\left\{ \left( -2,1 \right),\left( 2,-1 \right) \right\}$
The given equations are, \begin{align} & 2{{x}^{2}}+xy=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & {{x}^{2}}+2xy=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\ \end{align} By multiplying equation (I) by $-2$ and adding equation (I) and equation (II): we get, \begin{align} & -4{{x}^{2}}-2xy=-12 \\ & \underline{{{x}^{2}}+2xy=0} \\ & -3{{x}^{2}}=12 \\ & x=\pm 2 \\ \end{align} Substitute $x=-2$ in equation (II): \begin{align} & {{\left( -2 \right)}^{2}}+2\left( -2 \right)y=0 \\ & 4-4y=0 \\ & y=1 \end{align} Substitute $x=2$ in equation (II): \begin{align} & {{\left( 2 \right)}^{2}}+2\left( 2 \right)y=0 \\ & 4+4y=0 \\ & y=-1 \end{align} Thus, the solution is $\left\{ \left( -2,1 \right),\left( 2,-1 \right) \right\}$.