Precalculus (6th Edition) Blitzer

Solution set is, $\left\{ \left( -4,3 \right),\left( 0,-1 \right) \right\}$.
Let us consider the provided system: \begin{align} & {{x}^{2}}-4y=4 \\ & x+y=-1 \end{align} So, from the second equation: $y=-1-x$ Substitute $y=-1-x$ in the first equation and solve for x: \begin{align} & {{x}^{2}}-4\left( -1-x \right)=4 \\ & {{x}^{2}}+4+4x=4 \\ & {{x}^{2}}+4x=0 \\ & x\left( x+4 \right)=0 \end{align} It implies that: $x=0,-4$ Substitute $x=0$ in the second equation and find the corresponding value of y: \begin{align} & 0+y=-1 \\ & y=-1 \end{align} Substitute $x=-4$ in the second equation and find the corresponding value of y: \begin{align} & -4+y=-1 \\ & y=3 \end{align} Thus, that set is $\left\{ \left( -4,3 \right),\left( 0,-1 \right) \right\}$.