## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( 0,0 \right),\left( -1,1 \right) \right\}$
The given equations are, \begin{align} & {{x}^{3}}+y=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & {{x}^{2}}-y=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\ \end{align} For solving the equation use the addition method: Add equation (I) and (II) to eliminate y: \begin{align} & {{x}^{3}}+y=0 \\ & \underline{{{x}^{2}}-y=0} \\ & {{x}^{3}}+{{x}^{2}}=0 \\ \end{align} ${{x}^{2}}\left( x+1 \right)=0$ (III) ${{x}^{2}}=0$ or $x+1=0$ $x=0$ or $x=-1$ If $x=0$ then substitute $x=0$ in equation (I): \begin{align} & {{\left( 0 \right)}^{3}}+y=0 \\ & y=0 \end{align} If $x=-1$ then substitute $x=-1$ in equation (I): \begin{align} & {{\left( -1 \right)}^{3}}+y=0 \\ & y=1 \end{align} Therefore, the solution set is $\left\{ \left( 0,0 \right),\left( -1,1 \right) \right\}$ Hence, the solution set of the system is, $\left\{ \left( 0,0 \right),\left( -1,1 \right) \right\}$.