# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 30

The solution is, $\left\{ \left( 4,0 \right),\left( -3,\sqrt{7} \right)\text{ and }\left( -3,-\sqrt{7} \right) \right\}$

#### Work Step by Step

Let us consider the given equation: $x+{{y}^{2}}=4$ …… (I) ${{x}^{2}}+{{y}^{2}}=16$ …… (II) Substract equation (I) from equation (II): \begin{align} & {{x}^{2}}+{{y}^{2}}=16 \\ & x+{{y}^{2}}=4 \\ & \underline{-\text{ }-\text{ }-} \\ & {{x}^{2}}-x=12 \\ \end{align} Further simplify, \begin{align} & {{x}^{2}}-x-12=0 \\ & x\left( x-4 \right)+3\left( x-4 \right)=0 \\ & \left( x-4 \right)\left( x+3 \right)=0 \\ & x=4,-3\, \end{align} Substitute $x=4$ in equation (I): \begin{align} & 4+{{y}^{2}}=4 \\ & {{y}^{2}}=0 \\ & y=0 \end{align} Substitute $x=-3$ in equation (II): \begin{align} & -3+{{y}^{2}}=4 \\ & {{y}^{2}}=7 \\ & y=\pm \sqrt{7} \end{align} Thus, the solution is $\left\{ \left( 4,0 \right),\left( -3,\sqrt{7} \right)\text{ and }\left( -3,-\sqrt{7} \right) \right\}$.

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