## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 28

#### Answer

The solutions of the equations ${{x}^{2}}-2y=8$ and ${{x}^{2}}+{{y}^{2}}=16$ are $\left( \sqrt{12},2 \right),\left( -\sqrt{12},2 \right)$ and $\left( 0,-4 \right)$.

#### Work Step by Step

Let us consider the equations as follows: \begin{align} & {{x}^{2}}-2y=8 \\ & {{x}^{2}}+{{y}^{2}}=16 \\ \end{align} Also, consider these equations as (I) and (II). Now, demonstrate the steps as follows: Step 1: By eliminating the variable ${{x}^{2}}$, multiply $-1$ in equation (I) and using the addition method in equations (I) and (II) simplify as given below: \begin{align} & -{{x}^{2}}+2y+\left( {{x}^{2}}+{{y}^{2}} \right)=-8+16 \\ & {{y}^{2}}+2y=8 \\ & {{y}^{2}}+2y-8=0 \\ & {{y}^{2}}+4y-2y-8=0 \end{align} Simplify further, \begin{align} & y\left( y+4 \right)-2\left( y+4 \right)=0 \\ & \left( y-2 \right)\left( y+4 \right)=0 \end{align} Therefore, $y=2,-4$ are the solutions of both the equations. Step 2: Substitute the value of $y$ in equation (I) to find out the value of $x$ and simplify as follows: For $y=2$, \begin{align} & {{x}^{2}}-2y=8 \\ & {{x}^{2}}-2\left( 2 \right)=8 \\ & {{x}^{2}}-4=8 \\ & {{x}^{2}}=12 \end{align} Therefore, $x=\pm \sqrt{12}$ is the solution of this equation. For $y=-4$, \begin{align} & {{x}^{2}}-2y=8 \\ & {{x}^{2}}-2\left( -4 \right)=8 \\ & {{x}^{2}}+8=8 \\ & {{x}^{2}}=0 \end{align} Threfore, $x=0$ is the solution of this equation. Step 3: And verify the values of $x$ and $y$ in both of the equations. Now start by taking the pair $\left( \pm \sqrt{12},2 \right)$, and put $x=\pm \sqrt{12}$ and $y=2$. \begin{align} & {{x}^{2}}-2y=8 \\ & {{\left( \pm \sqrt{12} \right)}^{2}}-2\times 2=8 \\ & 12-4=8 \\ & 8=8 \end{align} Thus, the values satisfy the equation. And, \begin{align} & {{x}^{2}}+{{y}^{2}}=16 \\ & {{\left( \pm \sqrt{12} \right)}^{2}}+{{2}^{2}}=16 \\ & 12+4=16 \\ & 16=16 \end{align} Thus, the values satisfy the equation. Now, check by $\left( 0,-4 \right)$, and put $x=0$ and $y=-4$. \begin{align} & {{x}^{2}}-2y=8 \\ & {{\left( 0 \right)}^{2}}-2\times \left( -4 \right)=8 \\ & 0+8=8 \\ & 8=8 \end{align} Thus, the values satisfy the equation. And, \begin{align} & {{x}^{2}}+{{y}^{2}}=16 \\ & {{0}^{2}}+{{\left( -4 \right)}^{2}}=16 \\ & 0+16=16 \\ & 16=16 \end{align} Thus, the values satisfy the equation. Hence, the points $\left( \sqrt{12},2 \right),\ \left( -\sqrt{12},2 \right)$, and $\left( 0,-4 \right)$ are the required solutions of the system.

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