## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 25

#### Answer

The solutions of the stated algebraic equations are $\left( 3,4 \right)$ and$\left( 3,-4 \right)$.

#### Work Step by Step

The equations are as given below: ${{x}^{2}}+{{y}^{2}}=25$ (I) ${{\left( x-8 \right)}^{2}}+{{y}^{2}}=41$ (II) Now, demonstrate the steps as given below: Step 1: When necessary, multiply either of the equations or both equations by a significant number so that the sum of the coefficients of ${{x}^{2}}$ or the sum of the coefficients of ${{y}^{2}}$ is zero. Thus, multiply the equation (II) by the value $-1$ so that the sum of the coefficients of ${{y}^{2}}$ becomes zero. \begin{align} & -{{x}^{2}}-{{y}^{2}}+\left( {{\left( x-8 \right)}^{2}}+{{y}^{2}} \right)=-25+41 \\ & {{\left( x-8 \right)}^{2}}-{{x}^{2}}=16 \\ & {{x}^{2}}+64-16x-{{x}^{2}}=16 \end{align} \begin{align} & 48=16x \\ & 3=x \\ \end{align} Therefore, $x=3$ is a solution of both equations. Step 2: Substitute the value of $x$ in equation (I) to find out the value of y as given below: For $x=3$, \begin{align} & {{x}^{2}}+{{y}^{2}}=25 \\ & {{3}^{2}}+{{y}^{2}}=25 \\ & 9+{{y}^{2}}=25 \\ & {{y}^{2}}=16 \end{align} \begin{align} & {{y}^{2}}=16 \\ & {{y}^{2}}-16=0 \\ & (y-4)(y+4)=0 \end{align} Putting the factors as zero to get: \begin{align} & y-4=0 \\ & y=4 \end{align} Or \begin{align} & y+4=0 \\ & y=-4 \end{align} Therefore, $y=\pm 4$ are solutions of this equation. Step 3: And verify the values of $x$ and $y$ in both of the equations: start by taking the pair$\left( 3,4 \right)$ and put $x=+\sqrt{5}\text{or}-\sqrt{5}$ and $y=4$. \begin{align} & {{x}^{2}}+{{y}^{2}}=25 \\ & {{3}^{2}}+{{4}^{2}}=25 \\ & 9+16=25 \\ & 25=25 \end{align} So, the values satisfy the equation. And, \begin{align} & {{\left( x-8 \right)}^{2}}+{{y}^{2}}=41 \\ & {{\left( 3-8 \right)}^{2}}+{{4}^{2}}=41 \\ & 25+16=41 \\ & 41=41 \end{align} So, the values satisfy the equation. Step 4: Now, begin by checking$\left( 3,-4 \right)$, and replace $x=3$ and $y=4$ . \begin{align} & {{x}^{2}}+{{y}^{2}}=25 \\ & {{3}^{2}}+{{\left( -4 \right)}^{2}}=25 \\ & 9+16=25 \\ & 25=25 \end{align} Therefore, the values satisfy the equation. And, \begin{align} & {{\left( x-8 \right)}^{2}}+{{y}^{2}}=41 \\ & {{\left( 3-8 \right)}^{2}}+{{\left( -4 \right)}^{2}}=41 \\ & 25+16=41 \\ & 41=41 \end{align} Therefore, the values satisfy the equation. Hence, the points $\left( 3,4 \right)$ and $\left( 3,-4 \right)$ are a solution of the system.

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