## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 851: 42

#### Answer

The required solution is $\left\{ \left( -5,0 \right),\left( 4,3 \right) \right\}$

#### Work Step by Step

The given equations are, $x-3y=-5$ (I) And ${{x}^{2}}+{{y}^{2}}-25=0$ (II) From equation (I) $x=3y-5$ (III) Substitute $x=3y-5$ in equation (II) \begin{align} & {{\left( 3y-5 \right)}^{2}}+{{y}^{2}}-25=0 \\ & 9{{y}^{2}}-30x+25+{{y}^{2}}-25=0 \\ & 10\left( {{y}^{2}}-3y \right)=0 \\ & y\left( y-3 \right)=0 \end{align} Now, $y=0,3$ Substitute $y=0$ in equation (III): \begin{align} & x=3\left( 0 \right)-5 \\ & x=-5 \\ \end{align} Substitute $y=3$ in equation (III): \begin{align} & y=3\left( 3 \right)-5 \\ & y=4 \\ \end{align} Thus, the solution is $\left\{ \left( -5,0 \right),\left( 4,3 \right) \right\}$.

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