## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 33

#### Answer

The solution is $\left\{ \left( 2,2 \right),\left( 4,1 \right) \right\}$

#### Work Step by Step

The given equations are, \begin{align} & ~{{x}^{2}}+4{{y}^{2}}=20~~~~~~~~~~~~~~~~~~\left( \text{I} \right) \\ & x+2y=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} From equation (II): $x=6-2y$ Substitute $x=6-2y$ in equation (I): \begin{align} & {{\left( 6-2y \right)}^{2}}+4{{y}^{2}}=20 \\ & 8{{y}^{2}}-24{{y}^{{}}}+16=0 \\ & \left( y-2 \right)\left( y-1 \right)=0 \\ & y=1,2\, \end{align} Substitute $y=2$ in equation (II): \begin{align} & x+2\left( 2 \right)=6 \\ & x=2 \end{align} Substitute $x=1$ in equation (II): \begin{align} & x+2\left( 1 \right)=6 \\ & x=4 \end{align} Thus, the solution is $\left\{ \left( 2,2 \right),\left( 4,1 \right) \right\}$.

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