Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Concept and Vocabulary Check - Page 850: 6


When solving ${{x}^{2}}+4{{y}^{2}}=20\text{ and }xy=4$ by the substitution method, one can eliminate $y$ by solving the second equation for $y$ to obtain $y=\frac{4}{x}$; then we substitute $\frac{4}{x}$ for $y$ in the first equation.

Work Step by Step

Let us consider the provided system: $\begin{align} & {{x}^{2}}+4{{y}^{2}}=20 \\ & xy=4 \end{align}$ So, from the second equation: $y=\frac{4}{x}$ Substitute $y=\frac{4}{x}$ in the first equation to obtain, $\begin{align} & {{x}^{2}}+4{{\left( \frac{4}{x} \right)}^{2}}=20 \\ & {{\left( {{x}^{2}} \right)}^{2}}+64=2{{x}^{2}} \\ & {{\left( {{x}^{2}} \right)}^{2}}-2{{x}^{2}}+64=0 \end{align}$ Thus, $y=\frac{4}{x}$ was obtained from the second equation and the quadratic expression was obtained by substituting $\frac{4}{x}$ for $y$ in the first equation.
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