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 - Exercise Set - Page 850: 36


The solution is, $\left\{ \left( 0,0 \right),\left( -2,8 \right) \right\}$

Work Step by Step

The given equations are, ${{x}^{3}}+y=0$ (I) And $2{{x}^{2}}-y=0$ (II) For solving this equation use the addition method: Add equation (i) and (ii) to eliminate y: $\begin{align} & {{x}^{3}}+y=0 \\ & 2\underline{{{x}^{2}}-y=0} \\ & {{x}^{3}}+2{{x}^{2}}=0 \\ \end{align}$ ${{x}^{2}}\left( x+2 \right)=0$ (III) ${{x}^{2}}=0$ or $ x+2=0$ $ x=0$ or $ x=-2$ If $ x=0$ then substitute $ x=0$ in equation (I): $\begin{align} & {{\left( 0 \right)}^{3}}+y=0 \\ & y=0 \end{align}$ If $ x=-2$ then substitute $ x=-2$ in equation (I): $\begin{align} & {{\left( -2 \right)}^{3}}+y=0 \\ & y=8 \end{align}$ Therefore, the solution set is $\left\{ \left( 0,0 \right),\left( -2,8 \right) \right\}$ Hence, the solution set of the system is, $\left\{ \left( 0,0 \right),\left( -2,8 \right) \right\}$.
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