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 852: 66


The solution set for the system of nonlinear equations is $\left\{ \left( 3,-2 \right),\left( 3,2 \right),\left( -3,2 \right),\left( -3,-2 \right) \right\}$.

Work Step by Step

The addition method is used to eliminate all the other variables to get the value of a different variable so that the equation found is an equation in one variable that can be solved. In the addition method, the equations in the system of equations are added after multiplying some equations with significant numbers so that it becomes easy to eliminate a variable to get the value of the other variable Example: Let us consider an example and use the addition method to solve the provided system of equations: ${{x}^{2}}-{{y}^{2}}=5$ …. (1) $3{{x}^{2}}-2{{y}^{2}}=19$ ….. (2) Multiply equation (I) by 2, so that the coefficient of ${{y}^{2}}$ in both equations becomes the same. $\begin{align} & 2\left( {{x}^{2}}-{{y}^{2}} \right)=2\left( 5 \right) \\ & 2{{x}^{2}}-2{{y}^{2}}=10 \\ \end{align}$ Now, multiply -2 to equation (I) and add to equation (II) to eliminate the variable term ${{y}^{2}}$. $\begin{align} & \underline{\begin{align} & -2{{x}^{2}}+2{{y}^{2}}=-10 \\ & 3{{x}^{2}}-2{{y}^{2}}=19 \end{align}} \\ & \,\,\,\,\,\,{{x}^{2}}+0y\,\,\,\,=9 \\ \end{align}$ Therefore, y is eliminated from the equation and simplifying the equation for the values of variable $ x $, we get: $\begin{align} & {{x}^{2}}+0y=9 \\ & {{x}^{2}}=9 \\ & {{x}^{2}}={{3}^{2}} \\ & x=\pm 3 \end{align}$ So, the solution to the equation is either $ x=3$ or $ x=-3$. Substitute these values of x in equation (I) to find values for y: For $ x=3$: $\begin{align} & {{x}^{2}}-{{y}^{2}}=5 \\ & {{\left( 3 \right)}^{2}}-{{y}^{2}}=5 \\ & {{y}^{2}}=4 \\ & y=\pm 2 \end{align}$ And, when $ x=3$ then $ y=2$ or $ y=-2$. For $ x=-3$: $\begin{align} & {{x}^{2}}-{{y}^{2}}=5 \\ & {{\left( -3 \right)}^{2}}-{{y}^{2}}=5 \\ & {{y}^{2}}=4 \\ & y=\pm 2 \end{align}$ Thus, for $ x=-3$ then $ y=2$ or $ y=-2$. Hence, the solution set for the system of equations is $\left\{ \left( 3,-2 \right),\left( 3,2 \right),\left( -3,2 \right),\left( -3,-2 \right) \right\}$
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