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: 65


The solution to the system of the nonlinear system of equations is the set $\left\{ \left( 0,-3 \right),\left( \frac{12}{5},\frac{9}{5} \right) \right\}$.

Work Step by Step

Let us consider the substitution method. So, in this method, a single equation is transformed into an equation with one variable by putting the values of other variables in the form of the required variable in this equation. And the values of the other variable are calculated with the use of other equation in the system of equations. Now, consider an example, $\begin{align} & 2x-y=3 \\ & y=2x-3 \end{align}$ Now, put this value of y in the first equation to convert the first equation into an equation with a single variable x. $\begin{align} & {{x}^{2}}+{{y}^{2}}=9 \\ & {{x}^{2}}+{{\left( 2x-3 \right)}^{2}}=9 \\ & {{x}^{2}}+4{{x}^{2}}-12x+9=9 \\ & 5{{x}^{2}}-12x=0 \end{align}$ So, this equation can be solved by taking factors to get the value of x. $\begin{align} & 5{{x}^{2}}-12x=0 \\ & x\left( 5x-12 \right)=0 \end{align}$ And the values of this equation is $ x=0$ or $ x=\frac{12}{5}$. Now, after putting this value of x in the second equation, one can calculate the value of y For $ x=0$: $\begin{align} & y=2x-3 \\ & y=2\left( 0 \right)-3 \\ & y=3 \end{align}$ Therefore, for $ x=0$ and $ y=-3$. For $ x=\frac{12}{5}$: $\begin{align} & y=2x-3 \\ & y=2\left( \frac{12}{5} \right)-3 \\ & y=\frac{24}{5}-3 \\ & y=\frac{9}{5} \end{align}$ Thus, for $ x=\frac{12}{5}$ the value of $ y=\frac{9}{5}$. Hence, the solution to the system of nonlinear equations after using the substitution method is the set $\left\{ \left( 0,-3 \right),\left( \frac{12}{5},\frac{9}{5} \right) \right\}$.
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