Chapter 11, Problem 7CLR

To determine

To calculate:

The system of the following equations

  $\begin{array}{l}2x^2+y^2=6\\ x-y=3\end{array}$

Answer to Problem 7CLR

The solutions to the system of the equations are $x=1,y=-2$ .

Explanation of Solution

Given information:

  $\begin{array}{l}2x^2+y^2=6\\ x-y=3\end{array}$

Calculation:

Consider the equation

  $\begin{array}{l}2x^2+y^2=6\mathrm{...}\left(1\right)\\ x-y=3\mathrm{...}\left(2\right)\end{array}$

Put value of $y$ in equation (2)

  $\begin{array}{l}x-y=3\\ x=y+3\end{array}$

Put value of $x$ in equation (1)

  $\begin{array}{l}2x^2+y^2=6\\ 2{\left(y+3\right)}^2+y^2=6\\ 2\left(y^2+6y+9\right)+y^2=6\\ 2y^2+12y+18+y^2=6\\ 3y^2+12y+12=0\end{array}$

By using quadratic equation formula:

  $\begin{array}{l}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\left[a=3,b=12,c=12\right]\\ y=\frac{-12\pm \sqrt{{12}^2-4\left(3\right)\left(12\right)}}{2\left(3\right)}\\ y=\frac{-12\pm \sqrt{0}}{2.3}\\ y=\frac{-12}{2\left(3\right)}\\ y=-2\end{array}$

Put the value in equation (2):

  $\begin{array}{l}x=y+3\\ x=-2+3\\ x=1\end{array}$

The solutions to the system of the equations are $x=1,y=-2$ .