Chapter 11, Problem 39RE

To determine

To find: The solution of system of equations.

Answer to Problem 39RE

The solution of the system of equations can be $\left(-2,1\right)$ and $\left(\frac{-2}{5},\frac{-11}{5}\right)$

Explanation of Solution

Given information:

The equations are $2x+y+3=0$ and $x^2+y^2=5$

Calculation:

Consider the equations.

  $2x+y+3=0$ …… (1)

  $x^2+y^2=5$ …… (2)

Add $-y-3$ both sides of Equation (1) .

  $\begin{array}{l}2x+y+3-y-3=0-y-3\\ y=-3-2x\end{array}$ …… (3)

Substitute $-3-2x$ in Equation (2).

  $\begin{array}{l}{x}^2+{\left(-3-2x\right)}^2=5\\ 5x^2+12x+4=0\end{array}$

Solve for $x$ .

  $\begin{array}{c}x=\frac{-12\pm \sqrt{{12}^2-4\left(5\right)\left(4\right)}}{2\left(5\right)}\\ =\frac{12\pm 8}{10}\\ =\frac{-2}{5}\text{Or}-2\end{array}$

Substitute $\frac{-2}{5}$ for $x$ in equation (3).

  $\begin{array}{l}y=-3-2\left(\frac{-2}{5}\right)\\ y=\frac{-11}{5}\end{array}$

Substitute $-2$ for $x$ in equation (3).

  $\begin{array}{l}y=-3-2\left(-2\right)\\ y=1\end{array}$

Therefore, the solution of the system of equations can be $\left(-2,1\right)$ and $\left(\frac{-2}{5},\frac{-11}{5}\right)$