Q. 39 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 39–43, solve each system of equations.

$\left\{\begin{array}{l}2x+y+3=0\\ x^2+y^2=5\end{array}\right.$

Step-by-Step Solution

Answer

Solutions of the system of equations$\left\{\begin{array}{l}2x+y+3=0\\ x^2+y^2=5\end{array}\right.$ are$\left(\frac{-2}{5},\frac{-11}{5}\right) \& \left(-2,1\right)$

Step 1. Given data

The given system of equation is

$\left\{\begin{array}{l}2x+y+3=0\\ x^2+y^2=5\end{array}\right.$

Step 2. Formation of the single-variable equation

Rearrange equation $2x+y+3=0$

$$\begin{gathered} 2x+y+3=0 \\ y=-2x-3 \end{gathered}$$

Substitute expression of y in equation $x^2+y^2=5$

$$\begin{gathered} x^2+y^2=5 \\ {x}^2+{(-2x-3)}^2=55x^2+12x+9=5 \end{gathered}$$

Step 3. Solution of the single variable equation

Use the quadratic formula to solve $5x^2+12x+9=5$

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-12\pm \sqrt{{12}^2-4\left(5\right)\left(9\right)}}{2\left(5\right)} \\ x=\frac{-12\pm 8}{10} \end{gathered}$$

Separate the solutions

$$\begin{gathered} x=\frac{-12+8}{10} \\ x=\frac{-2}{5} \end{gathered}$$

and

$$\begin{gathered} x=\frac{-12-8}{10} \\ x=-2 \end{gathered}$$

Step 4. Solution of the system of equations

Substitute $x=\frac{-2}{5}$in equation $y=-2x-3$

$$\begin{gathered} y=-2x-3 \\ y=-2\left(\frac{-2}{10}\right)-3 \\ y=\frac{4}{5}-3 \\ y=\frac{-11}{5} \end{gathered}$$

Substitute $x=-2$in equation$y=-2x-3$

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

So solutions of the system of equations are $\left(\frac{-2}{5},\frac{-11}{5}\right) \& \left(-2,1\right)$