Q. 19 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 5–24, graph each equation of the system. Then solve the system to find the points of intersection.

$$\begin{gathered} xy=4; \\ {x}^2+y^2=8 \end{gathered}$$

Step-by-Step Solution

Answer

The graph of the system of equations $xy=4; x^2+y^2=8$ is:

Step 1. Given

The system of non-linear equation:

$$\begin{gathered} xy=4; \\ {x}^2+y^2=8 \end{gathered}$$

To graph the equation and to find the point of intersection.

Step 2. Graph the equations

Graph the equations in the same plane. 

Step 3. To find the point of intersection.

$$\begin{gathered} xy=4 \\ y=\frac{4}{x} \end{gathered}$$

Substitute the above value in the second equation,

$$\begin{gathered} x^2+y^2=8 \\ x^2+{\left(\frac{4}{x}\right)}^2=8 \\ x^2+\frac{16}{x^2}=8 \\ x^4+16=8x^2 \\ {x}^4-8x^2+16=0 \end{gathered}$$

Step 4. Solve the equation

$$\begin{gathered} x^4-8x^2+16=0 \\ {(x^2-4)}^2=0 \\ x^2=4 \\ x=\pm 2 \end{gathered}$$

Step 5. Find y

When $x=-2$,

$$\begin{gathered} xy=4 \\ -2y=4 \\ y=-2 \end{gathered}$$

When $x=2$,

$$\begin{gathered} 2y=4 \\ y=2 \end{gathered}$$

The points of intersection are $(-2,-2),(2,2)$

Step 6. Plot the point of intersection

The graph of the system of non-linear equations with the point of intersection is: