Q. 7 - Precalculus Enhanced with Graphing Utilities

Question

Graph each equation of the system. Then solve the system to find the points of intersection. 

$\left\{\begin{array}{l}y=\sqrt{36-x^2}\\ y=8-x\end{array}\right.$

Step-by-Step Solution

Answer

Solutions of system of equations$\left\{\begin{array}{l}y=\sqrt{36-x^2}\\ y=8-x\end{array}\right.$ are $(4+\sqrt{2},4-\sqrt{2}) \& (4-\sqrt{2},4+\sqrt{2})$and the graph is  

Step 1. Given data

The given system of equations is  

$\left\{\begin{array}{l}y=\sqrt{36-x^2} \cdots \left(i\right)\\ y=8-x \cdots \left(ii\right)\end{array}\right.$

Step 2. Graph of the system of equations

Plot the graph of $y=\sqrt{36-x^2} \& y=8-x$on same cartesian plane  

Step 3. Solution of system

Equate both equations

$$\begin{gathered} \sqrt{36-x^2}=8-x \\ 36-x^2={\left(8-x\right)}^2 \\ 36-x^2=x^2-16x+64 \\ 2x^2-16x+28=0 \end{gathered}$$

use quadratic formula

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-16)\pm \sqrt{{(-16)}^2-4\left(2\right)\left(28\right)}}{2\left(2\right)} \\ x=\frac{16\pm \sqrt{32}}{4} \\ x=4\pm \sqrt{2} \end{gathered}$$

Step 4. Solution of system

Substitute $x=4+\sqrt{2}$ in equation ii

$$\begin{gathered} y=8-x \\ y=8-\left(4+\sqrt{2}\right) \\ y=4-\sqrt{2} \end{gathered}$$

Substitute$x=4-\sqrt{2}$ in equation ii

$$\begin{gathered} y=8-x \\ y=8-\left(4-\sqrt{2}\right) \\ y=4+\sqrt{2} \end{gathered}$$

So solutions of the system are $(4+\sqrt{2},4-\sqrt{2}) \& (4-\sqrt{2},4+\sqrt{2})$

Step 6. Point of intersection

Locate $(4+\sqrt{2},4-\sqrt{2}) \& (4-\sqrt{2},4+\sqrt{2})$ for point of interception in the graph of a system of equations