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Answers without the blur. Sign up and see all textbooks for free! Q. 7

Expert-verified Found in: Page 772 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # 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\$

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

## 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

$\sqrt{36-{x}^{2}}=8-x\phantom{\rule{0ex}{0ex}}36-{x}^{2}={\left(8-x\right)}^{2}\phantom{\rule{0ex}{0ex}}36-{x}^{2}={x}^{2}-16x+64\phantom{\rule{0ex}{0ex}}2{x}^{2}-16x+28=0$

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\phantom{\rule{0ex}{0ex}}x=\frac{-\left(-16\right)±\sqrt{{\left(-16\right)}^{2}-4\left(2\right)\left(28\right)}}{2\left(2\right)}\phantom{\rule{0ex}{0ex}}x=\frac{16±\sqrt{32}}{4}\phantom{\rule{0ex}{0ex}}x=4±\sqrt{2}$

## Step 4. Solution of system

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

$y=8-x\phantom{\rule{0ex}{0ex}}y=8-\left(4+\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}y=4-\sqrt{2}$

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

$y=8-x\phantom{\rule{0ex}{0ex}}y=8-\left(4-\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}y=4+\sqrt{2}$

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

## Step 6. Point of intersection

Locate $\left(4+\sqrt{2},4-\sqrt{2}\right)&\left(4-\sqrt{2},4+\sqrt{2}\right)$ for point of interception in the graph of a system of equations  ### Want to see more solutions like these? 