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Answers without the blur. Sign up and see all textbooks for free! Q. 6

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={x}^{2}+1\\ y=4x+1\end{array}\right\$

Solutions of system of equations$\left\{\begin{array}{l}y={x}^{2}+1\\ y=4x+1\end{array}\right\$ are $\left(0,1\right)&\left(4,17\right)$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={x}^{2}+1\cdots \left(i\right)\\ y=4x+1\cdots \left(ii\right)\end{array}\right\$

## Step 2. Graph of the system of equations

Plot the graph of $y={x}^{2}+1&y=4x+1$on same cartesian plane ## Step 3. Solution of system

subtract equation ii from equation i

$y-y=\left({x}^{2}+1\right)-\left(4x+1\right)\phantom{\rule{0ex}{0ex}}0={x}^{2}-4x\phantom{\rule{0ex}{0ex}}0=x\left(x-4\right)$

By zero product property

$x-4=0\phantom{\rule{0ex}{0ex}}x=4$

or

$x=0$

So $x=0&4$

## Step 4. Solution of system

Substitute $x=0$ in equation ii

role="math" localid="1646869492730" $y=4x+1\phantom{\rule{0ex}{0ex}}y=4\left(0\right)+1\phantom{\rule{0ex}{0ex}}y=1$

Substituterole="math" localid="1646869479187" $x=4$ in equation ii

role="math" localid="1646869502982" $y=4x+1\phantom{\rule{0ex}{0ex}}y=4\left(4\right)+1\phantom{\rule{0ex}{0ex}}y=17$

So solutions of the system are role="math" localid="1646869524607" $\left(0,1\right)&\left(4,17\right)$

## Step 5. Verification

Substitute $x=0&y=1$in the system

$\left\{\begin{array}{l}1={0}^{2}+1\\ 1=4\left(0\right)+1\end{array}\right\\to \left\{\begin{array}{l}1=1\\ 1=1\end{array}\right\$

Substitute $x=4&y=17$ in system

$\left\{\begin{array}{l}17={\left(4\right)}^{2}+1\\ 17=4\left(4\right)+1\end{array}\right\\to \left\{\begin{array}{l}17=17\\ 17=17\end{array}\right\$

The system of equations is satisfied by both solutions

so solutions $\left(0,1\right)&\left(4,17\right)$ are correct

## Step 6. Point of intersection

Locate the$\left(0,1\right)&\left(4,17\right)$ for point of interception in the graph of a system of equations  ### Want to see more solutions like these? 