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Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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![**Title: Solving Linear Equations for Rollercoaster Line Predictions**
**Overview:**
An amusement park features two new rollercoasters. The number of people in line for each rollercoaster, denoted as \( P \), varies with time \( t \) (in hours since the park opened). The lines are modeled by the following linear equations:
\[
\begin{align*}
1. \quad & P = 6t + 4 \\
2. \quad & P = 10t
\end{align*}
\]
**Problem Statement:**
Solve the system of equations graphically to determine:
1. How long after the park opens will there be an equal number of people in each rollercoaster line?
2. How many people will be in each line at this time?
**Solution Approach:**
- **Graphing the Equations:**
- **Equation 1**: \( P = 6t + 4 \)
- Y-intercept (when \( t=0 \)): \( P = 4 \)
- Slope: 6 (line rises 6 units for every 1 unit increase in \( t \))
- **Equation 2**: \( P = 10t \)
- Y-intercept (when \( t=0 \)): \( P = 0 \)
- Slope: 10 (line rises 10 units for every 1 unit increase in \( t \))
- **Finding the Intersection:**
By setting the equations equal to each other to find when the lines intersect:
\[
6t + 4 = 10t
\]
\[
4 = 4t
\]
\[
t = 1
\]
After 1 hour, the number of people in each line will be equal.
- **Calculating the Number of People:**
Substitute \( t = 1 \) back into either equation to find \( P \):
Using \( P = 10t \):
\[
P = 10 \times 1 = 10
\]
Therefore, 1 hour after the park opens, there will be 10 people in each line.
**Conclusion:**
After solving graphically, we determine that precisely 1 hour after opening, both rollercoaster lines will have an equal number of](https://content.bartleby.com/qna-images/question/bc9d167a-3919-4c8c-a43c-665a91017611/96013a88-2197-4401-824d-eaf543dd88e4/dwsgjwl_thumbnail.png)
Transcribed Image Text:**Title: Solving Linear Equations for Rollercoaster Line Predictions**
**Overview:**
An amusement park features two new rollercoasters. The number of people in line for each rollercoaster, denoted as \( P \), varies with time \( t \) (in hours since the park opened). The lines are modeled by the following linear equations:
\[
\begin{align*}
1. \quad & P = 6t + 4 \\
2. \quad & P = 10t
\end{align*}
\]
**Problem Statement:**
Solve the system of equations graphically to determine:
1. How long after the park opens will there be an equal number of people in each rollercoaster line?
2. How many people will be in each line at this time?
**Solution Approach:**
- **Graphing the Equations:**
- **Equation 1**: \( P = 6t + 4 \)
- Y-intercept (when \( t=0 \)): \( P = 4 \)
- Slope: 6 (line rises 6 units for every 1 unit increase in \( t \))
- **Equation 2**: \( P = 10t \)
- Y-intercept (when \( t=0 \)): \( P = 0 \)
- Slope: 10 (line rises 10 units for every 1 unit increase in \( t \))
- **Finding the Intersection:**
By setting the equations equal to each other to find when the lines intersect:
\[
6t + 4 = 10t
\]
\[
4 = 4t
\]
\[
t = 1
\]
After 1 hour, the number of people in each line will be equal.
- **Calculating the Number of People:**
Substitute \( t = 1 \) back into either equation to find \( P \):
Using \( P = 10t \):
\[
P = 10 \times 1 = 10
\]
Therefore, 1 hour after the park opens, there will be 10 people in each line.
**Conclusion:**
After solving graphically, we determine that precisely 1 hour after opening, both rollercoaster lines will have an equal number of
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