Algebra and Trigonometry (6th Edition)
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
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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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