Question

Sara went to meet her friend Rianna in Rome, but Rianna left Rome on a train 15 minutes earlier. The distance traveled by the train, in miles, with respect to time can be modeled by the equation y=40x+10
. Sara can buy a ticket on a faster train that will leave in 10 minutes and follow the same route. If the distance of the faster train with respect to time can be modeled by the equation y=60x−10
, which two lines on the graph show the solution to when Sara’s train catches up to Rianna’s train?

Answer 1

To determine when Sara's train catches up to Rianna's train, we'll first analyze the two distance equations provided:

1. Rianna's train: \(y = 40x + 10\)
2. Sara's train: \(y = 60x - 10\)

**Understanding the variables:**
- \(y\) represents the distance traveled (in miles).
- \(x\) represents time (in hours).

**Finding the point where Sara catches up to Rianna:**
First, we need to express how long each train has been traveling.

- **Rianna's train:** Since Rianna left 15 minutes earlier (which is \( \frac{15}{60} = \frac{1}{4}\) hours), we can write the time variable for Rianna as \(x + \frac{1}{4}\):

\(y = 40\left(x + \frac{1}{4}\right) + 10\)

Simplifying this:

\(y = 40x + 10 + 10 = 40x + 20\)

So, the equation for Rianna's train with respect to Sara's time starts at 10 miles and increases as per the new time variable.

**Setting up the equations:**
Now we equate the two expressions for distance to find when Sara catches up to Rianna:

\[
60x - 10 = 40x + 20
\]

**Solving for \(x\):**
1. Subtract \(40x\) from both sides:
\[
20x - 10 = 20
\]

2. Add 10 to both sides:
\[
20x = 30
\]

3. Divide both sides by 20:
\[
x = \frac{30}{20} = \frac{3}{2} \text{ hours or } 1.5 \text{ hours}
\]

**Finding the distances:**
Now, substitute \(x = \frac{3}{2}\) back into either equation to find out how far both trains have traveled at that time:

Using Sara's equation:
\[
y = 60\left(\frac{3}{2}\right) - 10 = 90 - 10 = 80 \text{ miles}
\]

Using Rianna's equation:
Rearranging Rianna's train's time as \(x + \frac{1}{4}\):
\[
y = 40\left(\frac{3}{2} + \frac{1}{4}\right) + 10
\]

First, convert \( \frac{1}{4} \) to \(\frac{1}{4} = \frac{2}{8}\) = \(\frac{0.25}{1} \).

Next, we can combine the fractions:
\[
\frac{3}{2} + \frac{1}{4} = \frac{6}{4} + \frac{1}{4} = \frac{7}{4}
\]

Now, substitute this back:
\[
y = 40 \cdot \frac{7}{4} + 10 = 70 + 10 = 80 \text{ miles}
\]

**Conclusion:**
Sara's train catches up to Rianna's train after 1.5 hours, at a distance of 80 miles from the starting point.

**Graphically, the two lines would intersect at the point (1.5, 80).** The solution represents the time near 1.5 hours and the distance being traveled by both trains at that point being the same (80 miles).