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

Question

. 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).