How do you set up this?I am lost

Find an equation for the linear model of the situation below and use it to make a prediction. A train is traveling north at a constant rate. At 3:00 P.M. it is 55 miles north of a city. At 4:15 P.M. it is 80 miles north of the city. If d represents the distance in miles, and t represents the time in hours, how many miles north of the city will the train be at 5:45 P.M.?
MY answer book has - I don'tknow how they came up with these #'s
d=64t+55 d=231 miles

Beats me. I looked at Damon's solution, and 110 miles looks right to me.

Don't know where the 64 comes from, as that does not fit the data.

To set up the linear model for this situation, you need to use the given information to determine the equation that relates the distance traveled by the train (d) to the time (t) in hours.

Let's start by understanding the given information. At 3:00 P.M., the train is 55 miles north of the city. At 4:15 P.M., it is 80 miles north of the city.

From 3:00 P.M. to 4:15 P.M., the time difference is 1 hour and 15 minutes, which can be expressed as 1.25 hours. In this time, the train travels an additional distance of 80 - 55 = 25 miles.

So, we have the following information:
- Initial distance (d₁) at 3:00 P.M. = 55 miles
- Time (t) from 3:00 P.M. to 4:15 P.M. = 1.25 hours
- Additional distance (d₂) traveled in this time = 25 miles

Now we can find the equation for the linear model.

We know that the linear equation has the form: d = mt + b, where m is the slope and b is the y-intercept.

To find the slope (m), we can use the formula: m = (d₂ - d₁) / (t₂ - t₁), where d₂ is the final distance, d₁ is the initial distance, and t₂ - t₁ is the time difference.

Substituting the given values:
m = (25) / (1.25) = 20

Now that we have the slope, we can use any of the given points to find the y-intercept (b). Let's use the initial point at 3:00 P.M. (d = 55 miles, t = 0 hours).

Using the equation: d = mt + b,
55 = (20)(0) + b
b = 55

Now we have the slope (m = 20) and the y-intercept (b = 55), so the equation for the linear model is:
d = 20t + 55

To make a prediction for 5:45 P.M. (a time difference of 2.75 hours from 3:00 P.M.), substitute t = 2.75 into the equation:
d = 20(2.75) + 55
d = 55 + 55
d = 110

Therefore, according to the linear model, the train will be 110 miles north of the city at 5:45 P.M.

It's important to note that the answer book you mentioned has incorrect information. The correct equation for the linear model is d = 20t + 55, and the distance at 5:45 P.M. is 110 miles.