# Thread: Application - System of Linear Equations

1. ## Application - System of Linear Equations

A hotel 35 miles from an airport runs a shuttle service to and from the airport. The 9:00 A.M. bus leaves for the airport traveling at 30 miles per hour. The 9:15 A.M. bus leaves for the airport traveling at 40 miles per hour. Write a system of linear equations that represents distance as a function of time for each bus. Graph and solve the system. How far from the airport will the 9:15 A.M. bus catch up to the 9:00 A.M. bus?

Okay I kind of understand the problem, but I just don't get how to set up a system of linear equations. From the application problem, I can only think of one. Help?

BTW, I'll graph the system of equations myself. I also attached a picture of the problem for reference.

2. Originally Posted by chrozer
A hotel 35 miles from an airport runs a shuttle service to and from the airport. The 9:00 A.M. bus leaves for the airport traveling at 30 miles per hour. The 9:15 A.M. bus leaves for the airport traveling at 40 miles per hour. Write a system of linear equations that represents distance as a function of time for each bus. Graph and solve the system. How far from the airport will the 9:15 A.M. bus catch up to the 9:00 A.M. bus?

Okay I kind of understand the problem, but I just don't get how to set up a system of linear equations. From the application problem, I can only think of one. Help?

BTW, I'll graph the system of equations myself. I also attached a picture of the problem for reference.
Setting up the system is easy.

For the first, it starts at the origin, so just remember, $\displaystyle d = tv$.

In this case, the first equation is $\displaystyle d = 30t$.

In the second, it leaves 15 minutes later, i.e. 0.25 of an hour later.

So point 1 is $\displaystyle (t, d) = (0.25, 0)$.

If it travels at 40miles/hour, then $\displaystyle m = 40$.

Plug them all into the equation $\displaystyle y = mx + c$ to get c. (in this case, y = d = 0, x = t= 0.25, m = v = 40).

You should find that $\displaystyle c = -10$

So the second equation is $\displaystyle d = 40t - 10$.

Therefore the system of equations is

$\displaystyle d = 30t$
$\displaystyle d = 40t - 10$.

Can you solve them to see where they meet?

3. Originally Posted by chrozer
I just don't get how to set up a system of linear equations. From the application problem, I can only think of one.
What equation did you get? And by what reasoning?

Thank you!

4. Originally Posted by Prove It
Setting up the system is easy.

For the first, it starts at the origin, so just remember, $\displaystyle d = tv$.

In this case, the first equation is $\displaystyle d = 30t$.

In the second, it leaves 15 minutes later, i.e. 0.25 of an hour later.

So point 1 is $\displaystyle (t, d) = (0.25, 0)$.

If it travels at 40miles/hour, then $\displaystyle m = 40$.

Plug them all into the equation $\displaystyle y = mx + c$ to get c. (in this case, y = d = 0, x = t= 0.25, m = v = 40).

You should find that $\displaystyle c = -10$

So the second equation is $\displaystyle d = 40t - 10$.

Therefore the system of equations is

$\displaystyle d = 30t$
$\displaystyle d = 40t - 10$.

Can you solve them to see where they meet?
Ah...Okay I see. I could only get the equation $\displaystyle d = 30t$.

So I solved it for $\displaystyle t$, and it came out to be that $\displaystyle t=1$, which means 1 hour. But do I plug this in the 2nd equation to find how far from the airport will the 9:15 A.M. bus catch up to the 9:00 A.M. bus?

And one more thing, the information about the hotel being 35 miles from an airport, does that have any importance in the problem?

5. Originally Posted by chrozer
Ah...Okay I see. I could only get the equation $\displaystyle d = 30t$.

So I solved it for $\displaystyle t$, and it came out to be that $\displaystyle t=1$, which means 1 hour. But do I plug this in the 2nd equation to find how far from the airport will the 9:15 A.M. bus catch up to the 9:00 A.M. bus?

And one more thing, the information about the hotel being 35 miles from an airport, does that have any importance in the problem?
Before solving the equations you need to think carefully about what d and t represent (the argument for equation (1) gives you the clue). Then you will know the answer to what you're asking.

6. Well I know that "d" represents distance and "t" represents time, with 30 and 40 being the velocity. I solved the system of equations of "t" and got 1, indicating to me that it is 1 hour.

I just don't get what it means when the question asks "How far from the airport will the 9:15 A.M. bus catch up to the 9:00 A.M. bus?".

If I plug it in both equations I would get 30 as the distance, of course, but is that the right answer?

Or is it 35 - 30, thus the answer being 5 miles that he 9:15 A.M. bus catch up to the 9:00 A.M. bus?

7. Originally Posted by chrozer
Well I know that "d" represents distance and "t" represents time, with 30 and 40 being the velocity. I solved the system of equations of "t" and got 1, indicating to me that it is 1 hour.

I just don't get what it means when the question asks "How far from the airport will the 9:15 A.M. bus catch up to the 9:00 A.M. bus?".

If I plug it in both equations I would get 30 as the distance, of course, but is that the right answer?

Or is it 35 - 30, thus the answer being 5 miles that he 9:15 A.M. bus catch up to the 9:00 A.M. bus?
The hotel bus is a distance d1 = 30t from the hotel at time t. Therefore it's a distance d2 = 35 - 30t from the airport.

The airport bus is a distance d2 = 40t - 10 from the airport.

The two buses meet when d2 = d3. Solve for t. Hence solve for the distance from the airport at which they meet.

8. Originally Posted by mr fantastic
The hotel bus is a distance d1 = 30t from the hotel at time t. Therefore it's a distance d2 = 35 - 30t from the airport.

The airport bus is a distance d2 = 40t - 10 from the airport.

The two buses meet when d2 = d3. Solve for t. Hence solve for the distance from the airport at which they meet.
So solving for $\displaystyle t$:
$\displaystyle 40t - 10 = 35 - 30t$
$\displaystyle 70t = 45$
$\displaystyle t = \frac {9}{14}$

Then I substitute "t" back into the eqation to find the distance, and I get approximately 15.71. Is that right?

9. Originally Posted by chrozer
So solving for $\displaystyle t$:
$\displaystyle 40t - 10 = 35 - 30t$
$\displaystyle 70t = 45$
$\displaystyle t = \frac {9}{14}$

Then I substitute "t" back into the eqation to find the distance, and I get approximately 15.71. Is that right?
Yes.