# Thread: Linear Equations Problem (Please Help!!!)

1. ## Linear Equations Problem (Please Help!!!)

A group of hikers is to travel xkm by bus at an average speed of 48km/hr to an unknown destination. They then plan to walk back along the same route at an average speed of 4.8km/hr and to arrive back 24hrs after setting out in the bus. If they allow 2hrs for lunch and rest, how far must the bus take them?

If you can help to solve this problem, it would be greatly appreciated
P.S Please show all your workings
Thanks xoxo

2. You need to setup two simultaneous equations to solve for the multiple variables in the problem. The variables are: the time spent on the bus, the time spent on foot and the distance traveled each way. The basic formula for motion like this is $\displaystyle d=rt$, which means distance equals rate times time. I'll get you started.

Let's look at the bus scenario. You travel some number of kilometers for some amount of hours at a rate of 48 km/hr. Put another way $\displaystyle D=48t_b$.

Now look at the on foot scenario. Same idea except you travel at a different rate and time. So $\displaystyle D=4.8t_f$.

The times are different but the distances are equal and you can set them equal to each other. The other key is that the total time of movement was 24 hours - 2 hours for rest = 22 hours. So we also know that $\displaystyle t_b+t_f=22$

Hope that gets you to the solution.

3. We're not here to solve you problems for you, we are here to help you solve them yourself. Here's another way to approach this problem. Since the question asks for a distance, let "x" be the answer you want, the distance they ride the bus which is also the distance they walk back. Speed= distance/time so time= distance/speed. That tells you that it will take x/48 hours to walk x miles at 48 miles per hour. Adding f2 hours for rest, that gives x/48+ 2 hours spent not walking at so 24- (x/48+ 2)= 22- x/48 hours to walk back at 4.8 miles per hour. Again using "time= distance/speed" we have x/4.8= 22- x/48. Solve that for x.

(That is, of course, a different way of getting Jameson's solution. His $\displaystyle t_b$ is my x/48 and his $\displaystyle t_f$ is my x/4.8.