## anonymous 5 years ago Please HELP! :)mr. Duncan traveled to a city 180 miles from him home to attend a meeting. Due to car trouble, his average speed returning was 13 miles less than his speed going. If the total time for the road trip was 7 hours, at what rate of speed did he travel to the city? (round to nearest tenth)

1. anonymous

Formula is d=rt. Make two equations and set them equal to each other to find the rate, and then find the average, which is the first plus the second divided by 2

2. anonymous

so 180=r7 so 173/2..so 86.5?

3. anonymous

distance equal rate x time there are two equations we can use here. Going and coming. let t= time going let 7-t= time coming let x = rate going let x-13 = rate coming distance going and coming is 90 (180/2) so remember d=rt coming 90 =(x-13)(7-t) going 90 = xt solve this system of equations.

4. anonymous

wait, did you just solve it? ha, i'm confused. That all kind of makes sense. I just am not sure what the last step is?

5. anonymous

@junwagh I think u r making a mistake The question says "Duncan traveled to a city 180 miles from his home" This means one way distance is 180 miles and not 90 miles

6. anonymous

I set up the problem. We set up two equations using d =rt. One for going and one for coming. Then you solve that system of equations. I won't do that for you, the hard part is setting up the equation. 90 = xt 90 = (x-13)(7-t)

7. anonymous

yep the total distance is 180

8. anonymous

yes the total distance is 180, which means the distance going there is 90 and the distance coming back is also 90.

9. anonymous

oh wait mi bad lol replace 90with 180 then.

10. anonymous

The total distance has to be 360 miles as the "city is 180 miles from his house" so he has to go 180 miles and come back 180 miles

11. anonymous

it says from his house to attend the meeting he drove 180 miles.. hmm yeah. you might be right. Ahh. does anyone have an answer. Now all the numbers are confusing me. want to make sure I know how to solve with the last steps

12. anonymous

someone suggested 59 mph. anyone else get that?

13. anonymous

Okay here is how u do it Let his speed while going = x miles /hour Then his speed while coming back = (x-13) miles/hr Distance each side is 180 miles we know distance = speed x time or time = distance/speed So time taken while going = 180/x and time taken while coming back = 180/(x-13)

14. anonymous

Now the total time taken is 7 hours. This means (180/x) + 180/(x-13) = 7 or 180(x-13) + 180 (x) ------------------ = 7 x(x-13) Simplifying the above we get 180x - 2340 + 180x = 7 (x^2 - 13x)

15. anonymous

so how do we find the rate of speed he traveled to the city Thank you for writing that all out for me. I just want to make sure I put in the correct answer.

16. anonymous

DOes anyone have a final answer for this problem? I have been trying to work it out for over an hour so I would really appreciate any help with the solution.

17. anonymous

18. anonymous

Further this gives us 360x - 2340 = 7x^2 - 91x Further simplifying we get 0 = 7x^2 - 91x - 360x + 2340 or 7x^2 - 451x + 2340 = 0 This is a quadratic eqn. U can resolve it and get value of x

19. anonymous

59 mph

20. anonymous

that got me 2 solutions. 5.69 and 58.74...Is there a way to make this into 1 answer?

21. anonymous

maybe i'll just try 58.74.. Think this will be right?

22. anonymous

i got it right. THanks everyone

23. anonymous

yes it is 59.1 mph

24. anonymous

58.74 is correct as your speed cannot be 13 miles less than 5.69!!

25. anonymous

i actually did 58.7 and this was correct.

26. anonymous

It is a long winded question for sure

27. anonymous

28. anonymous

x=speed out, t=time out, x-13=speed back, 7-t=time back {x t == 180, (x-13)*(7-t)==180} {{x->5.69119,t->31.6278},{x->58.7374,t->3.06449}} Using second pair of x, y values. Speed, Time Out = {58.7374,3.06449} Speed, Time Back = {45.7374,3.93551}

29. anonymous

@robtobey how did u get th values {{x->5.69119,t->31.6278},{x->58.7374,t->3.06449}}

30. anonymous

Running the Mathematica Solve function: Solve[ { x t == 180, (x-13)*(7-t)==180}, {x,t} ]

31. anonymous

The syntax is somewhat weird. Those are, two consecutive = symbols.

32. anonymous

I am not familiar with this... Anyway thanks for th reply!!

33. anonymous

$\left\{x\to \frac{1}{14} \left(451+\sqrt{137881}\right),t\to \frac{1}{26} \left(451-\sqrt{137881}\right)\right\}$ The above is the actual output returned from Mathematica for the x, t, values used. Placing "//N" at the end converts to decimal numbers.

34. anonymous

Is this Mathematica some program to calculate answer or is it a mathematical theory like quadratic equations etc The answer seems like solution to the quadratic eqn

35. anonymous

I calculated using the formula -b +- root b^2 - 4ac x = --------------------- 2a

36. anonymous

Harkirat, mathematica is a mathematics software tool which helps you do calculations.

37. anonymous

OK!!??!! Thanks dhatraditya !! Is it free and available on net??

38. anonymous

Mathematica is a very large application system. About 3Gb of data to install it. It has been years in development and is mathematically very comprehensive in scope. It is about $2500 to a the common mom and popper who belly up to the cash counter to buy. For students and teachers they almost give it away. On the order of$125 one time cost as long as the student remains a student. The teachers probably would not allow it to be used in a class test but it will always show the correct answers for home work as long as the input is correct. If you give me a minute I'll find their web site.

39. anonymous

sure , thanks

40. anonymous

The Mathematica site is: http://www.wolfram.com/ I just purchasee Mathematica version 8 Home edition. It cost me about \$280 for a box edition delivered to California. By the way the installation comes on a DVD not a CD. Because of low price, the use of the Home is very restricted. For instance I can not use in to make any money.

41. anonymous

Thanks robtobey, I am in India and this is beyond my reach

42. anonymous

They may ship to India or have an agent there. I have to break off now. Been good talking to you. Adios.

43. anonymous

thank friend n bye till we chat again