Question 1207768: Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist.After 6 hours they are 246 miles apart. How fast is each cyclist riding?
Westbound cyclist = x + 5
Eastward cyclist = x
After 6 six:
Westward cyclist = (x + 5) + 6
Eastward cyclist = x + 6
My equation is
(x + 5) + 6 + x + 6 = 246, which reduces to
2x + 17 = 246
Is this right?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! Opposite directions according to description
SPEEDS TIMES DISTANCES
To West x+5 6 (x+5)(6)
To East x 6 (x)(6)
Total dist. apart 246
 -----------You take it from here.
Answer by ikleyn(52810)  (Show Source):
You can put this solution on YOUR website! ..
It is FATALLY wrong.
In your setup, you write
After 6 six:
Westward cyclist = (x+5)+6
Eastward cyclist = x+6.
Here you add velocity, which is miles per hour, with hours.
It is out of logic, since adding quantities with different dimension units is nonsense.
It is the same as to add eggs with nails hoping to get square miles.
Therefore, your final equation is incorrect and makes no sense.
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| By the way, it is a powerful method of self-checking |
| to check on the way of making setup, |
| if you add quantities with the same dimension units. |
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It is a typical Travel & Distance problem for beginners.
For simple Travel & Distance problems, see introductory lessons
- Travel and Distance problems
- Travel and Distance problems for two bodies moving in opposite directions
- Travel and Distance problems for two bodies moving in the same direction (catching up)
in this site.
They are written specially for you.
You will find the solutions of many similar typical Travel & Distance problems there.
Read them and learn once and for all from these lessons on how to solve simple Travel and Distance problems.
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By the way, this problem was solved at this forum many years ago (perhaps, 20 years ago).
See the solution under this link
https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.96486.html
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
You (or someone) is submitting a lot of questions showing setups with equations that make no sense. This is a common mistake for students just beginning to study algebra.
To learn to avoid that problem, start your work on each problem by WRITING OUT CLEAR AND PRECISE definitions of the variable(s) you are going to use.
In the work you show, you start with this:
Westbound cyclist = x + 5
Eastward cyclist = x
Although we know from the statement of the problem that those expressions represent the speeds of the two cyclists, it will be much easier to write a sensible equation if instead you start your work something like this:
Let x = SPEED OF Eastbound cyclist
Then x+5 = SPEED OF Westbound cyclist
(My capital letters, for emphasis; you don't need to use capitals!)
Seeing on your paper that x and x+5 are speeds, you can then use distance = rate times time to write expressions for the distances the two cyclists travel and use those expressions and the given distance of 246 miles after 6 hours to write a correct equation.
So, continuing a good presentation of your work....
Then 6(x) = 6x is the distance the eastbound cyclist travels in 6 hours
and 6(x+5) = 6x+30 is the distance the westbound cyclist travels in 6 hours
(6x)+(6x+30) = 246
12x+30 = 246
12x = 216
x = 18
ANSWERS:
Eastbound cyclist speed: x = 18 (miles per hour)
Westbound cyclist speed: x+5 = 23 (miles per hour)
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