SOLUTION: supposs you have entered a 73-mile biathlon that consists of a tun and a bicycle race. during your run, your average velocity is 5 miles per hour, and during your bicycle race, you

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Question 1064448: supposs you have entered a 73-mile biathlon that consists of a tun and a bicycle race. during your run, your average velocity is 5 miles per hour, and during your bicycle race, your average velocity is 29 miles pet hour. you finish the race in 5 hours. what is the distance of the run? what is the distance of the bicycle race?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
                SPEED       TIME     DISTANCE
RUN              5           x        5x
BICYCLE         29           y        29y
TOTAL                        5        73

system%28x%2By=5%2C5x%2B29y=73%29

5%285-y%29%2B29y=73
25-5y%2B29y=73
25%2B24y=73
24y=73-25
24y=48
highlight%28y=2%29

DISTANCE OF THE BICYCLE
29y
29%2A2
highlight%2858%2Amiles%29---------the bicycle portion

DISTANCE OF THE RUN PORTION
73-58
highlight%2815%2Amiles%29--------running portion

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

supposs you have entered a 73-mile biathlon that consists of a tun and a bicycle race. during your run, your average velocity is 5 miles per hour, and during your bicycle race, your average velocity is 29 miles pet hour. you finish the race in 5 hours. what is the distance of the run? what is the distance of the bicycle race?
Let distance of the run be D

Then distance of the bicycle race = 73 - D

We then get the following TIME equation: D%2F5+%2B+%2873+-+D%29%2F29+=+5

Solve this equation for D, the distance of the run

Subtract value for D from 73 to get the distance of the bicycle race

That's it....nothing COMPLEX and/or CONFUSING!