Question 638073: Dave and Kyle live five miles apart. They decide to meet for dinner in one-half hour at a diner that is located between them. They leave their homes at the same time and Dave walks two miles per hour faster than Kyle. How fast does each walk if they both arrive at the diner in exactly the one-half hour?
Thanks
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! they're 5 miles apart.
they will meet for dinner in 1/2 hour.
the dinner place is located between them.
they leave their homes at the same time.
dave walks 2 miles per hour faster than kyle.
they both arrive at the dinner place in 30 minutes.
let the rate that kyle walks be x miles per hour.
this means the rate that dave walks is x+2 miles per hour.
the basic formula is rate * time = distance.
since 30 minutes is equal to .5 hours, we'll use rate in miles per hour and time in hours.
kyle walks x miles per hour for 30 minutes.
dave walks x+2 miles per hour for 30 minutes.
the diner is located somewhere between them but we don't know exactly where.
the distance between them is 5 miles.
assume kyle walks y miles.
this means that dave must walk 5-y miles.
the equation for kyle is based on rate * time = distance and becomes:
.5x = y
the equation for kyle is .5x = y
the equation for dave is .5(x+2) = 5-y
these are 2 equation that need to be solved simultaneously.
the easiest way to do that is the substitute for y from the first equation into the second equation.
after substituting .5x for y, the second equation becomes:
.5(x+2) = 5 - .5x
simplify this equation to get:
.5x + 1 = 5 - .5x
subtract 1 from both sides of the equation and add .5x to both sides of the equation to get:
x = 4
this means that x+2 = 6
kyle is walking 4 miles per hour and dave is walking 6 miles per hour.
kyle walks 4 miles per hour for 1/2 hour for a total of 2 miles.
dave walks 6 miles per hour for 1/2 hour for a total of 3 miles.
the diner is 2 miles from kyle's house and 3 miles from dave's house.
you had 2 equations with 2 unknown variables each.
substituting from 1 equation into the other equation left you with 1 equation with 1 unknown.
this could be solved and that led to the value of x.
once x was found, the rest could be found without too much difficulty.
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