algebra word problem: still stuck
posted by john on .
rachel allows herself 1 hour to reach a sales appt 50 miles away. after she has driven 30 miles, she realizes that she must increase her speed by 15mph in order to get there on time. what was her speed for the first 30 miles?
* college algebra word problem - Reiny, Tuesday, February 17, 2009 at 9:09am
let her speed for the first leg be x mph
let her speed for the second leg be x+15 mph
so her time for the first leg is 30/x
her time for the second leg is 20/(x+15)
but 30/x + 20/(x+15) = 1
multiplying both sides by x(x+15) and simplifying I got
x^2 + 35x - 450 = 0
(x+10)(x-45) = 0
x = -10, which is silly or
x = 45 mph
* college algebra word problem - john, Tuesday, February 17, 2009 at 4:04pm
But wouldnt that actually factor into
which would imply that the positive value for x is actually 10.
plugging that back into the problem, that means she can go 30 miles at 10 miles per hour and 20 miles at 25 miles per hour to arrive at 50 miles in an hour?
how can she go 30 miles at 10 miles per hour? the speed doesnt seem correct here.
* college algebra word problem....help - john, Tuesday, February 17, 2009 at 4:28pm
Is it possible that the equation should have been set up as:
30miles(xmiles/per 1 hour) + 20miles (x+15/per 1 hour) = ?
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