Question 213295
A salesman drives from Ajax to Barrington, a distance of 120 mi. at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 min. more time then the first leg, how fast was he driving between Ajax and Barrington? 
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You will need to apply the distance formula:
d = rt
where
d is distance
r is speed or rate
t is time (in hours)
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Start by converting 6 mins into hours:
6 * 1/60 = 0.1 hours
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Let x = speed driven from Ajax to Barrington
then
x+10 = speed driven from Barrington to Collins
.
d=rt
solving for t we get:
t = d/r
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From:"If the second leg of his trip took 6 min. more time then the first leg" we get:
120/x  = 150/(x+10) + .1
Multiply both sides by x(x+10) we get:
120(x+10) =  150x + .1x(x+10)
120x+1200 =  150x + .1x^2+1
1200 =  30x + .1x^2+1
0 =  30x + .1x^2 - 1199
0 =  .1x^2 + 30x - 1199
0 =  x^2 + 300x - 11990
Solve by applying the quadratic equation.  Doing so yields:
x = {35.71, -335.71}
We can throw out the negative solution (doesn't make sense) leaving us with:
x = 35.71 mph
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Details of quadratic to follow:
*[invoke quadratic "x", 1, 300, -11990 ]