SOLUTION: I am really having a hard time setting this equation up, please help, show steps to answer for future use. A salesman drives from Ajax to Barrington, a distance of 120 mi. at a

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Question 213295This question is from textbook College Algebra
: I am really having a hard time setting this equation up, please help, show steps to answer for future use.
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?
All help is greatly appreciated! This question is from textbook College Algebra

Found 3 solutions by stanbon, nerdybill, josmiceli:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
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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A to B DATA:
distance = 120 miles ; rate = r mph ; time = d/r = 120/r hrs.
================================
B to C DATA:
distance = 150 miles ; rate = (r+10) mph ; time = 150/(r+10) hrs.
================================
Equation:
B to C time - A to B time = 1/10 hr
---------
150/(r+10) - 120/r = 1/10
---
Multiply thru by 10r(r+10) to get:
1500r - 1200(r+10) = r(r+10)
300r - 12000 = r^2 + 10r
r^2 - 290r + 12000 = 0
(r-50)(r-240) = 0
Realistic solution:
r = 50 mph (His rate from A to B)
=====================================
Cheers,
Stan H.

Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website!
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?
.
You will need to apply the distance formula:
d = rt
where
d is distance
r is speed or rate
t is time (in hours)
.
Start by converting 6 mins into hours:
6 * 1/60 = 0.1 hours
.
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
.
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
.
Details of quadratic to follow:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=137960 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 35.7148351640224, -335.714835164022. Here's your graph:

Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website!
Write 2 equations:
1 for A-B and another for B-C
A-B
---

(1)
---------------
B-C
---

(2) (note that 6 min = .1 hrs)
--------------------------------

And substituting from (1)

Substituting from (1) again:

Solve by completing the square:

I'm stuck- If I take the square root
of both sides, I get an imaginary on the right.
Can you see where I goofed?