I need to know how to set up this problem:

A salesman drives from Ajax to Barrington, a distance of 120 miles, at a steady speed. He then increases his speed by 10 mph to drive the 150 miles from Barrington to Collins. If the second leg of his trip took 6 minutes more time than the first leg, how fast was he driving between Ajax and Barrington?

let his first steady speed by x mph

so the time for the first leg of the trip was 120/x hours

on the second leg of the trip his speed was x+10 mph
so the time taken for that was 150/(x+10)

but this took 6 min or 1/10 hour longer

so 150/(x+10) - 120/x = 1/10

multiply each term by x(x+10) to get a quadratic equation.

one of the roots should be extraneous, let me know what you got

Let the two speeds be V1 and V2. The times required to travel the two legs of the journey are

120/V1 and 150/V2 = 150/(V1+10)
The second leg took 0.1 hours (6 minutes) more, so
150/(V1+10) = (120/V1) + 0.1
That equation can be solved for V1.
150/(V1 + 10) = (120 + 0.1V1)/V1
150 V1 = 120 V1 + 1200 + V1 + 0.1V1^2
0.1V1^2 -29 V1 + 1200 = 0
V1 = [29 +/-19]/0.2 = 50 or 240 mph
The higher of the two roots is unrealistic. Go with 50 mph

To set up this problem, we can use the formula:

Time = Distance / Speed

First, let's find the time it took for the salesman to drive from Ajax to Barrington.

Let the speed between Ajax and Barrington be "x" mph.

Distance from Ajax to Barrington = 120 miles

Time taken from Ajax to Barrington = Distance / Speed
= 120 / x
= 120x^(-1)

Now, let's find the time it took for the salesman to drive from Barrington to Collins.

The speed between Barrington and Collins is 10 mph faster than the speed between Ajax and Barrington, so it is (x + 10) mph.

Distance from Barrington to Collins = 150 miles

Time taken from Barrington to Collins = Distance / Speed
= 150 / (x + 10)
= 150(x+10)^(-1)

According to the problem, the second leg of the trip took 6 minutes more time than the first leg. Since 1 hour = 60 minutes, we can convert 6 minutes into hours by dividing by 60.

6 minutes = 6/60 = 1/10 hour

So, the equation is:

Time taken from Barrington to Collins = Time taken from Ajax to Barrington + 1/10

Now, let's set up the equation:

150(x+10)^(-1) = 120x^(-1) + 1/10

Now you can solve this equation to find the value of "x".

To set up this problem, we need to use the formula for calculating time: time = distance/speed.

Let's assume the salesman's speed from Ajax to Barrington is x mph. As given, the distance is 120 miles. We can calculate the time for this leg of the trip as follows:

Time (Ajax to Barrington) = distance/speed
Time (Ajax to Barrington) = 120/x

According to the problem, the salesman then increases his speed by 10 mph to drive from Barrington to Collins, which is a distance of 150 miles. The time for this leg of the trip is 6 minutes longer than the time taken for the first leg. We know that 1 hour is equal to 60 minutes, so we need to convert the minutes to hours to match the units.

Time (Barrington to Collins) = distance/speed
Time (Barrington to Collins) = 150/(x + 10)

Since the second leg took 6 minutes more than the first leg:

Time (Barrington to Collins) = Time (Ajax to Barrington) + 6 minutes
150/(x + 10) = 120/x + 6/60

Now, we can solve this equation to find the value of x, the speed of the salesman between Ajax and Barrington.

Multiply both sides by (x)(x + 10) to eliminate the denominators:

150x = 120(x + 10) + (x)(x + 10)(6/60)

Simplify the equation:

150x = 120x + 1200 + (1/10)(x^2 + 10x)

Combine like terms:

150x = 120x + 1200 + (1/10)x^2 + x

Rearrange the terms:

(1/10)x^2 + 11x - 1200 = 0

At this point, you can solve the quadratic equation by factoring, completing the square, or using the quadratic formula. Once you find the values of x, you can determine the speed of the salesman between Ajax and Barrington.