It takes You, riding a motorcycle, 2 hours less time to travel 60 miles than it takes Marley to travel 50 miles riding a horse. You travel 10 miles per hour faster than Marley on a horse. Find yours and Marley’s times and rates of travel. Hint: Set up a chart using d= rt

.......... D ..R ...T

You 60 x+10 60/(x+10)
Marley 50 x 50/x

50/x - 60/(x+10) = 2
times x(x+10)
50(x+10 - 60x = 2x(x+10)
50x + 500 - 60x = 2x^2 + 20x
2x^2 + 30x - 500 = 0
x^2 + 15x - 250 = 0
(x+25)(x-10) = 0
x = 10 or x = -25, the latter being silly

Marley went 10 mph on a horse, you went 20 mph on a bike
Marley spent 50/10 or 5 hrs
You spent 60/20 or 3 hrs, which is 2 hours less as needed

if the horse has speed h, and the motorbike has speed m, then since time = distance/speed,

60/m = 50/h - 2
m = h+10

60/(h+10) = 50/h - 2
60h = 50(h+10) - 2h(h+10)
2h^2 + 30h -500 = 0

Now you can find h, and then m=h+10

To solve this problem, we can set up a chart using the formula d = rt, where d represents the distance, r represents the rate or speed, and t represents the time taken.

Let's assign variables:
Let "x" represent Marley's rate of travel on a horse (in miles per hour).
Let "y" represent your rate of travel on a motorcycle (in miles per hour).

We're given the following information:
1) It takes you 2 hours less time to travel 60 miles compared to Marley's 50 miles.
2) You travel 10 miles per hour faster than Marley on a horse.

Now, let's create two equations based on the given information:

Equation 1: Marley's time equation
Marley's time is determined by the distance traveled (50 miles) divided by his rate (x mph).
So, Marley's time is given by t1 = 50/x.

Equation 2: Your time equation
Your time is determined by the distance traveled (60 miles) divided by your rate (y mph).
So, your time is given by t2 = 60/y.

We're also told that your time is 2 hours less than Marley's time. So, we have the equation:
t2 = t1 - 2

Now, let's substitute the values of t1 and t2 from the above equations:
60/y = 50/x - 2

We also know that your rate y is 10 mph faster than Marley's rate x, so we can write:
y = x + 10

Now, we have a system of two equations:
Equation 1: 60/y = 50/x - 2
Equation 2: y = x + 10

We can solve this system of equations to find the values of x and y.

First, let's solve Equation 2 for x:
x = y - 10

Now, substitute the value of x into Equation 1:
60/y = 50/(y - 10) - 2

To eliminate the fractions, we can multiply both sides by y(y - 10):
60(y - 10) = 50y - 2y(y - 10)

Expanding and simplifying the equation:
60y - 600 = 50y - 2y^2 + 20y

Rearranging the equation:
2y^2 - 30y + 600 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.

By factoring, we can simplify the equation:
(y - 20)(2y - 30) = 0

So, either y - 20 = 0 or 2y - 30 = 0.

Solving each equation separately:
1) y - 20 = 0
y = 20

2) 2y - 30 = 0
2y = 30
y = 15

We have two possible solutions for the value of y: y = 20 or y = 15.

Since we're looking for your rate (y) and Marley's rate (x), we can use Equation 2 to find the corresponding values of x.

For y = 20:
x = y - 10
x = 20 - 10
x = 10

For y = 15:
x = y - 10
x = 15 - 10
x = 5

So, we have two possible solutions for the rates of travel:
1) Your rate (y) is 20 mph and Marley's rate (x) is 10 mph.
2) Your rate (y) is 15 mph and Marley's rate (x) is 5 mph.

Now that we know the rates of travel, we can calculate the times taken by substituting the values back into the original equations.

For the first solution (y = 20 and x = 10):
Your time t2 = 60/y = 60/20 = 3 hours.
Marley's time t1 = 50/x = 50/10 = 5 hours.

For the second solution (y = 15 and x = 5):
Your time t2 = 60/y = 60/15 = 4 hours.
Marley's time t1 = 50/x = 50/5 = 10 hours.

Therefore, the time and rates of travel are:
1) Your time is 3 hours and your rate is 20 mph. Marley's time is 5 hours, and Marley's rate is 10 mph.
2) Your time is 4 hours and your rate is 15 mph. Marley's time is 10 hours, and Marley's rate is 5 mph.