Question

Two cyclists leave towns
126mi
apart at the same time and travel toward each other. One cyclist travels
6

/mih
slower than the other. If they meet in
3
hours, what is the rate of each cyclist?

Answer 1

Cyclist A is slower.
He travels at s - 6 speed.
He travels x miles for 3 hours.

Cyclist B is faster.
He travels at s speed.
He travels 126 - x miles also for 3 hours.

Write two equations, and solve for x and s.

Answer 2

I don't get it @mathstudent55

Answer 3

Speed is distance divided by time.
That means that distance equals speed times time.

You can set up two equations, one for each cyclist, using the speed, distance and timer of each cyclist.
Then you solve the equations simultaneously.

Answer 4

Cyclist A:
speed = s - 6
distance = x
time = 3

distance = speed * time

x = (s - 6)(3)

x = 3(s - 6)

This is the equation for cyclist A.

Now we do the same for cyclist B using his speed, distance, and time.

Cyclist B:
speed = s
distance = 126 - x
time = 3

distance = speed * time

126 - x = s * 3

126 - x = 3s

This is the equation of cyclist B.

We have two equations in two unknowns, so we can solve for those unknowns.

Answer 5

x = 3(s - 6) -----> x = 3s - 18
126 - x = 3s -----> 126 - x = 3s
------------------ add the equations to eliminate x

126 = 6s - 18

6s - 18 = 126

6s = 144

s = 24

s is the speed of the faster cyclist.
The faster cyclist's speed is 24 mph.
The slower cyclist is 6 mph slower, so his speed is 18 mph.