## Problem:

At the beginning of a bicycle ride, Jose and Wally are 30 miles apart. If they leave at the same time and ride in the same direction, Jose overtakes Wally in 6 hours. If they ride toward each other, they pass each other in 1 hour. What are their speeds?

### Given:

First scenario, overtaking:

Where:

x – speed of Jose

y – speed of Wally

dt – total distance

d1 – initial distance between Jose’s starting point and Wally’s starting point = 30 miles

d2 – distance between Wally’s starting point and the overtake point (indicated by the red dot)

tx – time travelled by Jose to overtake point = 6 hours

ty – time travelled by Wally at overtake point

Using the two key points in dealing with distance problems (refer to Distance problem: Overtaking & Meet With)

We can establish that *tx = ty = 6 hours*

And use the second key point to establish our equation:

*d1 + d2 = dt*

30 miles + ( y (ty) ) = ( x (tx) )

30 miles + ( y (6 hours) ) = ( x (6 hours) )

30 + 6y = 6x

Simplifying:

30/6 + 6y/6 = 6x/6

**5 + y = x or x – y = 5** *(eq. 1)*

Second scenario, meet with:

Where:

x – speed of Jose

y – speed of Wally

dt – total distance apart = 30 miles

d1 – distance travelled by Jose to the meet point (indicated by the red dot)

d2 – distance travelled by Wally to the meet point

tx – time travelled by Jose to the meet point = 1 hour

ty – time travelled by Wally to the meet point

Using the two key points in dealing with distance problems (refer to Distance problem: Overtaking & Meet With)

We then establish that *tx = ty = 1 hour*

And use the second key point to establish our equation:

*d1 + d2 = dt*

( x (tx) ) + ( y (ty) ) = 30 miles

( x (1 hour) ) + ( y (1 hour) ) = 30 miles

**x + y = 30** *(eq. 2)*

Using eq. 1, substituting x on eq. 2, we get:

x + y = 30

(5 + y) + y = 30

y + y = 30 – 5

2y = 25

**y = 12.5 mph speed of Wally**

using eq. 1, we get Jose’s speed:

x = 5 + y

x = 5 + 12.5 mph

**x = 17.5 mph speed of Jose**

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