SOLUTION: <pre>Alphonse strats strats at point A and runs at a constant rate toward point C. At the same time, Brigitte strats at point B and runs toward point C also at a constant rate.Th

Alphonse starts at point A and runs at a constant rate toward point C. At the
same time, Brigitte starts at point B and runs toward point C also at a
constant rate.They arrive at C at exactly the same moment. I f they continue
running in the same directions, Alphonse arrives at B exactly 10 seconds before
Brigitte arrives at A. How fast was Brigitte running?

Brigitte arrives at A. How fast was Brigitte running?
                    60m                    40m
              ____________________________________________
              A                        C                 B

Let x = Alphonse's speed
Let y = Bridgitte's speed

Make this chart:

                                 D             R             T
Alphonse running from A to C                                     
Brigitte running from B to C                                   
Alphonse running from C to B                                    
Brigitte running from C to A                                    

Fill in the distances and the rates:

                                 D             R             T
Alphonse running from A to C     60            x                 
Brigitte running from B to C     40            y                
Alphonse running from C to B     40            x                
Brigitte running from C to A     60            y                

Fill in the times using T = D/R

                                 D             R             T
Alphonse running from A to C     60            x            60/x 
Brigitte running from B to C     40            y            40/y
Alphonse running from C to B     40            x            40/x
Briditte running from C to A     60            y            60/y

>>...They arrive at C at exactly the same moment...<<

That tells us that 

(Alphonse's time from A to C) = (Brigitte's time from B to C)

                                    60/x = 40/y


>>...Alphonse arrives at B exactly 10 seconds before Brigitte arrives at A...<<

This tells us that

(Alphonse's time from C to B) = (Brigitte's time from C to A) MINUS 10 seconds
             
                                    40/x = 60/y - 10

So we have this system of two equations and two unknowns:

60/x = 40/y
40/x = 60/y - 10

or

60/x - 40/y =   0
40/x - 60/y = -10

The first equation can be divided through by 20 and
The second equation can be divided through by 10

3/x - 2/y =  0
4/x - 6/y = -1

DO NOT clear of fractions!

Eliminate x
Multiply the first equation through by 4
Multiply the second equation through by -3

 12/x -  8/y = 0
-12/x + 18/y = 3                                           
                                           
Add the equations term by term

 12/x -  8/y = 0
-12/x + 18/y = 3 
————————————————
        10/y = 3

Multiply both sides by y

          10 = 3y
        10/3 =  y
           y = 10/3 = 3 1/3 meters/second 

That's all we were asked for, since y is Brigitte's speed. 

-------------------------------------
To check, we must also find x, Alphonse's speed = 5 meters/second

Now we can check:

Alphonse runs 60m from A to C at 5 meters/second. 
This takes her 60/5 or 12 seconds.

Brigitte runs 40m from B to C at 10/3 meters/second.
This takes her 40/(10/3) or 40(3/10) or 12 seconds also.

Brigitte runs 60m from C to A at 10/3 meters/second.
This takes her 60/(10/3) or 60(3/10) or 18 seconds.   

Alphonse runs 40m from C to B at 5 meters/second.
This takes her 40/5 or 8 seconds.  
This 8 seconds is exactly 10 seconds less than the 18 seconds 
it took Brigitte to run from C to A.  So the answers check.

Edwin
AnlytcPhil@aol.com