SOLUTION: Brian paid $50 for 12 of honeydews, coconut and watermelons. The unit price for honeydew, coconut and watermelon respectively is $2, $5, and $9. Suppose that Brian bought at least

Brian paid $50 for 12 of honeydews, coconut and watermelons. 
The unit price for honeydew, coconut and watermelon 
respectively is $2, $5, and $9. Suppose that Brian bought at 
least one fruit for each. How many coconuts did Brian buy?

I set up a system of equations
h + c + w = 12
2h + 5c + 9w = 50 

The problem is it keeps cancelling out completely. Where did 
I mess up or is there a better way?

1. Use the first equation to eliminate the h term in the 
   second equation.
2. Then, use the second equation to eliminate the c term in 
   the first equation.
3. Solve for h and c in terms of w
4. Use the fact that the variables must represent positive 
   numbers to define inequalities to find w's domain.
5. Try all positive integers for w in w's domain.  



================================

1. Use the first equation to eliminate the h term in the 
   second equation.

Add -2 times the first equarion to 1 times the second
equation, but keep the first equation as is:

-2[ h +  c +  w = 12]
 1[2h + 5c + 9w = 50] 
—————————————————————
        3c + 7w = 26

So our system is not

    h +  c +  w = 12
        3c + 7w = 26



2. Then, use the second equation to eliminate the c 
term in the first equation, but keep the second 
equation as is.

    -3[h +  c +  w =  12
     1[    3c + 7w =  26
   —————————————————————
     -3h      + 4w = -10 

So now our system is

     -3h      + 4w = -10
           3c + 7w =  26

3. Solve for h and c in terms of w

Solving the first for h and the second for c

        h = (10+4w)/3

        c  = (26-7w)/3

4. Use the fact that the variables must represent 
   positive numbers to find w's domain.

All the variables are positive, so

        w > 0
(10+4w)/3 > 0, which solves to give w > -10/4
(26-7w)/3 > 0, which solves to give w < 26/7 or 3 5/7

So we have 0 < w < 3 5/7

And since w must be an integer, w can only be 1, 2, or 3

We try w = 1.

        h = (10+4w)/3 = (10+4·1)/3 = 14/3, not an integer

        c = (26-7w)/3 = (26-7·1)/3 = 19/3, not an integer

We try w = 2.

        h = (10+4w)/3 = (10+4·2)/3 = 18/3 = 6, an integer,
        which is feasible.

        c = (26-7w)/3 = (26-7·2)/3 = 12/3 = 4, an integer, 
        which is feasible.

We try w = 3.

        h = (10+4w)/3 = (10+4·3)/3 = 22/3, not an integer

        c = (26-7w)/3 = (26-7·3)/3 = 5/3, not an integer
      
So the only solution is w = 2, h = 6, c = 4.

Edwin
AnlytcPhil@aol.com