SOLUTION: you have 20 dollars. you have to buy roosters, hens and chicks. roosters cost $3.00 a piece, $1.00 for hens, and $0.50 for chicks. you must buy 20 heads. how many for each? i ha

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: you have 20 dollars. you have to buy roosters, hens and chicks. roosters cost $3.00 a piece, $1.00 for hens, and $0.50 for chicks. you must buy 20 heads. how many for each? i ha      Log On


   



Question 253759: you have 20 dollars. you have to buy roosters, hens and chicks. roosters cost $3.00 a piece, $1.00 for hens, and $0.50 for chicks. you must buy 20 heads. how many for each?
i have 3 unknowns in here but i only can think of 2 equations:
20 = x+y+z
20 = 3x+y+0.5z
please help me solve this problem
thanks
adrian

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

system%2820+=+x%2By%2Bz%2C%0D%0A20+=+3x%2By%2B0.5z%29

This is called an "over-determined system,
so there are likely more than one solution.

Eliminate y and get

4x = z

Eliminate x and get

y+=+20-5%2F4z

4y = 80-5z

4y = 80-(4+1)z

4y = 80-4z-z

Divide through by 4

y+=+20-z-z%2F4

Isolate the fraction:

z%2F4+=+20-z-y

The right side is an integer, so the right side is too

Let that intger be A

then z%2F4+=+A and 20-z-y = A

z = 4A

Substitute in

4y = 80-5z

4y = 80-5(4A)

4y = 80-20A

y = 20-5A
 
Substitute that and z = 4A in

20 = x+y+z

20 = x+(20-5A)+4A
20 = x+20-5A+4A
20 = x+20-A
-x = -A
 x = A

So we have

system%28x=A%2C+y=20-5A%2C+z=4A%29

All must be > 0

so

system%28A+%3E+0%2C+20-5A+%3E+0%2C+4A+%3E+0%29

Solving the first and third give A > 0

Solving the second one:

20-5A > 0
  -5A > -20

Divide both sides by -5 reverses the inequality

    A < 4

0 < A < 4
  
So A is in the set {1, 2, 3}

system%28x=A%2C+y=20-5A%2C+z=4A%29

with A=1 becomes

system%28x=1%2C+y=15%2C+z=4%29

with A=2 

system%28x=2%2C+y=10%2C+z=8%29

with A=3

system%28x=3%2C+y=5%2C+z=12%29

The problem has 3 solutions.

Edwin