Question 253759
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{{{system(20 = x+y+z,
20 = 3x+y+0.5z)}}}

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 = 20-5/4z}}}

4y = 80-5z

4y = 80-(4+1)z

4y = 80-4z-z

Divide through by 4

{{{y = 20-z-z/4}}}

Isolate the fraction:

{{{z/4 = 20-z-y}}}

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

Let that intger be A

then {{{z/4 = 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(x=A, y=20-5A, z=4A)}}}

All must be > 0

so

{{{system(A > 0, 20-5A > 0, 4A > 0)}}}

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(x=A, y=20-5A, z=4A)}}}

with A=1 becomes

{{{system(x=1, y=15, z=4)}}}

with A=2 

{{{system(x=2, y=10, z=8)}}}

with A=3

{{{system(x=3, y=5, z=12)}}}

The problem has 3 solutions.

Edwin</pre>