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



4y = 80-5z

4y = 80-(4+1)z

4y = 80-4z-z

Divide through by 4



Isolate the fraction:



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

Let that intger be A

then  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



All must be > 0

so



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}



with A=1 becomes



with A=2 



with A=3



The problem has 3 solutions.

Edwin