If x, y, z are natural numbers, then how many number of different solutions of the equation x+y+z=6 are possible?

Question

How many integer solutions are there to the system of equations below and which are those solutions?

x^2+y-z = 42
x+y^2-z = 18

Answer 1

10 solutions

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x,y,z>0
x+y+z=6 => x,y,z<5 x=1=> y+z=5 => y,z=1,4/2,3/3,2/4,1 --> 4 solutions
x=2=> y+z=4 => y,z=1,3/2,2/3,1 --> 3 solutions
x=3=> y+z=3 => y,z=1,2/2,1 --> 2 solutions
x=4=> y+z=2 => y,z=1,1 --> 1 solution
4+3+2+1=10 solutions