let the numbers be x>y>z (largest to smallest)
x+y+z=23 eq1
3z=x-4
x=3z+4 eq2
x+z=16 eq3
substituting for x+z in eq1 we get
16+y=23
y=23-16
y=7
substituting for x and y in eq1 we get
3z+4+7+z=23
4z+11=23
4z=23-11=12
z=12/4
z=3
x=3z+4
x=3(3)+4=9+4
x=13
The three numbers are 13,7 and 3.