Patrick B. answered 10/15/20
Math and computer tutor/teacher
Per eq 3
x = z+5
eq 1 becomes: z+5 - y + 2z = 7
3z - y = 2
eq 2 becomes: 2(z+5) + y + z =8
2z + 10 + y + z = 8
3z + y = -2
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Then 3z = y+2 and 3z = -2 - y
so y+2 = -2 - y
2y = -4
y = -2
then 3z + -2 = -2
3z = 0
z=0
this forces x=5
{x=5, y = -2, z=0} is the solution