Three Unknowns

A system can have three unknowns and three equations. You solve it with the same two methods. First eliminate one variable to get a system with two unknowns. Then solve that smaller system. In $x+y+z = 6;\quad x-y = 1;\quad y+z = 4$, the third equation gives $z = 4-y$. Substitute this into the first equation: $x+y+4-y = 6$. The $y$ terms cancel, so $x+4 = 6$ and $x = 2$. The second equation then gives $2-y = 1$, so $y = 1$. Finally $z = 4-1$, so $z = 3$.

Substitution and elimination also work for three unknowns. Eliminate one unknown to obtain equations in two unknowns, solve those equations, and substitute back to find the remaining value.