Solving a system of equations with 3 variables in under a minute

Solution 1:

Add all three equations together. The left-hand side is $3(x+y+z)$. The right-hand side is $17$. Thus, $\displaystyle{x+y+z\over 3} = {17\over 9}<2$.

Solution 2:

The accepted answer is good enough as far as it goes. (It's the method I used to get a first cut at an answer.) However, it is incomplete -- it does not work for overconstrained or underconstrained systems. Consider \begin{align*} x &= 0 & & & x &= 1 \\ x &= 1 & & \text{or} & y + z &= 2 \\ x &= 2 & & & 2x &= 2 \text{.} \\ \end{align*} The overconstrained case is (implicitly) excluded by the range of answers to the problem: no inequality holds and we have enough information to say so. The underconstrained case is not so easily disposed. (Just comparing the number of equations with the number of variables is inadequate for both these examples.) In short we need to verify that there is a unique solution.

To see this in the given problem: It would be handy if we could just substitute the second equation into the first. So double the first and substitute: $4x+2y = 4 \rightarrow 4x+(8-z) = 4$, which says $4x-z = -4$. Likewise, we double again and substitute the third equation: $8x - 2z = -8 \rightarrow 8x - (7-x) = -16$, which says $9x = 9$.

So there is a solution and, for any solution, $x=1$. This fact with the first and third equations gives a unique choice for $y$ and for $z$, so the solution is unique and the system is not underconstrained. When we try the same thing with the underconstrained system above, we do not find a unique choice for $y$ and $z$, so there is insufficient information to determine the result.

Edit : For those convinced an underconstrained system always has a finite average... Find the average of $x$, $y$, and $z$ subject to $2x = 2$.