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

Question

2x + y = 2
2y + z = 8
2z + x = 7

Quantity I: The average value of x, y, and z.
Quantity II: 2

Which of the following is true:

A) I is bigger than II
B) II is bigger than I
C) I is equal to II
D) Insufficient information to determine

This is a question on a (home-made) SAT-like test where you have a minute per question. I solve it in the most basic way possible (exchanging one variable for another, solving for that, exchanging for the variable in the next equation, etc) but that is already cumbersome and the numbers you get to work with here are very unforgiving ($x = -\frac19$ , $y=2+\frac29$, $z=7+\frac {1}{18}$) for a test where you have very little time and no calculator.

Is there a faster way to solve it?

Answer 1

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

Answer 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 $2x = 2$.