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?

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    $\begingroup$ You don't have to solve it. It asks for the average so all you have to do is solve x+y+z. If you add all the terms you will have 3 of each variable so that's enough. 3x+3y+3z = 17 so the average is 17/9. $\endgroup$
    – fleablood
    Mar 23, 2016 at 8:33
  • 1
    $\begingroup$ @fleablood but you have to verify one solution exists $\endgroup$
    – ohno
    Mar 23, 2016 at 10:21
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    $\begingroup$ Um, no I don't. I just need to know what the average is which is equivalent to knowing what x+y+z is. Which, I'll admit, is not always possible but that is the goal. Not the individual solution, which need not be determinable. If you are given x + 2y + 2z = 26; 2x + y + z = 22; and 4x - y - y = 14 these terms are linearly dependent and can not be solved. The best you can do is determine x=6 and y+z = 10. But that's enough to know the average is 16/3. $\endgroup$
    – fleablood
    Mar 23, 2016 at 17:42
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    $\begingroup$ @fleablood suppose you're given the 2 by 3 system $x+y+z=1, x+y+z=2$. What's the average then? $\endgroup$ Mar 23, 2016 at 20:48
  • $\begingroup$ There is one easy way to settle the "no solution" case: using Sarrus' rule, it's immediate to see the determinant is $9$. The matrix of the system is $$\left(\begin{matrix}2 & 1 & 0\\0 & 2 & 1\\1 & 0 & 2\end{matrix}\right)$$ $\endgroup$ Mar 23, 2016 at 22:00

2 Answers 2


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


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 $2x = 2$.

  • 8
    $\begingroup$ The accepted answer correctly shows that from the given equations you can deduce that average < 2. It doesn't really matter whether the given equations have multiple solutions -- the deduction would still be valid and the answer would be the same. (E.g., suppose you had just the two equations x+2y = 8, x+2z = 3 -- then you'd conclude that 2(x+y+z)=11 and hence (x+y+z)/3=11/6 and hence I<II, even though there are infinitely many specific choices of x,y,z that satisfy the equations.) $\endgroup$ Mar 23, 2016 at 16:02
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    $\begingroup$ " In short we need to verify that there is a unique solution." No, we don't. If there is any solution unique or otherwise, we know that x+y+z = 17/3 and that is all we need to know. If we were given 1 equation: 5x + 5y + 5z = 45 there is absolutely no way to find a unique solution and absolutely no need to find a unique solution. The average is 3 because (x+y+z)/3 = 3. That's true no matter which of the infinite solutions we choose. $\endgroup$
    – fleablood
    Mar 23, 2016 at 17:47
  • $\begingroup$ @fleablood Tell me whether the arithmetic mean of $x$ and $y$ is greater or less than $2$: \begin{cases}x+y=2\\x+y=6\end{cases} $\endgroup$
    – egreg
    Mar 23, 2016 at 18:45
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    $\begingroup$ Those equations are inconsistant and none of the 4 answers are accurate. That is the "overconstrained" case. You should be asking that question of Eric Towers who was content in stating the question implies the overconstrained case need not be considered (which I agree with). I said *nothing about that. My objection is with " The underconstrained case is not so easily disposed". Yes, it is. The underconstrained case case is no hinderance whatsoever and needs absolutely no consideration. $\endgroup$
    – fleablood
    Mar 23, 2016 at 18:54
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    $\begingroup$ @fleablood Strictly speaking, it is mathematically correct to deduce from $x+y=2\land x+y=6$ that the average of $x,y$ is less than $2$, and greater than $2$ (and anything else). In practice, this means that unless the test specifically requests that you prove the assumptions are consistent, you can feel free to go with any piece of info you can find to solve the problem (like ignoring $x+y=6$ and jumping straight to $(x+y)/2=2/2=1<2$). Thus both the "overconstrained" and "underconstrained" cases require no extra work. $\endgroup$ Mar 24, 2016 at 0:36

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