6th Grade (12 Years Old) Mathematics - Triangle Inequality Question My son's math homework has this question which I find to be extremely confusing.  Neither my wife or I can wrap our heads around what they are expecting as an answer:

Two sides of a triangle have lengths $7$ and $9$.  The third side has length $x$.  Tell whether each of the following statements is always, sometimes but not always or never true.
a) $x = 12$
b) $x = 2$
c) $x < 2$
d) $x < 16$
e) $x < 15$

To my mind the answer is sometimes to all of them.  For $x = 12$ and $x = 2$, when considering all the possible values of $x$ it will sometimes be those values.  For $x < 2$, $x < 15$, and $x < 16$, it's sometimes because all values less than those make it true, until you hit $0$ which forms a line and no longer a triangle.
It looks like I've found the the teachers edition of the book (scroll to page 6 in the pdf; question 12), that just confuses me even further!
 A: From your pdf, just use the following:

If a triangle has sides of lengths $x$, $y$, and $z$, the following three inequalities must be true:
  \begin{align}
x+y&>z,\\
x+z&>y,\\
y+z&>x.
\end{align}

Here, $y=7$ and $z=9$.
So, for example, b. and c. can never happen since it would violate the first inequality.
A: Think of a "hinge" of two sides with fixed lengths 7 and 9, respectively. You can put it at any opening between $0^\circ$ and $180^\circ$, but not including the extreme values $0^\circ$ and $180^\circ$.
The third side will be given once you fix the degree of opening of your hinge. In the limit where the angle is extremely close to $0^\circ$, the third side tends to $9-7=2$. In the other limit, $180^\circ$, the third side tends to $9+7=16$.
So the third side can have all lengths between 2 and 16, not including those two extreme values.
(Once you guys have thought a bit about this, both you, your wife and your son will understand the triangle inequality well.)
A: I think you and your wife and your son can have fun figuring this out.
Even before you start thinking about triangles, you should be clear about distinguishing among "sometimes", "always" and "never". Each requires a different kind of argument. That's what I'll focus on in this answer.
I'd also recommend cutting out some strips of paper $7$ and $9$ inches long, and and a strip $16$ inches long marked off in inches. Then play around a bit before trying to think about the arithmetic. You'll notice that if you put the two shorter strips together at one end, the other ends will be at most $16$ inches apart and at least $2$ inches apart – the extreme cases when they're lined up.
You will have discovered something that you already know: the sum of any two sides of a triangle must be strictly greater than the third side. If the sum is equal, the "triangle" lies on one line.
Now you're ready for a careful reading, one statement at a time.
a) $x=12$.  There is a triangle with a third side of $12$, but the third side does not have to be $12$.  So the answer is that it's sometimes $12$.
My original answer was wrong. I'll leave it here since it's instructive to think about why. 
a) $\require{enclose}\enclose{horizontalstrike}{x=12}$. 
Since there's just one number to think about, the answer can't be "sometimes".
It has to be "always" or "never". 
Your strips of paper will tell you it's "always". 
The arithmetic is $\enclose{horizontalstrike}{9-7 < 12 < 9 + 7}$.
b) and c) Clearly "never", since $9-7 = 2$.
d) $x < 16$. Well, every triangle you managed to build had a third side less than $16$. The question starts with an actual triangle, so the answer is "always". That's not the same as "for every $x < 16$ you have a triangle". That's false, as you observed in a comment, since $2 < 16$ and there's no triangle with third side of $2$.
e) $x < 15$. OK, we have a triangle with third side $x$. Can it be less than $15$? Well, yes, obviously. Must it be less than $15$? No. It must be less than $16$, but it might be $15.5$.  So the answer to this one is "sometimes". 
(Even if $x$ had to be an integer (not specified in the question), it could be exactly $15$.)
Note: I've deliberately written a rather long winded answer. Thinking things out this way is the best way to learn, even if what your son ends up turning in to his teacher is closer to "just the answer".
A: The important thing here is to understand the bounds on the third side of the triangle. The triangle inequality tells us that the third side must have length $x\in (2,16)$, i.e. $2<x<16$ . This makes sense, because the sum of the lengths of the first two sides $A+B$ should never exceed the length $x$ of side $C$. Similarly, the difference between the lengths of the sides: $A-B$ should never be less than the length $x$ of the side $C$. 
$(a)$ This answer is sometimes true, since $2<12<16$.
$(b)$ This answer is never true, since $2\not<2<16.$
$(c)$ This answer is also never true, for the same reason as $(b)$.
$(d)$ This answer is always true, because any $x$ satisfying the constraint will satisfy $x<16$.
$(e)$ This is sometimes true, because a side of length $x=15.5$ could work, for example.
A: The answer is $d)$ as $ x < 7+9 = 16$ is always true.
