i) \(wy ≠0\)
This, in fact, MUST be true.
We know this, because if one variable (w or y) = 0, then the other variable must also be 0.
Since we're told all 4 integers are DIFFERENT, statement i MUST be true.
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ii) \(w + x = y + z\)
Key concept: If \(\frac{w}{x}= \frac{y}{z}\), then \(\frac{w}{x}\) and \(\frac{y}{z}\) are equivalent fractions.

This means that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)
IMPORTANT: Since w, x, y and z are DIFFERENT, we can be certain that k ≠ 1

So, we can take: \(w + x = y + z\)
And replace y and z to get: \(w + x = wk + xk\)
Factor to get: \(w + x = k(w + x)\)
This tells us that EITHER k = 1, OR (w+x) = 0
We already know that k ≠ 1, but it COULD be the case that (w+x) = 0
One possible case is that w = 1, x = -1, y = 2 and z = -2
In this case, \(w + x = y +z\)

So, statement ii CAN be true
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iii) \(w+y = x+z\)
We'll use (again) the fact that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)

So, we can take: \(w+wk = x+xk\)
Factor each side to get: \(w(1+k) = x(1+k)\)
Notice that this equation is true if EITHER \(w=x\) OR \((1+k)=0\)
We know that \(w≠k\), but what happens if \((1+k)=0\)?

If \((1+k)=0\), then \(k = -1\)
If \(k = -1\), then we get: \(y = w(-1)\) and \(z = x(-1)\)

So, one possible case is that w = 1, x = -2, y = -1 and z = 2

We get: \(\frac{w}{x}= \frac{y}{z}=\frac{1}{-2}= \frac{-1}{2}\). PERFECT!

In this case, \(w+y = x+z\)

So, statement iii CAN be true
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Answer: E

Cheers,
Brent