w, x, y, and z are different integers. If w/x = y/z, which

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.
-----------------

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
-----------------

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
----------------------------

Answer: E

Cheers,
Brent

Hi Brent,

Do you mean in the highlighted either w= x ? If no, can you elaborate how w = k?

Thanks

Even ii could be true.
Let w=2, x=-2, y=3 and z=-3....w+x=2-2=0=3-3=y+z.
w/x=y/z=-1..
So w+x=y+z

You're right. It should have read w = x.

Cheers,
Brent

You're absolutely right! I initially didn't account for the fact that (w+x) could equal 0

I've edited my response.

Kudos for everyone!

Cheers,
Brent

i) \(wy ≠0\)
Given \(\frac{w}{x}= \frac{y}{z}\), \(wy = 0\) only if \(w=y=0\).
Since w and y must be different integers, it is not possible that \(w=y=0\).
Thus, it must be true that \(wy ≠0\).
Eliminate D.

ii) \(w + x = y + z\)
\(\frac{w}{x}= \frac{y}{z}\)
\(\frac{w}{x} + \frac{x}{x}= \frac{y}{z} + \frac{z}{z}\)
\(\frac{w+x}{x}= \frac{y+z}{z}\)
The resulting equation is valid if \(w+x=0\) and \(y+z=0\).
Thus, it could be true that \(w+x = y+z\).
Eliminate A and C.

iii) \(w+y = x+z\)
\(\frac{w}{x}= \frac{y}{z}\)
\(\frac{w}{y}= \frac{x}{z}\)
\(\frac{w}{y} + \frac{y}{y}= \frac{x}{z} + \frac{z}{z}\)
\(\frac{w+y}{y}= \frac{x+z}{z}\)
The resulting equation is valid if \(w+y=0\) and \(x+z=0\).
Thus, it could be true that \(w+y = x+z\).
Eliminate B.

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