M06 #16

Because you can not tell from \(x(u-t)=y(t-u)\) that \(x=-y\) is necessarily true:
\(x(u-t)=y(t-u)\) --> \(x(u-t)+y(u-t)=(x+y)(t-u)\) --> either \(x=-y\) or \(t=u\).

Complete solution:
Given that: \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\).

So, \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{u} = \frac{y}{x}\) (from 2), which means that \(\frac{t}{u}\) and \(\frac{x}{y}\) equal to their reciprocals: \(\frac{t}{u}=\frac{u}{t}\) and \(\frac{x}{y}=\frac{y}{x}\) --> \(t^2=u^2\) and \(t^2=u^2\) --> \(|t|=|u|\) (or which is the same \(t = \pm u\)) and \(|x|=|y|\) (or which is the same \(x = \pm y\)).

Answer: D.

One more thing: notice that answer choices A and C are the same, since we can not have two correct answers than both are wrong (there are some types of questions for which more than one answer can be correct but this is not that type).

Hope it's clear.