Discussion
If \(\frac{x+1}{x-1} \lt 2\), then x lies in the interval
Answer: C
x lies in \((-\infty, 1) \cup (3, \infty)\).
Do not multiply through by \(x - 1\), whose sign is unknown. Bring everything to one side instead:
\[ \begin{aligned} \frac{x + 1}{x - 1} - 2 &\lt 0 \\ \frac{3 - x}{x - 1} &\lt 0 \\ \frac{x - 3}{x - 1} &\gt 0 \end{aligned} \]
The quotient is positive when numerator and denominator have the same sign: x greater than 3, or x less than 1. (x = 1 is excluded, where the expression is undefined.) The interval (1, 3) is where the inequality fails, and the options with -1 or -3 come from sign slips in the rearrangement.
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