\( \Rightarrow \frac { - x + 2 } { 1 - x } < 0 \)

To solve the inequality \( \frac{-x + 2}{1 - x} < 0 \), we want to determine when the fraction is negative. First, we can find the points where the numerator and denominator are zero. The numerator, \(-x + 2\), is zero when \(x = 2\), and the denominator, \(1 - x\), is zero when \(x = 1\). Now we'll analyze the intervals formed by these critical points: 1. \( (-\infty, 1) \) 2. \( (1, 2) \) 3. \( (2, \infty) \) Testing a point in each interval will reveal whether the fraction is negative or positive: 1. For \(x = 0\) in \((- \infty, 1)\): \( \frac{-0 + 2}{1 - 0} = \frac{2}{1} = 2 > 0 \) (not in solution set) 2. For \(x = 1.5\) in \( (1, 2) \): \( \frac{-1.5 + 2}{1 - 1.5} = \frac{0.5}{-0.5} = -1 < 0 \) (in solution set) 3. For \(x = 3\) in \( (2, \infty) \): \( \frac{-3 + 2}{1 - 3} = \frac{-1}{-2} = 0.5 > 0 \) (not in solution set) Thus, the inequality \( \frac{-x + 2}{1 - x} < 0 \) holds true in the interval \( (1, 2) \). The solution is: \( x \in (1, 2) \).