Solve each of the following system of equations in R.

27. $|\frac{2 x - 1}{x - 1}| > 2$

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Solution

Clearly, LHS of the inequality does not hold true for x = 1.

$|\frac{2 x - 1}{x - 1}| > 2 Clearly, LHS of the inequality is meaningful for x \neq \frac{1}{2} . \Rightarrow |\frac{2 x - 1}{x - 1}| - 2 > 0 Now, two cases arise . CASE 1 : \frac{2 x - 1}{x - 1} > 0 i . e . x \in (- \infty, \frac{1}{2}) \cup (1, \infty) Then, |\frac{2 x - 1}{x - 1}| = \frac{2 x - 1}{x - 1} ∴ |\frac{2 x - 1}{x - 1}| - 2 > 0 \Rightarrow \frac{2 x - 1}{x - 1} - 2 > 0 \Rightarrow \frac{2 x - 1 - 2 x + 2}{x - 1} > 0 \Rightarrow \frac{1}{x - 1} > 0 \Rightarrow x \in (1, \infty) . . . (i) CASE 2 : When, \frac{2 x - 1}{x - 1} < 0, i . e . x \in (\frac{1}{2}, 1) \Rightarrow |\frac{2 x - 1}{x - 1}| = \frac{- (2 x - 1)}{x - 1} ∴ |\frac{2 x - 1}{x - 1}| - 2 > 0 \Rightarrow \frac{- (2 x - 1)}{x - 1} - 2 > 0 \Rightarrow \frac{- 2 x + 1 - 2 x + 2}{x - 1} > 0 \Rightarrow \frac{- 4 x + 3}{x - 1} > 0 \Rightarrow x \in (\frac{3}{4}, 1) . . . (ii) Hence, the solution to the given set of inequalities is the union of (i) and (ii) . x \in (\frac{3}{4}, 1) \cup (1, \infty)$

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