To solve the inequality \(2x > |x - 1|\), we need to consider the definition of absolute value, which gives us two cases:
Case 1: \(x - 1 \\geq 0\), which implies \(x \\geq 1\).
In this case, \(|x - 1| = x - 1\), so the inequality becomes:
\(2x > x - 1\)
Solving for \(x\), we get:
\(2x - x > -1\)
\(x > -1\)
Since \(x \\geq 1\) is a condition for this case, the solution for this case is \(x \\geq 1\).
Case 2: \(x - 1 < 0\), which implies \(x < 1\).
In this case, \(|x - 1| = -(x - 1) = -x + 1\), so the inequality becomes:
\(2x > -x + 1\)
Solving for \(x\), we get:
\(2x + x > 1\)
\(3x > 1\)
\(x > \frac{1}{3}\)
Since \(x < 1\) is a condition for this case, the solution for this case is \(\frac{1}{3} < x < 1\).
Combining both cases, the overall solution is \(x > \frac{1}{3}\).
