To solve the inequality \(|x - 3a| > |x - a|\), we consider the critical points where the expressions inside the absolute values are zero. These points are \(x = 3a\) and \(x = a\).
We analyze the inequality in the intervals determined by these critical points: \((-\infty, a)\), \((a, 3a)\), and \((3a, \infty)\).
1. For \(x < a\):
\(|x - 3a| = 3a - x\) and \(|x - a| = a - x\).
The inequality becomes \(3a - x > a - x\), which simplifies to \(3a > a\), always true.
2. For \(a < x < 3a\):
\(|x - 3a| = 3a - x\) and \(|x - a| = x - a\).
The inequality becomes \(3a - x > x - a\), which simplifies to \(2a > 2x\) or \(x < 2a\).
3. For \(x > 3a\):
\(|x - 3a| = x - 3a\) and \(|x - a| = x - a\).
The inequality becomes \(x - 3a > x - a\), which simplifies to \(-3a > -a\), never true.
Thus, the solution is \(x < 2a\).
