To solve the inequality \(2|x - 2| > |3x + 1|\), we consider the critical points where each absolute value expression changes sign.
1. Solve \(2(x - 2) = 3x + 1\) and \(2(x - 2) = -(3x + 1)\).
2. For \(2(x - 2) = 3x + 1\):
\(2x - 4 = 3x + 1\)
\(-4 - 1 = 3x - 2x\)
\(-5 = x\)
3. For \(2(x - 2) = -(3x + 1)\):
\(2x - 4 = -3x - 1\)
\(2x + 3x = -1 + 4\)
\(5x = 3\)
\(x = \frac{3}{5}\)
4. The critical values are \(x = -5\) and \(x = \frac{3}{5}\).
5. Test intervals around the critical points to determine where the inequality holds:
- For \(x < -5\), choose \(x = -6\):
\(2|-6 - 2| = 16\) and \(|3(-6) + 1| = 17\), so \(16 \not> 17\).
- For \(-5 < x < \frac{3}{5}\), choose \(x = 0\):
\(2|0 - 2| = 4\) and \(|3(0) + 1| = 1\), so \(4 > 1\).
- For \(x > \frac{3}{5}\), choose \(x = 1\):
\(2|1 - 2| = 2\) and \(|3(1) + 1| = 4\), so \(2 \not> 4\).
6. The solution is \(-5 < x < \frac{3}{5}\).
