To solve the inequality \(|2x + 1| < 3|x - 2|\), we consider the critical points where each expression inside the absolute value is zero.
1. Solve \(2x + 1 = 0\):
\(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)
2. Solve \(x - 2 = 0\):
\(x - 2 = 0 \Rightarrow x = 2\)
These critical points divide the number line into intervals. We test each interval:
Interval 1: \(x < -\frac{1}{2}\)
Interval 2: \(-\frac{1}{2} < x < 2\)
Interval 3: \(x > 2\)
For each interval, remove the absolute values and solve the inequality:
1. For \(x < -\frac{1}{2}\), both expressions are negative:
\(-(2x + 1) < -3(x - 2)\)
\(2x + 1 > 3x - 6\)
\(1 + 6 > 3x - 2x\)
\(7 > x\)
2. For \(-\frac{1}{2} < x < 2\), \(2x + 1\) is positive and \(x - 2\) is negative:
\(2x + 1 < -3(x - 2)\)
\(2x + 1 < -3x + 6\)
\(2x + 3x < 6 - 1\)
\(5x < 5\)
\(x < 1\)
3. For \(x > 2\), both expressions are positive:
\(2x + 1 < 3(x - 2)\)
\(2x + 1 < 3x - 6\)
\(1 + 6 < 3x - 2x\)
\(7 < x\)
Combining the results, the solution is \(x < 1\) and \(x > 7\).
