To solve the inequality \(|2x - 1| > 3|x + 2|\), we first consider the non-modular inequality:
\((2x - 1)^2 > 3^2(x + 2)^2\)
Expanding both sides, we have:
\((2x - 1)^2 = 4x^2 - 4x + 1\)
\(9(x + 2)^2 = 9(x^2 + 4x + 4) = 9x^2 + 36x + 36\)
Setting up the inequality:
\(4x^2 - 4x + 1 > 9x^2 + 36x + 36\)
Rearranging terms gives:
\(4x^2 - 4x + 1 - 9x^2 - 36x - 36 > 0\)
\(-5x^2 - 40x - 35 > 0\)
Dividing the entire inequality by -1 (and reversing the inequality sign):
\(5x^2 + 40x + 35 < 0\)
Factoring the quadratic:
\((x + 7)(x + 1) < 0\)
The critical points are \(x = -7\) and \(x = -1\).
Testing intervals, we find the solution is:
-7 < x < -1
