To solve \(|2x + 3| > 3|x + 2|\), consider the non-modular inequality:

\((2x + 3)^2 > 3^2(x + 2)^2\)

Expanding both sides gives:

\((2x + 3)^2 = 4x^2 + 12x + 9\)

\(3^2(x + 2)^2 = 9(x^2 + 4x + 4) = 9x^2 + 36x + 36\)

Set up the inequality:

\(4x^2 + 12x + 9 > 9x^2 + 36x + 36\)

Simplify to:

\(4x^2 + 12x + 9 - 9x^2 - 36x - 36 > 0\)

\(-5x^2 - 24x - 27 > 0\)

Multiply through by -1 (reversing the inequality):

\(5x^2 + 24x + 27 < 0\)

Factor the quadratic:

\((5x + 9)(x + 3) < 0\)

Find critical points by setting each factor to zero:

\(5x + 9 = 0 \Rightarrow x = -\frac{9}{5}\)

\(x + 3 = 0 \Rightarrow x = -3\)

Test intervals around the critical points to determine where the inequality holds:

For \(-3 < x < -\frac{9}{5}\), the inequality holds.

For \(x > -3\) and \(x < -\frac{9}{5}\), the inequality holds.

Thus, the solution is:

\(-3 < x < \frac{9}{5} \text{ or } x > -3 \text{ and } x < \frac{9}{5}\)

To solve the inequality \(|2x - 3| < 3x + 2\), we consider two cases based on the definition of absolute value.

Case 1: When \(2x - 3 \\geq 0\), the inequality becomes:

\(2x - 3 < 3x + 2\)

Subtract \(2x\) from both sides:

\(-3 < x + 2\)

Subtract 2 from both sides:

\(-5 < x\)

Case 2: When \(2x - 3 < 0\), the inequality becomes:

\(-(2x - 3) < 3x + 2\)

\(-2x + 3 < 3x + 2\)

Add \(2x\) to both sides:

\(3 < 5x + 2\)

Subtract 2 from both sides:

\(1 < 5x\)

Divide by 5:

\(x > \frac{1}{5}\)

Combining both cases, the solution is \(x > \frac{1}{5}\).

The function \(y = |2x - 3|\) is an absolute value function, which creates a V-shaped graph.

1. The expression inside the absolute value, \(2x - 3\), is zero when \(x = \frac{3}{2}\). This is the vertex of the V-shape.

2. For \(x < \frac{3}{2}\), \(2x - 3\) is negative, so \(y = -(2x - 3) = -2x + 3\).

3. For \(x > \frac{3}{2}\), \(2x - 3\) is positive, so \(y = 2x - 3\).

4. The graph consists of two linear pieces: \(y = -2x + 3\) for \(x < \frac{3}{2}\) and \(y = 2x - 3\) for \(x > \frac{3}{2}\), meeting at the vertex \((\frac{3}{2}, 0)\).