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

To solve the inequality \(|x - 2| < 3x - 4\), we first consider the expression inside the absolute value and the linear expression on the right.

1. Find the x-coordinate of intersection by solving the equation \(x - 2 = 3x - 4\).

2. Rearrange to get \(x - 3x = -4 + 2\), which simplifies to \(-2x = -2\).

3. Solving for \(x\), we get \(x = 1\).

4. Consider the inequality \(3x - 4 > 2 - x\) to find the critical value.

5. Rearrange to get \(3x + x > 4 + 2\), which simplifies to \(4x > 6\).

6. Solving for \(x\), we get \(x > \frac{3}{2}\).

Thus, the solution to the inequality is \(x > \frac{3}{2}\).

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

The vertex of the graph is at the point \((2, 0)\) because the expression inside the absolute value, \(x - 2\), equals zero when \(x = 2\).

For \(x < 2\), the graph is a line with a negative slope, \(y = -(x - 2) = -x + 2\).

For \(x > 2\), the graph is a line with a positive slope, \(y = x - 2\).

The graph is symmetric about the vertical line \(x = 2\).

To solve the inequality \(|2x - 3| > 4|x + 1|\), we first consider the non-modular inequality:

\((2x - 3)^2 > 4^2(x + 1)^2\)

This simplifies to:

\((2x - 3)^2 > 16(x + 1)^2\)

Expanding both sides, we get:

\(4x^2 - 12x + 9 > 16(x^2 + 2x + 1)\)

\(4x^2 - 12x + 9 > 16x^2 + 32x + 16\)

Rearranging terms gives:

\(4x^2 - 12x + 9 - 16x^2 - 32x - 16 < 0\)

\(-12x^2 - 44x - 7 < 0\)

Solving the quadratic inequality \(12x^2 + 44x + 7 < 0\) gives the critical values:

\(x = -\frac{7}{2}\) and \(x = -\frac{1}{6}\)

The solution to the inequality is:

\(\frac{7}{2} < x < -\frac{1}{6}\)

To solve the inequality \(3|2x - 1| > |x + 4|\), we first consider the non-modular inequality \(3^2(2x - 1)^2 > (x + 4)^2\).

This simplifies to \(35x^2 - 44x - 7 = 0\).

We solve the quadratic equation \(35x^2 - 44x - 7 = 0\) by factoring or using the quadratic formula.

Factoring gives \((5x - 7)(7x + 1) = 0\), leading to critical values \(x = \frac{7}{5}\) and \(x = -\frac{1}{7}\).

The solution to the inequality is \(x > \frac{7}{5}\) or \(x < -\frac{1}{7}\).