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

\(2^2(2x - a)^2 < (x + 3a)^2\).

This simplifies to solving the quadratic inequality:

\(4(2x - a)^2 < (x + 3a)^2\).

Expanding both sides, we have:

\(4(4x^2 - 4ax + a^2) < x^2 + 6ax + 9a^2\).

Simplifying further, we get:

\(16x^2 - 16ax + 4a^2 < x^2 + 6ax + 9a^2\).

Rearranging terms gives:

\(15x^2 - 22ax - 5a^2 < 0\).

Solving the quadratic equation \(15x^2 - 22ax - 5a^2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the critical values:

\(x = \frac{5}{3}a\) and \(x = -\frac{1}{5}a\).

The solution to the inequality is the interval between these critical values:

\(-\frac{1}{5}a < x < \frac{5}{3}a\).

To solve the inequality \(|5x - 3| < 2|3x - 7|\), we first consider the non-modular inequality:

\((5x - 3)^2 < 2^2(3x - 7)^2\).

This simplifies to:

\(11x^2 - 138x + 187 > 0\).

We solve the quadratic equation \(11x^2 - 138x + 187 = 0\) to find the critical values:

\(x = \frac{17}{11}\) and \(x = 11\).

The solution to the inequality is:

\(x < \frac{17}{11}\) or \(x > 11\).

To solve the inequality \(|x - 3| < 3x - 4\), we consider two cases based on the definition of absolute value.

Case 1: Assume \(x - 3 \, \geq \, 0\), which implies \(x \, \geq \, 3\).

In this case, the inequality becomes:

\(x - 3 < 3x - 4\)

Solving for \(x\):

\(x - 3 < 3x - 4\)

\(x - 3x < -4 + 3\)

\(-2x < -1\)

\(x > \frac{1}{2}\)

Since \(x \, \geq \, 3\), the solution for this case is \(x > 3\).

Case 2: Assume \(x - 3 \, < \, 0\), which implies \(x \, < \, 3\).

In this case, the inequality becomes:

\(-(x - 3) < 3x - 4\)

\(-x + 3 < 3x - 4\)

Solving for \(x\):

\(-x + 3 < 3x - 4\)

\(-x - 3x < -4 - 3\)

\(-4x < -7\)

\(x > \frac{7}{4}\)

Since \(x \, < \, 3\), the solution for this case is \(\frac{7}{4} < x < 3\).

Combining both cases, the solution to the inequality is \(x > \frac{7}{4}\).

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 \(x > 7\).

Start by considering the inequality \(|x - 4| < 2|3x + 1|\).

Rewrite it as \((x - 4)^2 < (2(3x + 1))^2\).

This simplifies to \((x - 4)^2 < 4(3x + 1)^2\).

Expand both sides: \(x^2 - 8x + 16 < 36x^2 + 24x + 4\).

Rearrange to form a quadratic inequality: \(0 < 35x^2 + 32x - 12\).

Find the critical points by solving \(35x^2 + 32x - 12 = 0\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 35\), \(b = 32\), \(c = -12\).

Calculate the discriminant: \(b^2 - 4ac = 32^2 - 4 \times 35 \times (-12) = 1024 + 1680 = 2704\).

\(x = \frac{-32 \pm \sqrt{2704}}{70}\).

\(\sqrt{2704} = 52\), so \(x = \frac{-32 \pm 52}{70}\).

This gives critical points \(x = \frac{20}{70} = \frac{2}{7}\) and \(x = \frac{-84}{70} = -\frac{6}{5}\).

The solution to the inequality is \(x < -\frac{6}{5}\) or \(x > \frac{2}{7}\).