We are tasked with solving the inequality:

$|3x - 2| > |x + 2|$

Step 1: Consider the different cases

Since we are dealing with absolute values, we need to break the inequality into different cases based on the points where the expressions inside the absolute values change sign, i.e., where $3x - 2 = 0$ and $x + 2 = 0$.

Critical Points:

1. $3x - 2 = 0 \Rightarrow x = \frac{2}{3}$
2. $x + 2 = 0 \Rightarrow x = -2$

So, the critical points are $x = \frac{2}{3}$ and $x = -2$. These points divide the real line into three intervals:

1. $x < -2$
2. $-2 \leq x \leq \frac{2}{3}$
3. $x > \frac{2}{3}$

Step 2: Solve for each case

Case 1: $x < -2$

In this region, both expressions inside the absolute values are negative: $|3x - 2| = -(3x - 2) = -3x + 2 \quad \text{and} \quad |x + 2| = -(x + 2) = -x - 2$ The inequality becomes: $-3x + 2 > -x - 2$ Simplify: $-3x + 2 + x > -2$ $-2x + 2 > -2$ $-2x > -4$ Divide by $-2$ (remember to reverse the inequality): $x < 2$ Since we are in the region $x < -2$, this inequality holds for all $x < -2$.

Case 2: $-2 \leq x \leq \frac{2}{3}$

In this region, $3x - 2$ is negative, and $x + 2$ is non-negative: $|3x - 2| = -(3x - 2) = -3x + 2 \quad \text{and} \quad |x + 2| = x + 2$ The inequality becomes: $-3x + 2 > x + 2$ Simplify: $-3x > x$ $-4x > 0$ Divide by $-4$ (reverse the inequality): $x < 0$ So, in this interval, the solution is $x \in [-2, 0)$.

Case 3: $x > \frac{2}{3}$

In this region, both expressions inside the absolute values are positive: $|3x - 2| = 3x - 2 \quad \text{and} \quad |x + 2| = x + 2$ The inequality becomes: $3x - 2 > x + 2$ Simplify: $3x - x > 2 + 2$ $2x > 4$ $x > 2$ So, the solution in this region is $x > 2$.

Step 3: Combine the solutions

The solution is: $x < -2 \quad \text{or} \quad -2 \leq x < 0 \quad \text{or} \quad x > 2$

In interval notation, the solution is: $(-\infty, -2) \cup [-2, 0) \cup (2, \infty)$

Would you like more details on any step or have any questions?

5 Related Questions:

1. How do you solve absolute value equations involving two expressions?
2. What are some strategies for graphing absolute value inequalities?
3. How do critical points help in solving inequalities with absolute values?
4. Can an absolute value inequality have no solution?
5. How do you deal with compound inequalities?

Tip: Always remember to reverse the inequality when dividing or multiplying both sides of an inequality by a negative number!