To solve the inequality $|x - 3| - |2x + 1| < 0$, we need to analyze the behavior of the absolute value functions in different intervals based on the critical points where the expressions inside the absolute values are zero.

Step 1: Identify the critical points

The expressions inside the absolute values will change their signs at the following points:

• $|x - 3| = 0$ when $x = 3$,
• $|2x + 1| = 0$ when $x = -\frac{1}{2}$.

These critical points divide the number line into three intervals:

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

We will analyze the inequality in each of these intervals.

Step 2: Solve the inequality in each interval

Interval 1: $x < -\frac{1}{2}$

In this region, both $x - 3 < 0$ and $2x + 1 < 0$, so we have: $|x - 3| = -(x - 3) = 3 - x$ $|2x + 1| = -(2x + 1) = -2x - 1$ Substitute these into the inequality: $(3 - x) - (-2x - 1) < 0$ Simplifying: $3 - x + 2x + 1 < 0$ $x + 4 < 0$ $x < -4$ Thus, in the interval $x < -\frac{1}{2}$, the inequality holds for $x < -4$.

Interval 2: $-\frac{1}{2} \leq x < 3$

In this region, $x - 3 < 0$ and $2x + 1 \geq 0$, so: $|x - 3| = -(x - 3) = 3 - x$ $|2x + 1| = 2x + 1$ Substitute these into the inequality: $(3 - x) - (2x + 1) < 0$ Simplifying: $3 - x - 2x - 1 < 0$ $2 - 3x < 0$ $-3x < -2$ $x > \frac{2}{3}$ Thus, in the interval $-\frac{1}{2} \leq x < 3$, the inequality holds for $x > \frac{2}{3}$.

Interval 3: $x \geq 3$

In this region, both $x - 3 \geq 0$ and $2x + 1 \geq 0$, so: $|x - 3| = x - 3$ $|2x + 1| = 2x + 1$ Substitute these into the inequality: $(x - 3) - (2x + 1) < 0$ Simplifying: $x - 3 - 2x - 1 < 0$ $-x - 4 < 0$ $x > -4$ This inequality is always true in the interval $x \geq 3$.

Step 3: Combine the results

From the analysis of each interval, we have the following solutions:

• In $x < -\frac{1}{2}$, the solution is $x < -4$,
• In $-\frac{1}{2} \leq x < 3$, the solution is $\frac{2}{3} < x < 3$,
• In $x \geq 3$, the inequality holds for all $x \geq 3$.

Thus, the solution to the inequality $|x - 3| - |2x + 1| < 0$ is: $x \in (-\infty, -4) \cup \left( \frac{2}{3}, 3 \right)$

Would you like more details on any step?

Related Questions:

1. How do you solve inequalities involving absolute values?
2. What are the properties of absolute value functions?
3. Can inequalities with multiple absolute values have no solutions?
4. How do critical points help in solving absolute value inequalities?
5. Can inequalities involving absolute values have solutions on the entire real line?

Tip:

When solving absolute value inequalities, always break the problem into intervals based on where the expressions inside the absolute values change signs.