We are tasked with finding the set of values of $x$ that satisfy the inequality:

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

Step 1: Analyze the absolute value inequalities

To solve this inequality, we need to consider different cases based on the critical points of the absolute value expressions. These points occur when the arguments of the absolute values equal zero. Let's identify these points:

• $|x + 2a|$ will change at $x = -2a$.
• $|x - a|$ will change at $x = a$.

So, we need to examine the inequality in the following intervals based on these points:

• $x < -2a$
• $-2a \leq x < a$
• $x \geq a$

Step 2: Solve the inequality in each region

Case 1: $x < -2a$

For $x < -2a$, both $x + 2a < 0$ and $x - a < 0$, so the inequality becomes:

$-(x + 2a) > 3(-(x - a))$

Simplifying both sides:

$-x - 2a > -3x + 3a$

Move terms involving $x$ to one side and constants to the other side:

$2x > 5a$

Thus,

$x > \frac{5a}{2}$

But for this to hold, we need $x < -2a$ (from our assumption for Case 1). Hence, there are no solutions in this region.

Case 2: $-2a \leq x < a$

For $-2a \leq x < a$, we have:

• $x + 2a \geq 0$, so $|x + 2a| = x + 2a$
• $x - a < 0$, so $|x - a| = -(x - a)$

The inequality becomes:

$x + 2a > 3(-(x - a))$

Simplifying:

$x + 2a > -3x + 3a$

Move terms involving $x$ to one side and constants to the other side:

$4x > a$

Thus,

$x > \frac{a}{4}$

Since we are in the interval $-2a \leq x < a$, the solutions to this case are:

$\frac{a}{4} < x < a$

Case 3: $x \geq a$

For $x \geq a$, we have:

• $x + 2a \geq 0$, so $|x + 2a| = x + 2a$
• $x - a \geq 0$, so $|x - a| = x - a$

The inequality becomes:

$x + 2a > 3(x - a)$

Simplifying:

$x + 2a > 3x - 3a$

Move terms involving $x$ to one side and constants to the other side:

$-2x > -5a$

Thus,

$x < \frac{5a}{2}$

Since we are in the interval $x \geq a$, the solutions to this case are:

$a \leq x < \frac{5a}{2}$

Step 3: Combine the results

The solution to the inequality is the union of the solutions from Case 2 and Case 3:

$\left( \frac{a}{4}, a \right) \cup \left[ a, \frac{5a}{2} \right)$

Final Answer:

The set of values of $x$ that satisfy the inequality $|x + 2a| > 3|x - a|$ is:

$\boxed{\left( \frac{a}{4}, \frac{5a}{2} \right)}$

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Would you like further details or clarifications on this solution?

Additional questions:

1. How does the solution change if we replace $|x + 2a|$ with $|x - 2a|$?
2. What happens to the inequality when $a = 0$?
3. How can we visualize this inequality on a number line?
4. What is the solution if we instead solve the inequality $|x + 2a| < 3|x - a|$?
5. How can this method be generalized to inequalities with different coefficients in front of the absolute values?

Tip:

When solving absolute value inequalities, always break them into different cases based on the critical points where the arguments of the absolute values are zero. This ensures that you handle each scenario properly.