The inequality provided is:

$|x + 10| \leq 3x - 2$

To solve this inequality, we need to split it into two cases based on the properties of absolute values. The absolute value function can be rewritten as two separate inequalities:

Case 1: $x + 10 \geq 0$ (i.e., $x \geq -10$)

In this case, the inequality becomes: $x + 10 \leq 3x - 2$

Subtract $x$ from both sides: $10 \leq 2x - 2$

Add 2 to both sides: $12 \leq 2x$

Divide by 2: $x \geq 6$

Thus, for $x \geq -10$, the solution is $x \geq 6$.

Case 2: $x + 10 < 0$ (i.e., $x < -10$)

In this case, the absolute value inequality becomes: $-(x + 10) \leq 3x - 2$

Simplify: $-x - 10 \leq 3x - 2$

Add $x$ to both sides: $-10 \leq 4x - 2$

Add 2 to both sides: $-8 \leq 4x$

Divide by 4: $x \geq -2$

However, we have a contradiction here because $x < -10$ in this case, and $x \geq -2$ is not possible under the condition $x < -10$. Therefore, there is no solution from this case.

Final Answer:

The solution to the inequality is $x \geq 6$.

Do you have any questions or would you like more details?

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Here are 5 follow-up questions based on this inequality:

1. What are the key properties of absolute values when solving inequalities?
2. How do we approach cases where there is a contradiction, as seen in Case 2?
3. How would the solution change if the inequality was $|x + 10| < 3x - 2$ instead of $\leq$?
4. What happens if we modify the inequality to $|x + 10| \geq 3x - 2$?
5. Can we solve this inequality graphically to verify the solution?

Tip: When solving absolute value inequalities, always check the conditions in both cases to avoid contradictions.