We are tasked with solving the inequality:

$|x - 10| < 3|2 - x|$

Step 1: Consider two cases for the absolute value of $|x - 10|$ and $|2 - x|$.

For absolute value expressions, we need to break the equation into different cases depending on whether the expressions inside the absolute values are positive or negative.

Case 1: $x \geq 10$

Here, both expressions inside the absolute values are non-negative, so we can remove the absolute value symbols directly:

$x - 10 < 3(2 - x)$

Expanding the right-hand side:

$x - 10 < 6 - 3x$

Now, let's solve for $x$:

$x + 3x < 6 + 10$ $4x < 16$ $x < 4$

But in this case, we assumed $x \geq 10$, and since $x < 4$ contradicts this, there is no solution in this case.

Case 2: $x < 10$

Now, we need to consider two further subcases: $x \geq 2$ and $x < 2$.

Subcase 2.1: $x \geq 2$

For $x \geq 2$, $x - 10$ is negative and $2 - x$ is non-negative. So:

$-(x - 10) < 3(2 - x)$

This simplifies to:

$- x + 10 < 6 - 3x$

Solving for $x$:

$- x + 3x < 6 - 10$ $2x < -4$ $x < -2$

But we assumed $x \geq 2$, so there is no solution in this subcase.

Subcase 2.2: $x < 2$

For $x < 2$, both expressions inside the absolute values are negative. Hence:

$-(x - 10) < 3(-(2 - x))$

Simplifying:

$- x + 10 < 3(x - 2)$ $- x + 10 < 3x - 6$

Now solve for $x$:

$- x - 3x < -6 - 10$ $- 4x < -16$ $x > 4$

But we assumed $x < 2$, so there is no solution in this subcase either.

Conclusion:

The inequality $|x - 10| < 3|2 - x|$ has no solution because in every case, the conditions contradict the assumed ranges of $x$.

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the general method for solving absolute value inequalities?
2. How would the solution change if one side of the inequality were just a constant?
3. Can you apply this method to other inequalities involving absolute values?
4. What happens if we change the inequality from "<" to "≤"?
5. How can graphing help in understanding absolute value inequalities?

Tip: When dealing with absolute value inequalities, always consider different cases based on the sign of the expressions inside the absolute value functions!