Let us solve the given inequality:

$|x| < 4|x-3|$

Step 1: Break into Cases Based on Absolute Values

We analyze the inequality $|x| < 4|x-3|$ by considering the possible cases of $x$ being positive, negative, or within certain key intervals. Remember that the definition of absolute value depends on whether the expression inside the absolute value is positive or negative.

Case 1: $x \geq 3$

• When $x \geq 3$, both $|x| = x$ and $|x-3| = x-3$.
• Substituting these into the inequality: $x < 4(x-3)$
• Simplify: $x < 4x - 12$ $-3x < -12 \quad \implies \quad x > 4$
• Combine with $x \geq 3$: $x \in (4, \infty)$.

Case 2: $0 \leq x < 3$

• When $0 \leq x < 3$, $|x| = x$ and $|x-3| = -(x-3) = -x+3$.
• Substituting these into the inequality: $x < 4(-x+3)$
• Simplify: $x < -4x + 12$ $5x < 12 \quad \implies \quad x < \frac{12}{5}$
• Combine with $0 \leq x < 3$: $x \in [0, \frac{12}{5})$.

Case 3: $x < 0$

• When $x < 0$, $|x| = -x$ and $|x-3| = -(x-3) = -x+3$.
• Substituting these into the inequality: $-x < 4(-x+3)$
• Simplify: $-x < -4x + 12$ $3x < 12 \quad \implies \quad x < 4$
• Combine with $x < 0$: $x \in (-\infty, 0)$.

Step 2: Combine All Cases

From all the cases:

1. Case 1 gives $x \in (4, \infty)$.
2. Case 2 gives $x \in [0, \frac{12}{5})$.
3. Case 3 gives $x \in (-\infty, 0)$.

Combining these intervals, the solution is: $x \in (-\infty, \frac{12}{5})$

Final Answer:

$x \in (-\infty, \frac{12}{5})$

Would you like me to explain any step in more detail or solve a similar inequality?

Related Questions:

1. How do you graph the solution $x \in (-\infty, \frac{12}{5})$?
2. What happens if the inequality is $|x| > 4|x-3|$?
3. How can we confirm the solution using test points in each interval?
4. What are the practical applications of absolute value inequalities like this one?
5. How does the solution change if the coefficients of $|x|$ or $|x-3|$ are different?

Tip:

When solving absolute value inequalities, always identify critical points where the expressions inside the absolute values switch between positive and negative, as these define the key cases to consider!