1. **State the problem:** Solve the equation $$\frac{|33x - 15|}{6} = \frac{|9x - 1|}{3}$$ for $x$. 2. **Understand the equation:** We have absolute values on both sides divided by constants. To solve, we can cross-multiply to eliminate denominators: $$6 \times \frac{|9x - 1|}{3} = 6 \times \frac{|33x - 15|}{6}$$ which simplifies to: $$2|9x - 1| = |33x - 15|$$ 3. **Rewrite the equation:** $$|33x - 15| = 2|9x - 1|$$ 4. **Recall the property of absolute values:** For any real numbers $A$ and $B$, if $|A| = k|B|$ where $k > 0$, then either $A = kB$ or $A = -kB$. So here: $$33x - 15 = 2(9x - 1) \quad \text{or} \quad 33x - 15 = -2(9x - 1)$$ 5. **Solve the first equation:** $$33x - 15 = 18x - 2$$ Bring all terms to one side: $$33x - 18x = -2 + 15$$ $$15x = 13$$ $$x = \frac{13}{15}$$ 6. **Solve the second equation:** $$33x - 15 = -18x + 2$$ Bring all terms to one side: $$33x + 18x = 2 + 15$$ $$51x = 17$$ $$x = \frac{17}{51}$$ 7. **Check solutions:** Both values satisfy the original equation because absolute values ensure non-negativity. **Final answer:** $$x = \frac{13}{15} \quad \text{or} \quad x = \frac{17}{51}$$