When we have the equation $|3x - 20| + |3x - 40| = 20$, we solve this using the case method for absolute value equations. When we have absolute value expressions, we need to consider when each expression inside is positive or negative. The critical points where the expressions change sign are: $3x - 20 = 0 \rightarrow x = \frac{20}{3}$ $3x - 40 = 0 \rightarrow x = \frac{40}{3}$ These critical points divide the number line into three regions where we can remove the absolute value signs.

When $x < \frac{20}{3}$: $3x - 20 < 0$, so $|3x - 20| = -(3x - 20) = 20 - 3x$ $3x - 40 < 0$, so $|3x - 40| = -(3x - 40) = 40 - 3x$ $(20 - 3x) + (40 - 3x) = 20$ $60 - 6x = 20$ $6x = 40$ $x = \frac{40}{6} = \frac{20}{3}$ We found $x = \frac{20}{3}$, but this contradicts our assumption that $x < \frac{20}{3}$. Therefore, no solution exists in this case.

When $\frac{20}{3} \leq x < \frac{40}{3}$: $3x - 20 \geq 0$, so $|3x - 20| = 3x - 20$ $3x - 40 < 0$, so $|3x - 40| = -(3x - 40) = 40 - 3x$ $(3x - 20) + (40 - 3x) = 20$ $3x - 20 + 40 - 3x = 20$ $20 = 20$ This is always true! Every value of $x$ in the interval $\frac{20}{3} \leq x < \frac{40}{3}$ satisfies our equation.

When $x \geq \frac{40}{3}$: $3x - 20 > 0$, so $|3x - 20| = 3x - 20$ $3x - 40 \geq 0$, so $|3x - 40| = 3x - 40$ $(3x - 20) + (3x - 40) = 20$ $6x - 60 = 20$ $6x = 80$ $x = \frac{80}{6} = \frac{40}{3}$ Since $\frac{40}{3} \geq \frac{40}{3}$, this solution is valid in this case.

Combining the valid cases: From Case 2: $\frac{20}{3} \leq x < \frac{40}{3}$ From Case 3: $x = \frac{40}{3}$ Therefore, the condition $|3x - 20| + |3x - 40| = 20$ holds when: $\frac{20}{3} \leq x \leq \frac{40}{3}$ Converting to decimals: $6.67 \leq x \leq 13.33$ Geometrically, this represents the distance from point $3x$ to points $20$ and $40$ on the number line. The sum of these distances equals $20$ (which is exactly the distance between $20$ and $40$) when $3x$ lies between these two points.

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