1. **State the problem:** We need to find the time $t$ when the blue car and the purple truck are at the same position on the road. The positions are given by the expressions $18t - 4$ for the blue car and $12t + 20$ for the purple truck. 2. **Set up the equation:** Since the positions are equal when they meet, we set: $$18t - 4 = 12t + 20$$ 3. **Solve for $t$:** Subtract $12t$ from both sides: $$18t - \cancel{12t} - 4 = \cancel{12t} + 20 - 12t$$ $$6t - 4 = 20$$ Add 4 to both sides: $$6t - 4 + 4 = 20 + 4$$ $$6t = 24$$ Divide both sides by 6: $$\frac{6t}{\cancel{6}} = \frac{24}{\cancel{6}}$$ $$t = 4$$ 4. **Interpretation:** The vehicles will be at the same position after 4 units of time (seconds, minutes, etc., depending on the context). 5. **Answer:** $$\boxed{t = 4}$$