1. **State the problem:** We are given six equations representing times $t$ when two vehicles meet. We need to sort these equations based on how often the vehicles meet: always, once, or never. 2. **Analyze each equation:** - Equation 1: $12 - t = t - 12$ Simplify: $$12 - t = t - 12$$ Add $t$ to both sides: $$12 = 2t - 12$$ Add 12 to both sides: $$12 + 12 = 2t$$ $$24 = 2t$$ Divide both sides by 2: $$\cancel{2}t = \frac{24}{\cancel{2}}$$ $$t = 12$$ This gives a single solution, so vehicles meet **once**. - Equation 2: $2t = 8t$ Subtract $2t$ from both sides: $$0 = 6t$$ So, $$t = 0$$ Single solution, vehicles meet **once**. - Equation 3: $8(t + 1) = 8t - 8$ Expand left side: $$8t + 8 = 8t - 8$$ Subtract $8t$ from both sides: $$8 = -8$$ This is false, so no solution. Vehicles **never** meet. - Equation 4: $t + 1 = t + 1$ This is true for all $t$, so vehicles meet **always**. - Equation 5: $2t + 6 = 2(t + 3)$ Expand right side: $$2t + 6 = 2t + 6$$ Subtract $2t + 6$ from both sides: $$0 = 0$$ True for all $t$, vehicles meet **always**. - Equation 6: $t = t + 2$ Subtract $t$ from both sides: $$0 = 2$$ False, no solution, vehicles **never** meet. 3. **Sort based on frequency:** - Always: $t + 1 = t + 1$, $2t + 6 = 2(t + 3)$ - Once: $12 - t = t - 12$, $2t = 8t$ - Never: $8(t + 1) = 8t - 8$, $t = t + 2$ Final sorted list by frequency: **Always:** - $t + 1 = t + 1$ - $2t + 6 = 2(t + 3)$ **Once:** - $12 - t = t - 12$ - $2t = 8t$ **Never:** - $8(t + 1) = 8t - 8$ - $t = t + 2$