1. **State the problem:** Alicia and Greg are running with different starting distances and speeds. Alicia starts at 2 miles and runs at 3 mph. Greg starts at 5 miles and runs at 2 mph. We want to model their distances over time, find when they have run the same distance, and analyze Tracy's catching time. 2. **Write the system of equations:** Let $t$ be the time in hours after the start of this observation. - Alicia's distance: $$d_A = 2 + 3t$$ - Greg's distance: $$d_G = 5 + 2t$$ 3. **Graph the system:** The equations are linear with $t$ on the x-axis (time in hours) and $d$ on the y-axis (distance in miles). Alicia's line starts at 2 and rises with slope 3. Greg's line starts at 5 and rises with slope 2. 4. **Find the solution to the system:** Set distances equal to find when they have run the same distance: $$2 + 3t = 5 + 2t$$ Subtract $2t$ from both sides: $$2 + \cancel{3t} - \cancel{2t} = 5 + \cancel{2t} - \cancel{2t} \Rightarrow 2 + t = 5$$ Subtract 2 from both sides: $$\cancel{2} + t - \cancel{2} = 5 - 2 \Rightarrow t = 3$$ Plug $t=3$ back into Alicia's equation: $$d_A = 2 + 3(3) = 2 + 9 = 11$$ So, the solution is $t=3$ hours and distance $11$ miles. 5. **Meaning of the solution:** After 3 hours, Alicia and Greg have both run 11 miles. This means they are at the same distance point on the route at the same time. 6. **Tracy catching Alicia:** Tracy runs at 5 mph and starts running at some time after Alicia started. To find how long it takes Tracy to catch Alicia, assume Tracy starts at time $t=0$ at distance 0. Set Tracy's distance equal to Alicia's distance: $$5t = 2 + 3t$$ Subtract $3t$ from both sides: $$5t - 3t = 2 + 3t - 3t \Rightarrow 2t = 2$$ Divide both sides by 2: $$\cancel{2}t / \cancel{2} = 2 / 2 \Rightarrow t = 1$$ So, Tracy will catch Alicia after 1 hour of running. **Final answers:** - System of equations: $d_A = 2 + 3t$, $d_G = 5 + 2t$ - Solution: $t=3$ hours, distance = 11 miles - Tracy catches Alicia after 1 hour