Math problem set Grade 8 Math
Marlbrook's Leaking Main: A System of Equations for Repair
A grade 8 math problem set where students model a fictional town's leak-loss and repair costs as two equations and solve the system to find the payback month.
Objectives
- Students will write two linear equations to model competing cost options over time.
- Students will solve a system of equations and interpret the intersection as a break-even point.
Vocabulary
Key words: system of equations, break-even point, intersection, cumulative cost
Teacher overview
Fictional Marlbrook has an aging water main that leaks, wasting treated water and money every month it goes unrepaired. The town can keep paying for the lost water or pay to fix the main and take on a small monthly loan payment instead. This problem set has students model each choice as a linear equation, solve the system to find the month when the two options cost the same, and interpret what happens before and after that point. A common misconception the set corrects is setting the two costs equal but leaving out the fixed repair amount, so the equation loses its constant term.
Problems
Suppose the fictional town of Marlbrook compares two choices for its leaking main. Doing nothing wastes treated water that costs $250 each month. Repairing the main costs $3,000 up front, plus a $50 monthly loan payment, and stops the water loss.
- Write an equation for the cumulative cost C, in dollars, of doing nothing after m months.
- Write an equation for the cumulative cost C, in dollars, of repairing the main after m months.
- Solve the system to find the month when the two choices cost the same amount.
- What is the total cost of each choice at that break-even month?
- After 24 months, find each choice's cost, then state which is cheaper and by how much.
- Explain what the point where the two lines cross means for the town's decision.
- If you graphed cost against months, what are the coordinates of the point where the two lines intersect?
- Over 36 months, how much would the town save by repairing the main at the start instead of never repairing it?
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: C = 250m. Doing nothing adds $250 of wasted water for each month m, with no starting cost.
- Answer: C = 3,000 + 50m. Repair starts at $3,000 and adds $50 each month for the loan payment.
- Answer: 15 months. Set 250m = 3,000 + 50m, so 200m = 3,000 and m = 15 months.
- Answer: $3,750. Doing nothing costs 250 × 15 = $3,750, and repair costs 3,000 + 50 × 15 = 3,000 + 750 = $3,750, which confirms the solution.
- Answer: Repair is cheaper by $1,800. At 24 months, doing nothing costs 250 × 24 = $6,000 and repair costs 3,000 + 50 × 24 = $4,200, and 6,000 − 4,200 = $1,800.
- Answer: The crossing point is the break-even month. Before it, doing nothing costs less; after it, repairing costs less, so waiting past that month makes the leak the more expensive choice.
- Answer: (15, 3750). The lines meet at 15 months and $3,750, the same solution found in problems 3 and 4.
- Answer: $4,200. Over 36 months, doing nothing costs 250 × 36 = $9,000 and repair costs 3,000 + 50 × 36 = $4,800, so repairing saves 9,000 − 4,800 = $4,200.
Differentiation
Support
Give students a completed cost table for months 0, 5, 10, 15, and 20 for both choices, so the break-even month appears in the table before they solve it algebraically. Provide the two equations C = 250m and C = 3,000 + 50m already written.
Extension
Ask students to find a monthly loan payment that would push the break-even month to exactly 20 months, keeping the repair cost and the monthly water loss the same, and to show the algebra that produces it.
ELL supports
Pre-teach "cumulative," "break-even," and "intersection" with a simple two-line graph showing where the lines cross. Provide the sentence frame "Before month ___, ___ is cheaper; after month ___, ___ is cheaper" for problem 6.