Worksheet Grade 9 Math

Marlbrook Lead Service Lines: A Replacement-Schedule Worksheet

A grade 9 worksheet where students model Marlbrook's 10-year lead service line replacement with a linear equation and check homes against the lead action level.

Grade: Grade 9
Subject: Math
Material: Worksheet
Time: 35 min
Prep: No prep
Materials: None

Standards

HSA-CED.A.2 Common Core Math

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

How this material aligns: Students write and use the linear equation L = 4800 - 500t in items 1 through 7 to relate lead lines remaining to years elapsed, and interpret its slope.

All materials for HSA-CED.A.2 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will write a linear equation modeling how many lead service lines remain over time.
Students will use the equation to plan a replacement schedule and check it against the 10-year rule.
Students will compare a home's test result to the lead action level that applies in a given year.

Vocabulary

Service line: A service line is the small pipe that connects the water main under the street to the plumbing inside one home or building. It is the last leg of the trip water makes to your tap.
Action level: A trigger number in lead and copper testing. If too many homes test above this level, the utility must take steps like improving treatment and telling the public.

Also: slope, linear equation, replacement rate

Teacher overview

The fictional town of Marlbrook is replacing the lead service lines that connect its water mains to homes. Federal rules ask water systems to replace nearly all lead service lines within about 10 years, and the lead action level falls from 15 ppb to 10 ppb in 2027. Across the country, EPA estimates that about 4 million lead service lines still remain. This worksheet has students build a linear equation for the number of lead lines left as the work goes on, then use it to plan the schedule and estimate the budget.

Marlbrook Lead Service Lines: A Replacement-Schedule Worksheet — Grade 9 Math

Name: ____________________________ Date: ______________

Student Worksheet

Marlbrook (a fictional town) begins 2026 with 4,800 lead service lines. The utility plans to replace 500 lines each year. Let t stand for the number of years after the start of 2026.

Write an equation for L, the number of lead lines remaining, after t years. Use the starting count and the yearly replacement rate.
How many lead lines remain after 3 years?
How many lead lines remain after 7 years?
Set your equation equal to 0. After how many years would every lead line be replaced?
Federal rules ask systems to replace nearly all lead lines within 10 years. Does 500 lines per year meet that goal? Use your answer to item 4 to explain.
Identify the slope of your equation. Explain what the slope means in this scenario.
Marlbrook now wants the job done in exactly 8 years. What yearly replacement rate is needed? Write the new equation.
The lead action level is 15 ppb now, and it drops to 10 ppb in 2027. A home tests at 12 ppb in 2028. Is that result above or below the action level that applies in 2028?
Each line costs $6,000 to replace. What is the total cost to replace all 4,800 lines?
Explain why replacing lead service lines is treated as a public health priority.

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: L = 4800 - 500t. Start with 4,800 lines, then subtract 500 for each year t that passes.
Answer: 3,300 lines. L = 4800 - 500(3) = 4800 - 1500 = 3,300.
Answer: 1,300 lines. L = 4800 - 500(7) = 4800 - 3500 = 1,300.
Answer: 9.6 years. Set 0 = 4800 - 500t, so 500t = 4800 and t = 9.6 years.
Answer: Yes. Every line is replaced by 9.6 years, which is inside the 10-year window, so 500 per year meets the goal.
Answer: Slope = -500. The town removes 500 lead lines each year, so the count drops by 500 for every one-year increase in t.
Answer: 600 lines per year. 4,800 lines divided by 8 years is 600 per year, so L = 4800 - 600t.
Answer: Above. In 2028 the action level is 10 ppb, and 12 ppb is greater than 10 ppb. (Strictly, the action level is compared with the 90th percentile of a system's samples rather than one home, and EPA's health goal for lead is zero, so no result is considered safe.)
Answer: $28,800,000. 4,800 lines times $6,000 per line equals $28,800,000.
Answer: Lead can harm health, especially in young children, and old service lines can add lead to tap water, so replacing them lowers that risk.

Differentiation

Support

Give students the equation frame L = ___ - ___ t and have them fill the two blanks from the starting count and the yearly rate before solving anything. Allow a calculator for items 2 through 9, and provide a number line to show what "after t years" means.

Extension

Ask students to write a second equation for a neighboring system that starts with 7,200 lines and replaces 800 lines per year. Have them find the year when both systems have the same number of lines remaining, and interpret what that shared value means.

ELL supports

Pre-teach "service line," "replace," "rate," and "slope" with a simple labeled picture of a pipe running from a water main to a house. Provide a sentence frame for item 10: "Replacing lead lines matters because ___."