Lesson plan Grade 10 Math

Fennmoor Lead Service Line Replacement: A Linear Model Lesson

A 50-minute grade 10 lesson where students build a linear function for a fictional city's lead service line replacement and check it against the 10-year federal deadline.

Grade: Grade 10
Subject: Math
Material: Lesson plan
Time: 50 min
Prep: Low prep
Materials: Projector or board, Printed handout, Graph paper, Calculators, Index cards

Standards

HSF-LE.A.2 Common Core Math

Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs.

How this material aligns: In guided and independent practice students construct a linear function for lead lines remaining over time from a stated starting count and yearly replacement rate, exactly this standard's build-a-function skill.

All materials for HSF-LE.A.2 →

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 graph the two-variable replacement equation with labeled and scaled axes for years and lines remaining, the coordinate-graphing this standard requires for a relationship between quantities.

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 construct a linear function for the number of lead service lines remaining over time.
Students will graph the replacement function with labeled, scaled axes.
Students will use the function to find when replacement meets a 10-year deadline and its total cost.

Vocabulary

Lead service line: A lead service line is an old water pipe made of lead metal that connects a house to the water main. Lead can get into the drinking water, so these pipes are being dug up and replaced across the country.
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: linear function, slope, replacement rate

Teacher overview

Water utilities across the country are replacing lead service lines under the federal Lead and Copper Rule Improvements, which lower the lead action level from 15 parts per billion to 10 parts per billion in 2027 and require replacing nearly all lead lines within 10 years. This lesson has students model one fictional city's replacement program as a linear function and test whether its yearly rate meets the deadline. A common misconception is that a program removing a fixed number of lines each year is exponential; students should see that a constant yearly rate produces a straight-line, linear model.

Procedure

Hook (5 min): Tell students that EPA estimates about 4 million lead service lines still carry drinking water in the United States, based on the inventories utilities reported in 2025, and that a recent federal rule requires utilities to replace nearly all of them within 10 years. Ask students to estimate how many lines a mid-sized city would have to replace each year to finish on time, and record a few estimates on the board.

Direct instruction (12 min): Introduce the fictional city of Fennmoor, which has 6,000 lead service lines and plans to replace 600 each year. Model the linear function on the board: let L be the lines remaining after t years, so L = 6,000 − 600t. Identify 6,000 as the starting value and −600 as the slope (lines removed per year). Evaluate L at t = 4: L = 6,000 − 600 × 4 = 3,600 lines remaining. Then set L = 0 and solve for t to show the program finishes in 10 years.

Guided practice (12 min): With the class, graph L = 6,000 − 600t on graph paper, labeling the horizontal axis "years" from 0 to 10 and the vertical axis "lines remaining" from 0 to 6,000, with a sensible scale. Plot the intercepts (0, 6,000) and (10, 0) and connect them, then read the graph at t = 3 to confirm it matches the equation.

Independent practice (16 min): Students complete the Student Handout, writing the function, evaluating it, finding when it reaches zero, computing the program's total and yearly cost, and building a faster plan. Circulate and confirm students treat the constant yearly rate as a linear slope, not a percentage.

Closure and exit ticket (5 min): Ask students to answer one exit-ticket question on an index card: a city has 4,000 lead lines and replaces 500 per year; how many remain after 5 years? The answer is 4,000 − 500 × 5 = 1,500 lines. Collect and scan for the correct linear setup before dismissal.

Fennmoor Lead Service Line Replacement: A Linear Model Lesson — Grade 10 Math

Name: ____________________________ Date: ______________

Student Handout

The fictional city of Fennmoor has 6,000 lead service lines and pays 12,000 dollars to replace each line. Use this information for every question.

Fennmoor plans to replace 600 lines per year. Write a linear function for the number of lines remaining, L, after t years.
Using your function, how many lines remain after 3 years?
Set your function equal to zero and solve for t to find how many years the plan takes to remove all the lines.
At 12,000 dollars per line, what is the total cost to replace all 6,000 lead service lines?
If Fennmoor spreads the total cost evenly across the years found in item 3, what is the yearly replacement budget?
A faster plan replaces 750 lines per year. Write the new function and find how many years it takes to reach zero lines remaining.
The lead action level is 15 parts per billion, dropping to 10 parts per billion in 2027. Explain why replacing the lead service lines, not just testing the water, is the long-term fix for lead in drinking water.

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 = 6,000 − 600t.
The starting count is 6,000 lines and the plan removes 600 lines per year, a constant rate, so the number remaining after t years is L = 6,000 − 600t.
Answer: 4,200 lines.
L = 6,000 − 600 × 3 = 6,000 − 1,800 = 4,200 lines remaining after 3 years.
Answer: 10 years.
Set L = 0: 0 = 6,000 − 600t, so 600t = 6,000 and t = 10 years. This meets the federal 10-year deadline exactly.
Answer: 72,000,000 dollars.
6,000 lines × 12,000 dollars per line = 72,000,000 dollars total.
Answer: 7,200,000 dollars.
72,000,000 dollars ÷ 10 years = 7,200,000 dollars per year.
Answer: 8 years.
The new function is L = 6,000 − 750t. Set L = 0: 750t = 6,000, so t = 8 years. Replacing 750 per year finishes two years sooner than the 600-per-year plan.
Answer: Testing finds lead, but only replacement removes the source.
The lead service line is usually the largest source of lead where one is present, so water testing can show a problem but cannot stop it. Replacing the line removes the lead pipe entirely, which is why the rule requires replacement to bring long-term exposure down rather than relying on testing alone.

Differentiation

Support

Provide the function L = 6,000 − 600t already written so students can begin at item 2, and give a printed graph grid with the axes already labeled and scaled. Work item 3 through together, showing how setting L equal to zero isolates t.

Extension

Ask students to find the yearly replacement rate Fennmoor would need to finish in exactly 6 years instead of 10, then explain how that faster rate changes the slope of the line and the yearly budget from item 5.

ELL supports

Pre-teach "service line," "remaining," and "replacement rate" with a simple diagram of a pipe from a water main to a house. Provide a sentence frame for item 7: "Replacing the line is the fix because ___." Allow students to label the graph on graph paper with numbers and short phrases rather than full sentences.