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Lesson plan Grade 7 Math

Replacing Quillfield's Water Mains: A Grade 7 Inequality Lesson

A 50-minute grade 7 lesson where students write and solve an equation and an inequality to find how many miles of water main a fictional town can replace on its budget.

Objectives

  1. Students will write an equation for total cost from a fixed fee and a per-mile rate.
  2. Students will solve an equation and an inequality to find how many miles fit a budget.

Vocabulary

Water main
A big underground pipe that carries drinking water beneath the streets to a neighborhood. Smaller service lines branch off it to reach each building.

Also: equation, inequality, fixed fee

Teacher overview

Water mains are the buried pipes that carry water through a town, and many were laid generations ago and now break as they age. A utility cannot replace them all at once, so it plans a few miles each year against a fixed budget. This lesson uses the fictional town of Quillfield, where replacement costs a flat yearly design fee plus a set amount per mile, to give students a real reason to write and solve a two-step equation and its matching inequality. The misconception to head off is dropping the fixed fee: students often model total cost as only the per-mile rate times miles, so the lesson keeps the flat fee visible at every step.

Procedure

Hook (5 min): Ask students how a town should decide how much old pipe to replace when it cannot afford to replace all of it in one year. Take a few answers, then tell them that in the fictional town of Quillfield, the water department gets a set budget each year and has to figure out how many miles of main that budget buys.

Direct instruction (12 min): Present Quillfield's two costs on the board: a flat design and setup fee of $50,000 each year, plus $600,000 for every mile of new main. Define the variable m as the number of miles replaced in a year. Build the total-cost expression together, 600,000m + 50,000, and stress that the fee is added once, not once per mile.

Worked example (10 min): Quillfield's budget this year is $2,450,000. Write the equation 600,000m + 50,000 = 2,450,000. Solve it as a class: subtract 50,000 from both sides to get 600,000m = 2,400,000, then divide both sides by 600,000 to get m = 4 miles. Then rewrite the problem as the inequality 600,000m + 50,000 ≤ 2,450,000 and show that staying within budget means m ≤ 4.

Guided and independent practice (15 min): Hand out the student page. Work problem 1 together, then have students complete problems 2 through 5 in pairs. Circulate and check that each pair keeps the fixed fee in the equation.

Closure and exit ticket (8 min): As an exit ticket, students answer: "If the yearly fee stayed at $50,000 but the cost per mile fell to $500,000, write the new equation for a $2,550,000 budget and solve for m." Collect and scan for the correct setup (500,000m + 50,000 = 2,550,000, giving m = 5) before students leave.

Replacing Quillfield's Water Mains: A Grade 7 Inequality Lesson — Grade 7 Math
Name: ____________________________ Date: ______________

Student Handout

Quillfield is a fictional town. Its water department replaces old water mains each year. Replacing main costs a flat fee of $50,000 each year, plus $600,000 for every mile. Let m stand for the number of miles replaced in a year.

  1. Write an equation for the total yearly cost C of replacing m miles.
  2. This year's budget is $2,450,000. Set the total cost equal to the budget and solve for m.
  3. Write an inequality for staying within the $2,450,000 budget, then state the greatest number of whole miles the town can replace.
  4. Quillfield has 60 miles of old main. At the number of miles per year from problem 2, how many years will it take to replace all of it?
  5. Next year the budget rises to $3,050,000, with the same fee and per-mile cost. Solve for the new number of miles.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: C = 600,000m + 50,000.
    The per-mile cost 600,000 is multiplied by the miles m, and the flat $50,000 fee is added once.
  2. Answer: 4 miles.
    600,000m + 50,000 = 2,450,000; subtract 50,000 to get 600,000m = 2,400,000; divide by 600,000 to get m = 4 miles.
  3. Answer: 4 miles.
    The inequality is 600,000m + 50,000 ≤ 2,450,000, which solves to m ≤ 4, so the town can replace at most 4 whole miles.
  4. Answer: 15 years.
    60 miles ÷ 4 miles per year = 15 years.
  5. Answer: 5 miles.
    600,000m + 50,000 = 3,050,000; subtract 50,000 to get 600,000m = 3,000,000; divide by 600,000 to get m = 5 miles.

Differentiation

Support

Give students the equation already set up for problem 2, with only the two solving steps left blank, and provide a printed reminder that subtraction undoes the added fee and division undoes the per-mile multiplication. Allow a calculator so students can focus on the structure of the two-step solution.

Extension

Ask students to find the exact budget that would let Quillfield replace 8 miles in a year, then explain in a sentence why doubling the miles does not double the budget. Have them connect the leftover difference to the fixed fee that is paid only once.

ELL supports

Pre-teach "fee," "budget," "per mile," and "at most" with a simple labeled diagram of one mile of pipe and a price tag. Provide the sentence frame "the cost is ___ for each mile plus ___ once" so students can describe the equation in words before writing it in symbols.

Extensions

Have students look up how many miles of water main their own city or water district reports replacing in a recent year, using only the figure the utility itself publishes, and compare it with Quillfield's plan. A short class discussion on why real replacement often falls behind need connects the arithmetic to the aging-pipe problem many communities face.