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Math problem set Grade 8 Math

Ashbourne PFAS Treatment Choice: A System of Equations

A grade 8 problem set where students solve systems to compare two PFAS treatment costs and to split flow between two wells in a fictional town.

Objectives

  1. Students will write a system of linear equations from a treatment cost description.
  2. Students will solve a system algebraically and explain what the solution point means.
  3. Students will solve a blending system for the flow contributed by each of two wells.

Vocabulary

Parts per trillion
A unit for measuring extremely small amounts of a chemical in water. PFAS limits are set in parts per trillion because these chemicals matter even in tiny doses.

Also: system of equations, break-even, blend, solution

Teacher overview

The town of Ashbourne is a fictional community that found PFOA in two of its wells. Ashbourne has two ways to respond, and both are used by real systems: build treatment, or blend water from a high-PFOA well with water from a cleaner one until the mixture meets the limit. Each choice turns into a pair of linear equations, and the answer sits where the two equations agree. Teaching note: students often set the two yearly costs equal to each other and ignore the upfront cost, which produces no meaningful year at all. Ask them to name what each side of the equation stands for before they solve problem 3, and expect some students to assume that the option costing less in the first few years must also cost less over the whole period.

Ashbourne PFAS Treatment Choice: A System of Equations — Grade 8 Math
Name: ____________________________ Date: ______________

Problems

The town of Ashbourne is fictional, and so are the costs and test results below. The federal limit for PFOA in drinking water is 4.0 parts per trillion, written ppt.

Ashbourne priced two treatment options. Option A costs $480,000 to build and $95,000 each year to run. Option B costs $760,000 to build and $60,000 each year to run. Let y stand for the number of years since the plant opened.

  1. Write an equation for the total cost a of Option A after y years.
  2. Write an equation for the total cost b of Option B after y years.
  3. Solve the system to find the year when the two options have cost the same.
  4. What is the total cost of either option in that year?
  5. Over 12 years, which option costs less?
  6. How much less does that option cost over 12 years?
  7. Ashbourne also studied blending. Well 3 tests at 12 ppt PFOA and Well 5 tests at 2 ppt. The town wants 1.2 million gallons per day of blended water at 3.0 ppt. Let x be the flow from Well 3 and let z be the flow from Well 5, both in million gallons per day. Write the system of two equations.
  8. Solve the system for x, the flow from Well 3.
  9. Solve the system for z, the flow from Well 5.
  10. Does the blended water meet the 4.0 ppt limit? Explain how you know.

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: a = 480,000 + 95,000y.
    The build cost is paid once, and the running cost is paid each year, so it is multiplied by y.
  2. Answer: b = 760,000 + 60,000y.
    Option B has the larger build cost and the smaller yearly cost.
  3. Answer: 8 years.
    Set the totals equal: 480,000 + 95,000y = 760,000 + 60,000y. Subtract 60,000y from both sides: 480,000 + 35,000y = 760,000. Subtract 480,000: 35,000y = 280,000. Divide by 35,000: y = 8 years.
  4. Answer: $1,240,000.
    Option A: 480,000 + 95,000(8) = 480,000 + 760,000 = 1,240,000. Option B: 760,000 + 60,000(8) = 760,000 + 480,000 = 1,240,000. The two agree, which confirms problem 3.
  5. Answer: Option B.
    Option A over 12 years: 480,000 + 95,000(12) = 480,000 + 1,140,000 = 1,620,000. Option B over 12 years: 760,000 + 60,000(12) = 760,000 + 720,000 = 1,480,000. Option B is the smaller total.
  6. Answer: $140,000.
    1,620,000 - 1,480,000 = 140,000 less for Option B across 12 years.
  7. Answer: x + z = 1.2 and 12x + 2z = 3.6.
    The first equation sets the total flow. The second sets the total amount of PFOA, since 3.0 ppt across 1.2 million gallons per day gives 3.0 x 1.2 = 3.6.
  8. Answer: 0.12 million gallons per day.
    From the first equation, z = 1.2 - x. Substitute: 12x + 2(1.2 - x) = 3.6, so 12x + 2.4 - 2x = 3.6. Combine: 10x + 2.4 = 3.6. Subtract 2.4: 10x = 1.2. Divide by 10: x = 0.12.
  9. Answer: 1.08 million gallons per day. z = 1.2 - 0.12 = 1.08. Check the PFOA total: 12(0.12) + 2(1.08) = 1.44 + 2.16 = 3.6, which matches the second equation.
  10. Answer: Yes.
    The blend is 3.0 ppt, and 3.0 is less than the 4.0 ppt limit, so the blended water meets the standard. The margin is small, so Ashbourne would need to keep testing both wells.

Differentiation

Support

For this fictional scenario, give students the two cost equations from problems 1 and 2 already written, and provide a table of total costs at years 5, 8, and 12 so they can see the crossing before they solve for it. For problem 7, label one equation "total water" and the other "total PFOA" on the page.

Extension

Ask students to find the yearly running cost that would make the two fictional options break even at exactly 5 years instead of 8, and to explain which number in the system they changed and why the break-even year moved earlier.

ELL supports

Pre-teach blend, break-even, and parts per trillion with a two-jar diagram showing a strong mixture and a weak mixture pouring into one tank. Provide the sentence frame for problem 3: "The two options cost the same in year ___ because ___ equals ___."

Extensions

Have students graph both cost equations from this fictional scenario on the same axes with years on the horizontal axis, mark the crossing point, and shade the region where each option is cheaper. Ask them to write a short recommendation to the Ashbourne town council that names the number of years the council should plan for and explains how the break-even year and the twelve-year totals support that number.