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Worksheet Grade 8 Math

Thornbury Treatment Plant: Systems of Equations for Chlorine Blending

A grade 8 worksheet where students set up and solve systems of two linear equations to blend chlorine stock solutions to a target strength at a fictional treatment plant.

Objectives

  1. Students will write a system of two linear equations modeling a mixture of two solutions with different strengths.
  2. Students will solve a system of two linear equations by substitution or elimination.
  3. Students will interpret a solution to a mixture system in the context of available supply.

Vocabulary

Key words: system of equations, stock solution, concentration, residual

Teacher overview

Thornbury Water Treatment Plant (a fictional treatment plant used for this worksheet) keeps two chlorine stock solutions on hand at different strengths and blends them to prepare test batches before diluting the blend far further for use in the distribution system. This worksheet turns that blending decision into a system of two linear equations: one equation balances total volume, and the other balances the amount of chlorine each stock contributes. A common misconception is writing the second equation using the stocks' percentages as if they were whole gallons rather than as a rate multiplied by the gallons of each stock; students need to multiply each stock's percent strength by its own volume, not just add the two percentages together.

Thornbury Treatment Plant: Systems of Equations for Chlorine Blending — Grade 8 Math
Name: ____________________________ Date: ______________

Student Worksheet

Suppose Thornbury's plant keeps an 8% chlorine stock and a 2% chlorine stock on hand. Staff blend the two stocks to prepare test batches at a specific strength.

  1. Plant staff need to prepare 100 gallons of a 3.8% chlorine solution by blending the 8% stock and the 2% stock. Let x be the gallons of 8% stock and y be the gallons of 2% stock. Write the system of two equations for this batch: one equation for total gallons, and one equation for total chlorine strength (8x + 2y equals the strength percent times the total gallons).
  2. Solve your system from problem 1 for x, the gallons of 8% stock needed.
  3. Solve your system from problem 1 for y, the gallons of 2% stock needed.
  4. Check your solution by substituting x and y back into both equations from problem 1. State whether both equations balance.
  5. Staff want a second batch: 50 gallons of a 5% chlorine solution, blended from the same 8% and 2% stocks. Write the system of two equations for this new batch (let x = gallons of 8% stock, y = gallons of 2% stock).
  6. Solve your system from problem 5 for x.
  7. Solve your system from problem 5 for y.
  8. The plant has only 20 gallons of 8% stock left today. Based on your answer to problem 6, can staff complete the 50-gallon, 5% batch with today's supply? Explain.
  9. Real distribution systems typically maintain a chlorine residual of 0.2 to 2.0 mg/L, well under the maximum allowed residual of 4.0 mg/L. Convert the 3.8% batch from problem 1 to mg/L, using the fact that 1% chlorine equals 10,000 mg/L, and find about how many times stronger that batch is than the top of the 0.2 to 2.0 mg/L residual range.
  10. If Thornbury needed 200 gallons total of the 3.8% blend instead of 100 gallons, how many gallons of the 8% stock would be required? (Hint: the ratio of 8% stock to total gallons stays the same as in problem 2.)

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: x + y = 100 and 8x + 2y = 380.
    Total volume gives x + y = 100. Total chlorine strength gives 8x + 2y = 3.8 × 100 = 380.
  2. Answer: 30 gal.
    From x + y = 100, y = 100 − x. Substituting: 8x + 2(100 − x) = 380 → 8x + 200 − 2x = 380 → 6x = 180 → x = 30.
  3. Answer: 70 gal. y = 100 − x = 100 − 30 = 70.
  4. Answer: yes, both equations balance. x + y = 30 + 70 = 100, which matches. 8(30) + 2(70) = 240 + 140 = 380, which also matches. Both equations check.
  5. Answer: x + y = 50 and 8x + 2y = 250.
    Total volume gives x + y = 50. Total chlorine strength gives 8x + 2y = 5 × 50 = 250.
  6. Answer: 25 gal.
    From x + y = 50, y = 50 − x. Substituting: 8x + 2(50 − x) = 250 → 8x + 100 − 2x = 250 → 6x = 150 → x = 25.
  7. Answer: 25 gal. y = 50 − x = 50 − 25 = 25. This makes sense because 5% is exactly halfway between the 2% and 8% stocks, so the batch needs equal parts of each.
  8. Answer: no.
    The batch requires 25 gallons of 8% stock, but only 20 gallons are on hand today, so staff cannot complete the full 50-gallon batch without more stock.
  9. Answer: about 19,000 times stronger.
    3.8% × 10,000 mg/L per 1% = 38,000 mg/L. Dividing by the top of the residual range, 38,000 ÷ 2.0 = 19,000, so the stock batch is about 19,000 times stronger than the highest typical distribution residual.
  10. Answer: 60 gal.
    Problem 2 found that 30 out of 100 gallons, or 30%, is 8% stock. For 200 gallons at the same ratio, 8% stock needed is 0.30 × 200 = 60 gallons.

Differentiation

Support (below grade level)

Provide the total-volume equation (x + y = 100) already written for problem 1, so students focus on writing and solving only the chlorine-strength equation. Offer a printed substitution walkthrough using a different example batch (not the 100-gallon or 50-gallon batch) before problem 2.

Extension (above grade level)

Ask students to solve problem 1's system using elimination instead of substitution, showing every step, and to explain one advantage elimination has over substitution for this particular system.

ELL supports

Pre-teach "stock solution," "concentration," and "residual" with a simple diagram showing two labeled jugs (8% and 2%) being poured into one larger labeled container. Provide a sentence frame for problem 8: "Staff ___ complete the batch because they need ___ gallons but only have ___ gallons."