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Lesson plan Grade 9 Math

Garrowby Water Authority: Two Entry Paths as a System of Equations

A 55-minute grade 9 lesson where students model cumulative earnings for two routes into a water utility and solve the system for the year the totals meet.

Objectives

  1. Students will write a linear equation for the cumulative earnings of each entry path.
  2. Students will solve a system of two linear equations both algebraically and graphically.
  3. Students will interpret the solution of the system as a year, not as a judgment about either path.

Vocabulary

Key words: cumulative earnings, system of equations, intersection, licensure, tuition

Teacher overview

People reach water-sector work through several routes: hiring straight out of high school with on-the-job training and a state operator license earned by exam, a certificate or associate program, a four-year engineering degree, or graduate study in hydrology or public health. The Bureau of Labor Statistics projects about 10,000 openings each year for water and wastewater operators specifically. This lesson uses two of those routes at the fictional Garrowby County Water Authority as the setting for a system of equations, and the mathematical answer is a year, not a verdict about which route is better. The misconception to watch for is that students set the two annual salaries equal to each other, which has no solution, instead of setting the two running totals equal. A second error is forgetting that the second path earns nothing during its two years of school. The model holds each salary flat to keep the equations linear; in practice, pay on both paths usually rises as operators gain experience and higher license classes.

Procedure

Launch (8 min): Describe the two fictional entry paths at the Garrowby authority and ask students to guess, before any algebra, how many years pass before the running totals are equal. Record the guesses. Make clear that both paths lead to licensed positions and that the calculation compares totals over time, nothing else.

Direct instruction (14 min): Define t as years since high school graduation. Build the first path's cumulative earnings with the class, arriving at 40,000t for the fictional starting salary of $40,000 a year. Then build the second path carefully, walking through the two years with no earnings and $8,000 of tuition, so students see why the equation is 56,000(t - 2) - 8,000, which simplifies to 56,000t - 120,000. Ask why the second equation applies only for t of at least 2.

Guided practice (10 min): With the class, graph both equations on axes labeled in years and dollars, scaling the vertical axis in increments of fifty thousand. Estimate the intersection from the graph, then solve the system algebraically and compare the two results. Ask students what the intersection point means in plain words.

Independent practice (18 min): Students complete questions 1 through 10 on the handout, which uses the same two fictional paths.

Closure and formative check (5 min): Students answer handout question 10 on an exit slip, finding the new crossover year when the fictional tuition doubles. Collect the slips and check the direction of the change. Any student whose answer moved earlier rather than later has misread how tuition enters the model, and that misreading is worth addressing at the start of the next session.

Garrowby Water Authority: Two Entry Paths as a System of Equations — Grade 9 Math
Name: ____________________________ Date: ______________

Student Handout

The two paths below describe the fictional Garrowby County Water Authority. Both lead to licensed operator positions. Let t be the number of years since high school graduation.

Path A: hired immediately as an operator trainee at $40,000 per year, with no tuition cost.

Path B: two years in an associate program with no earnings and $8,000 in total tuition, then hired at $56,000 per year.

  1. Find Path A's cumulative earnings 5 years after graduation.
  2. Write an equation for Path A's cumulative earnings after t years.
  3. Find Path B's cumulative earnings, after subtracting tuition, 5 years after graduation.
  4. Write an equation for Path B's cumulative earnings after t years, for t of at least 2.
  5. Solve the system of your two equations to find the year the two running totals are equal.
  6. Find the cumulative amount both paths have reached at the year you found in question 5.
  7. Find Path B's cumulative earnings 10 years after graduation.
  8. Find how much more Path B has accumulated than Path A 10 years after graduation.
  9. The median annual wage for water and wastewater treatment plant and system operators was $60,020 in May 2025. Find the difference between that median and Path A's starting salary of $40,000.
  10. Suppose Path B's total tuition were $16,000 instead of $8,000, with everything else unchanged. Find the new year at which the two running totals are equal.

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: $200,000.
    Path A earns $40,000 every year from the start, so 5 years gives 40,000(5) = $200,000.

  2. Answer: E = 40,000t.
    Cumulative earnings grow by the same $40,000 each year with no cost to subtract, which makes the relationship linear through the origin.

  3. Answer: $160,000.
    Path B works 3 of the 5 years at $56,000, giving 56,000(3) = $168,000, then subtracts $8,000 of tuition for $160,000.

  4. Answer: E = 56,000t - 120,000.
    Starting from 56,000(t - 2) - 8,000 and distributing gives 56,000t - 112,000 - 8,000 = 56,000t - 120,000.

  5. Answer: 7.5 years.
    Setting 40,000t = 56,000t - 120,000 gives 120,000 = 16,000t, so t = 7.5 years after graduation.

  6. Answer: $300,000.
    Substituting into either equation gives the same total: 40,000(7.5) = $300,000, and 56,000(7.5) - 120,000 = 420,000 - 120,000 = $300,000.

  7. Answer: $440,000.
    Substituting gives 56,000(10) - 120,000 = 560,000 - 120,000 = $440,000.

  8. Answer: $40,000.
    Path A reaches 40,000(10) = $400,000 at the same point, and 440,000 - 400,000 = $40,000.

  9. Answer: $20,020.
    Subtracting gives 60,020 - 40,000 = $20,020 per year.

  10. Answer: 8 years.
    The new equation is 56,000(t - 2) - 16,000 = 56,000t - 128,000. Setting 40,000t = 56,000t - 128,000 gives 128,000 = 16,000t, so t = 8 years, half a year later than before.

Differentiation

Support

Provide these students with a year-by-year table for both fictional paths, already labeled for years 0 through 10, and have them fill in running totals before writing any equation. Once the table is complete, ask them to locate the row where the two columns are closest, then check that year against the algebraic solution.

Extension

Have students add a third fictional path to the Garrowby comparison: four years of school with $12,000 of total tuition, followed by $72,000 per year. Ask them to write its equation, find where it crosses each of the first two paths, and describe what the three-line graph shows about which running total is largest at different times.

ELL supports

Pre-teach "cumulative," "tuition," "licensure," and "intersection," and provide a labeled timeline showing the two fictional paths side by side with the school years shaded. Give a sentence frame for question 5, such as "The two totals are equal after ___ years because ___."

Extensions

Students can look up the operator certification requirements published by their own state's environmental agency and list the license classes, the exam topics, and the experience each class requires. A short follow-up writing task is to describe one water-sector role reached through a four-year or graduate program and one reached through licensure and on-the-job training, without ranking them.