Water for Schools
Menu

Lesson plan Grade 8 Math

Kelvedge Main Repairs: One, No, or Many Solutions

A 55-minute grade 8 lesson where students solve cost equations for fictional pipe-repair contracts and sort them into one solution, no solution, and infinitely many.

Objectives

  1. Students will solve linear equations that have the variable on both sides.
  2. Students will classify an equation as having one solution, no solution, or infinitely many solutions.
  3. Students will explain what each type of solution means for two repair contracts.

Vocabulary

Key words: variable, no solution, infinitely many solutions, equivalent equations

Teacher overview

The fictional village of Kelvedge has to repair a run of failing water main and has collected four bids, each written as a fixed mobilization charge plus a daily rate. Setting two bids equal to each other produces a linear equation with the variable on both sides, and the three possible outcomes of that equation have plain meanings in the village's situation: the bids cross once, they never cross, or they are the same bid written two different ways. The misconception this lesson corrects is the belief that every equation has a numerical answer, which leads students to report a number of days even when the variable cancels and the statement left behind is false.

Procedure

Hook (5 min): Write two fictional Kelvedge bids on the board: Contract A charges a $600 mobilization fee plus $250 for each day of work, and Contract B charges $250 for each day with no fee. Ask the class after how many days the two bids cost the same. Let students argue briefly before telling them the algebra will settle it.

Direct instruction (12 min): Set up 250d plus 600 equals 250d for the fictional bids in the hook and subtract 250d from both sides, leaving 600 equals 0. Name the outcome: the statement is false for every value of d, so the equation has no solution and the two bids never cost the same. Then show a second pair that reduces to 600 equals 600, a statement true for every value of d, and name that outcome as infinitely many solutions. Finish by showing an equation that reduces to a single value of d and naming it one solution.

Guided practice (13 min): Work through 250d plus 600 equals 150d plus 900 as a class, ending at d equals 3. Ask students to state in a full sentence what the 3 means for the village. Then pose the check-for-understanding question: "When you subtract the variable term from both sides and are left with 600 equals 600, how many days make the two bids equal?" Expect students to answer that every number of days works. If fewer than three-quarters of the class answers correctly, redo the board example before releasing the handout.

Independent practice (20 min): Students complete the ten items on the Student Handout, showing each step and labeling every answer as one solution, no solution, or infinitely many solutions.

Closure and exit ticket (5 min): On an index card, students solve 4x plus 9 equals 4x plus 2 and name the type of solution. Collect the cards and check for "no solution." If more than a third of the class writes a number, reteach the cancelling step at the start of the next session.

Kelvedge Main Repairs: One, No, or Many Solutions — Grade 8 Math
Name: ____________________________ Date: ______________

Student Handout

Every contract and price below belongs to the fictional village of Kelvedge. Let d stand for the number of days of work. Show each step, and label your answer as one solution, no solution, or infinitely many solutions.

  1. Contract A charges a $600 mobilization fee plus $250 for each day. Write an equation for the total cost C after d days.
  2. Solve 250d + 600 = 150d + 900.
  3. In one sentence, explain what your answer to item 2 means about the two contracts.
  4. Solve 250d + 600 = 250d.
  5. In one sentence, explain why the equation in item 4 turns out the way it does.
  6. Solve 250d + 600 = 2(125d + 300).
  7. Contract C charges $150 for each day plus a $900 fee. Solve 150d + 900 = 2100 to find the number of days that costs that much.
  8. Contract B charges $250 for each day and no fee. Solve 250d = 3000 to find the number of days that costs that much.
  9. Solve 250d + 600 = 150d + 600.
  10. Contract C charges $150 for each day plus a $900 fee. What is the total cost of a 10-day job?

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 = 250d + 600. The daily rate of $250 is multiplied by the number of days d, and the one-time $600 fee is added once.
  2. Answer: 3 days. Subtract 150d from both sides to get 100d + 600 = 900, subtract 600 from both sides to get 100d = 300, then divide both sides by 100 to get d = 3. This equation has one solution.
  3. Answer: The two contracts cost the same for a 3-day job. For jobs shorter than 3 days the contract with the smaller fee costs less, and for longer jobs the contract with the smaller daily rate costs less.
  4. Answer: No solution. Subtracting 250d from both sides leaves 600 = 0, which is false for every value of d, so no number of days makes the two totals equal.
  5. Answer: The two contracts charge the same $250 for each day, so the cost lines rise at the same rate and stay exactly $600 apart. Lines that never meet give an equation with no solution.
  6. Answer: Infinitely many solutions. Distributing on the right gives 250d + 600 = 250d + 600.
    Subtracting 250d from both sides leaves 600 = 600, which is true for every value of d, so the two expressions describe the same contract written two ways.
  7. Answer: 8 days. Subtract 900 from both sides to get 150d = 1200, then divide both sides by 150 to get d = 8. This equation has one solution.
  8. Answer: 12 days. Divide both sides by 250 to get d = 12. This equation has one solution.
  9. Answer: 0 days. Subtract 150d from both sides to get 100d + 600 = 600, subtract 600 from both sides to get 100d = 0, then divide by 100 to get d = 0. The two contracts cost the same only before any work starts, so this equation has one solution.
  10. Answer: $2,400. Multiply the daily rate by the days, 150 times 10, to get 1500, then add the $900 fee to get $2,400.

Differentiation

Support

Give these students the first algebraic step already written for items 2, 4, 6, and 9, so they begin from the point where the variable terms are lined up. Provide a three-box sorting card labeled one solution, no solution, and infinitely many solutions, with a short worked example of each pinned inside the box.

Extension

Ask students to write their own pair of fictional contracts whose equation has infinitely many solutions, and a second pair whose equation has no solution, then trade with a partner and defend each classification. Have them graph one pair to show why parallel cost lines never meet.

ELL supports

Pre-teach "fee," "daily rate," and "solution" using a labeled diagram of a single bid, with the fixed charge and the per-day charge circled in different colors. Provide the sentence frames "The equation has ___ solution because ___" and "The two contracts cost the same after ___ days."

Extensions

Have students research how a public agency near them advertises a construction bid, then rewrite one advertised price structure as an equation and explain which parts of it are fixed and which grow with time.