Lesson plan Grade 6 Math

Tarrowford Farm's New Well: One-Step Equations for Watering

A 50-minute grade 6 lesson in which students write and solve one-step equations for pumping minutes, gallons added, and rows watered on a berry farm.

Objectives

  1. Students will write a one-step equation that represents a pumping or storage situation on a farm.
  2. Students will solve equations of the form px equals q and x plus p equals q with nonnegative rational numbers.
  3. Students will explain which quantity the variable stands for in each equation they write.

Vocabulary

Aquifer
An aquifer is an underground layer of rock, sand, or gravel that holds water in the tiny spaces between its grains. Wells reach down into aquifers to bring water up to the surface.

Also: equation, variable, solution, unit rate

Teacher overview

A farm well delivers water at a steady rate, so many of the farm's planning questions can be written as a one-step equation before anything is computed. This lesson uses the fictional Tarrowford Farm, whose new well draws from a sand and gravel aquifer and pumps 25 gallons each minute into a storage tank. The misconception this lesson addresses is that a word problem is solved by grabbing the two numbers and multiplying them; students who write the equation first can see that 25 times an unknown number of minutes equals a known number of gallons, which calls for division rather than multiplication.

Procedure

Hook (5 min): Ask students how long a hose would need to run to fill a container the size of a classroom recycling bin, and ask what single piece of information would settle the question. Draw out that a rate in gallons each minute is the missing piece. Introduce the fictional Tarrowford Farm, whose new well pumps 25 gallons each minute.

Direct instruction (12 min): On chart paper, write the situation in words before any symbols: 25 gallons each minute, times some number of minutes, equals 3,000 gallons. Name the unknown m and write 25m equals 3,000. Solve by dividing both sides by 25 and label the answer with its unit, 120 minutes. Then model the second form with the farm's storage tank, which held 1,200 gallons and finished at 5,000 gallons: write 1,200 plus x equals 5,000, and solve by subtracting 1,200 from both sides.

Guided practice (13 min): Pose a third situation aloud, in which the farm plans to use 7,500 gallons in a week and has already used 2,850 gallons. Have pairs decide which of the two equation forms fits before they compute anything, then write and solve it. Pause for a formative check: ask every student to hold up one finger for the times form and two fingers for the plus form, and re-teach the difference with the chart-paper examples if more than a quarter of the class holds up the wrong number of fingers.

Independent practice (15 min): Students complete items 1 through 6 on the handout, writing the equation on the first line and the solution on the second. Remind them that an answer without a unit is incomplete, since minutes, gallons, and rows are all possible answers on this page.

Closure (5 min): Students complete item 7 as an exit ticket. Collect the handouts and sort them into three piles, correct equation with correct solution, correct equation with an arithmetic slip, and wrong equation, which tells you whether tomorrow's warm-up should target the arithmetic or the set-up.

Answer Key

  1. Answer: 120 minutes.
    The equation is 25m equals 3,000. Divide both sides by 25, and 3,000 divided by 25 is 120 minutes.
  2. Answer: 180 minutes.
    The equation is 25m equals 4,500. Divide both sides by 25, and 4,500 divided by 25 is 180 minutes.
  3. Answer: 3,800 gallons.
    The equation is 1,200 plus g equals 5,000. Subtract 1,200 from both sides, and 5,000 minus 1,200 is 3,800 gallons.
  4. Answer: 4,650 gallons.
    The equation is 2,850 plus g equals 7,500. Subtract 2,850 from both sides, and 7,500 minus 2,850 is 4,650 gallons.
  5. Answer: 75 rows.
    The equation is 60r equals 4,500. Divide both sides by 60, and 4,500 divided by 60 is 75 rows.
  6. Answer: 360 minutes.
    The equation is 25m equals 9,000. Divide both sides by 25, and 9,000 divided by 25 is 360 minutes. That is 6 hours of pumping.
  7. Answer: 4,500 gallons.
    Multiply the rate by the minutes: 25 times 180 is 4,500 gallons, which checks the answer to item 2.

Differentiation

Support

Give these students a two-column organizer for the fictional farm, with the left column headed "the rate times the unknown" and the right column headed "a starting amount plus the unknown," and have them place each item in a column before writing anything. Assign items 1, 2, 3, and 7, and provide item 1 with the equation already written so the work starts at the solving step.

Extension

Ask these students to write their own pair of problems for the fictional Tarrowford Farm, one that needs the times form and one that needs the plus form, using a second well that pumps 40 gallons each minute. They should exchange problems with a partner, solve, and then explain how a reader could tell which form each problem needs without solving it.

ELL supports

Pre-teach equation, variable, and solve with a balance-scale sketch showing the same amount on both sides. Provide these sentence frames: "The rate is ___ gallons each minute, so the equation is ___ times m equals ___." and "The tank started with ___ gallons, so the equation is ___ plus g equals ___."

Extensions

Have students time a classroom faucet or a hose with a measuring bucket and a stopwatch to find its own rate in gallons each minute, then rewrite items 1, 2, and 6 using that measured rate in place of the fictional farm's 25 gallons each minute. Comparing the two sets of answers shows the class that the equation form does not change when the rate does, which is the point of writing the equation before computing.