Plan two-step solutions and test reasonableness

Learning goal: Solve whole-number stories requiring two connected operations, represent the unknown with a letter or symbol and assess whether the answer makes sense. Start by describing the question and the relationship between quantities. A keyword alone is not a solution plan. The word each can describe groups, but you still need to identify what is known and what is being asked.

A class has six packets with eight stickers in each packet. The teacher gives away nineteen stickers. How many remain? First find the starting total: 6 × 8 = 48 stickers. Next subtract the amount given away: 48 − 19 = 29 stickers. Forty-eight is an intermediate result. Stopping there answers how many were available, not how many remain.

Write the two equations with a letter for the final unknown: 6 × 8 = 48 and 48 − 19 = r. You may also write (6 × 8) − 19 = r. The parentheses identify the group total that must be found before subtracting. This lesson uses the story and explicit grouping to show order; it does not assume knowledge of an algebraic order-of-operations rule.

A different story can require addition before division. There are eighteen pencils in one container and fourteen in another. All pencils are shared equally among four tables. First combine the pencils: 18 + 14 = 32. Then find 32 ÷ 4 = 8 pencils per table. Dividing only fourteen by four ignores the first container and does not represent the complete story.

Estimation helps check an answer without replacing the exact calculation. In the sticker problem, nineteen is close to twenty. Start with the exact forty-eight and estimate 48 − 20 = 28. The exact answer twenty-nine is one more, which makes sense because one fewer sticker was removed. An answer such as sixty-seven is greater than the starting total even though stickers were given away, so it fails a basic relationship check.

Choose estimates that retain useful information. Rounding both small group counts to a nearby ten can change a multiplication situation too much. For six packets of eight, the known product forty-eight is easy enough to keep exact. For a collection of 198 beads plus 203 beads, rounding to 200 + 200 gives a useful estimate of four hundred. Explain what was rounded and why that estimate is informative.

Some problems involve a difference followed by grouping. A craft club has fifty-four beads, saves eighteen for tomorrow and puts the rest into bags of six. First 54 − 18 = 36 beads are available today. Then 36 ÷ 6 = 6 bags. The two sixes in the last equation have different roles: six beads per bag and six bags.

Worked example: Four boxes contain nine notebooks each. A donation adds sixteen more notebooks. Find the final total. The boxes hold 4 × 9 = 36 notebooks, then 36 + 16 = n gives n = 52. Use the estimate 36 + 20 = 56; since sixteen is four less than twenty, the exact total should be four less than fifty-six. Fifty-two passes that check.

Compare plans: For the pencil-sharing story, (18 + 14) ÷ 4 matches combining then sharing. The expression 18 + (14 ÷ 4) does not describe sharing all the pencils. A correct plan accounts for every quantity and the final unit, even before arithmetic is performed.

Guided practice: Retell each story, draw labeled bars or groups, identify the intermediate quantity and explain which operation finds it. Then calculate the final quantity and substitute it into the situation. Use problems with exact whole-number answers, including examples with all four operations across the sequence.

Independent performance: Write a two-step story that another learner can solve, provide two equations and justify an estimate. Exchange solutions and check whether the reported number answers the original question. If a learner gives an intermediate result, ask what action in the story has not yet been represented. If arithmetic is wrong but the plan is sound, correct the calculation without discarding the useful model.