One-Step Equations with an Unknown Number
Warm-up
Show a balance scale with 5 cubes on each side: balanced. Add 3 cubes to the left: unbalanced. What must we add to the right to rebalance? (3 cubes.) This is the equation: 5 + 3 = 5 + n. n = 3. Every equation is a balance problem.
Explore
Equation sorting: each pair receives 12 equation cards (4 of each type: start, change, result unknown). Sort by type. Solve each. Verify each. Record in a table: equation, type, solution, check.
Formalize
Even/odd investigation: if n is even, is n+1 odd? Always? Build several examples: 4+1=5 (odd), 6+1=7 (odd), 10+1=11 (odd). Write as an equation: n+1 = ? What can you say about the result? Generalize: even + 1 = odd. Always. This is algebraic reasoning without letters.
One-Step Equations with an Unknown Number
Word problem to equation: a baker made some muffins in the morning. She made 18 more in the afternoon. She has 31 muffins now. Write the equation and solve. n + 18 = 31. n = 31 - 18 = 13. Check: 13 + 18 = 31. The baker made 13 muffins in the morning.
Practice
Students solve 9 equations (3 of each type), verify each, and write one word problem for each type. Exit ticket: solve and verify n + 24 = 61.
Exit ticket
Students solve 9 equations (3 of each type), verify each, and write one word problem for each type. Exit ticket: solve and verify n + 24 = 61.
Way 1 — count on: start at 8, climb to 15: 9, 10, 11, 12, 13, 14, 15 — seven climbs. □ = 7.
Way 2 — fact family: 8, 7, 15 form a triangle; if 8 + □ = 15 then □ = 15 − 8 = 7. The unknown-addend equation and the subtraction are the same fact.
Way 3 — balance reasoning: the pans hold 8-and-□ on the left, 15 on the right. Remove 8 from both pans (legal — it keeps the balance level): □ alone = 7.
Step 4: Check by substitution — the step that makes it EQUATION work rather than answer-getting: 8 + 7 = 15 ✓. Say it as "seven makes the sentence TRUE."
Vocabulary to install now: solving an equation = finding the value that makes it true. That definition survives unchanged through Grade 12 algebra.
Step 1: Read the story shape: some mystery amount lost 6 and landed at 9. The unknown is the STARTING amount — usually the hardest position for students.
Step 2: Undo the story: if losing 6 left 9, then putting the 6 BACK restores the start: 9 + 6 = 15. So □ = 15.
Step 3: Show it on the bar model: whole bar □, split into parts 6 (removed) and 9 (left). The whole is the sum of its parts: 15.
Step 4: Substitute to verify: 15 − 6 = 9 ✓.
Step 5: Contrast with 6 − □ = 2 — visually similar, structurally different (now the unknown is what LEFT, not the start): □ = 4 because 6 − 4 = 2.
Watch for: students who grab both visible numbers and apply a favourite operation (9 − 6 = 3, so "3"). Substituting the claim back into the sentence — 3 − 6 = 9?? — is the self-correcting habit that catches it.
The story: "Jo had a bag of marbles. She won 8 more at recess. Now she has 21. How many were in the bag this morning?"
Step 1: Choose a symbol for the mystery: let □ be the morning marbles. (Any symbol works — a box, a heart, later a letter. The symbol is a NAME for a number we don't know yet.)
Step 2: Translate the story's events in order, resisting any urge to solve: started with □, won 8 → □ + 8, result is 21. Equation: □ + 8 = 21.
Step 3: NOW solve, any honest way: 21 − 8 = 13, or climb from 8 to 21.
Step 4: Answer the STORY, not the equation: "Jo had 13 marbles this morning." Check inside the story: 13, win 8, have 21 ✓.
The order matters pedagogically: represent FIRST, solve SECOND. Students who leap to arithmetic get this one right but fail the two-step stories coming in Grade 4; students who write the equation first have a machine that scales.