LESSON PLAN

Algebra — Expressions and Equations

Apothem Team

Grade 6 · Algebra & Patterning

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

Play three rounds of "guess my number" out loud: "I'm thinking of a number. I doubled it and added 5, and got 17." Students solve it however they like (most un-do: 17, back off 5 → 12, halve → 6). Round two: "tripled it, subtracted 4, got 20." (8.)

Then the reveal: write what you were saying as $2n + 5 = 17$ . They've been doing algebra by ear; today it gets a written form — and the un-doing they invented is THE solving method.

Explore

Expression-building stations: (1) pattern blocks — the toothpick-squares rule $3n+1$ revisited, now evaluating for given $n$ and solving for $n$ given a total; (2) story-to-expression matching — eight stories, eight expressions, four decoys that differ by order ( $2n+5$ vs $2(n+5)$ ); (3) balance puzzles — solve $3x + 2 = 14$ on a drawn pan balance by legal moves.

Harvest at the board: the two meanings of a letter (a specific unknown to FIND in an equation; a variable that ranges over many values in an expression like $3n+1$ ) — same symbol, different jobs, both legitimate.

Formalize

Formalize vocabulary and the solving principle. An expression is a phrase ( $2n+5$ ); an equation is a sentence claiming two expressions are equal ( $2n+5 = 17$ ). Solving = finding values making the sentence true, by doing the same legal move to both sides:

$2n + 5 = 17 \;\xrightarrow{-5}\; 2n = 12 \;\xrightarrow{\div 2}\; n = 6$

The undo-order observation students should own: the expression was built "double, then add 5," and solving peeled it "subtract 5, then halve" — reverse order of operations. Substitution closes every solve: $2(6)+5 = 17$ ✓ makes it true.

Practice

Practice: evaluate expressions for given values (including $4a - 3$ at $a = 2.5$ ); translate four stories to expressions or equations; solve six one- and two-step equations with the both-sides notation; one "find the error" solve.

Exit ticket: solve $\frac{m}{3} - 2 = 4$ and verify by substitution. ( $m = 18$ .)

Exit ticket

Exit ticket: solve $\frac{m}{3} - 2 = 4$ and verify by substitution. ( $m = 18$ .)

TIP Require the substitution check on every solved equation this unit — it converts "I finished" into "I'm right," and it quietly reteaches evaluating expressions every single time.

WORKED EXAMPLES

Example 1 — From story to solution: the bus fare

The story: a bus card starts with \$20. Each ride costs \$2.50. After some rides the card holds \$7.50. How many rides were taken?

Step 1: Declare the variable: let $r$ = number of rides.

Step 2: Translate the story's events: start 20, spend $2.50r$ , end 7.50 → $20 - 2.50r = 7.50$ .

Step 3: Solve with both-sides moves: subtract 20: $-2.50r = -12.50$ … smoother route: add $2.50r$ to both sides and subtract 7.50: $12.50 = 2.50r$ → divide: $r = 5$ .

Step 4: Substitute into the ORIGINAL: $20 - 2.50(5) = 20 - 12.50 = 7.50$ ✓.

Step 5: Answer the story in its own words: five rides. (And note the flexibility shown in step 3 — when a negative coefficient looms, moving the variable term to the other side first keeps everything positive and friendly.)

Example 2 — Two expressions, one table: are

2(n+3)

and

2n+6

the same?

Step 1: Test numerically — build a two-column table: $n = 1$ : $2(4) = 8$ and $2+6 = 8$ ✓. $n = 5$ : $16$ and $16$ ✓. $n = 10$ : $26$ and $26$ ✓. Every input agrees.

Step 2: Explain WHY with the picture: $2(n+3)$ is two rows of ( $n$ and 3) — an $n$ -by-2 slab plus a 3-by-2 slab: $2n + 6$ . Doubling a sum doubles each part.

Step 3: Contrast with the imposter: $2n + 3$ . Test $n = 1$ : gives 5 ≠ 8. One failed input kills equivalence — while no number of passed inputs PROVES it (the picture does that).

Step 4: Name the idea for later: expressions that agree for every input are equivalent — the distributive picture is the Grade 6 proof. This is the same "different stories, same rule" discovery from the pattern unit, now in pure symbols.

Example 3 — Error autopsy: the solve that subtracted from one side

The submitted work: $3x + 4 = 19$ ; line 2: $3x = 19$ ; line 3: $x = 19 \div 3 = 6.33\ldots$ The student is suspicious — "the answer's ugly" — and that instinct deserves praise.

Step 1: Audit line 2: the $+4$ vanished from the left but 19 never paid — the balance was broken (4 removed from one pan only).

Step 2: Repair: $3x + 4 = 19$ → subtract 4 from BOTH sides → $3x = 15$ → $x = 5$ .

Step 3: Substitute: $3(5) + 4 = 19$ ✓.

Step 4: Harvest both lessons: (a) every move hits both sides — no exceptions; (b) the "ugly answer" smell test is legitimate mathematics: textbook equations usually resolve cleanly, so a sudden $6.33\ldots$ is a prompt to re-audit. (Real-world equations get ugly — but then the CHECK, not the aesthetics, is the referee.)

MATERIALS

•Pan balance mats and chips

•Story-expression match cards

•Pattern blocks / toothpicks

•Practice set (PDF)

WATCH FOR

!The equals sign as "answer coming" resurfaces with letters: students write

2n+5 = 17 = 12 = 6

run-on chains. Each line must be a TRUE sentence.

!Letters as objects ("

a

stands for apples") instead of numbers OF apples — leads to nonsense like

3a

meaning "three apples" that can't be computed with.

!Both-sides moves applied to one side only, or the wrong inverse chosen first. The pan-balance image plus reverse-order-undo names the fix.