LESSON PLAN

Solving Equations

Apothem Team

Grade 7 · 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

Ask students to solve, mentally, "I think of a number, multiply by 4, subtract 7, and get 29." Nearly everyone un-does: $29 + 7 = 36$ , $36 \div 4 = 9$ . Then hand them $4x - 7 = 29$ and watch the recognition land: they already solve equations — today formalizes the record-keeping for when mental undoing runs out.

Set the stakes honestly: two-step equations are the last size mental math handles; the both-sides method being formalized today is the one that scales to everything after.

Explore

Balance-mat progression: (1) two-step solves with chips and bags — $3x + 2 = 14$ : remove 2 from both pans, split 12 among 3 bags. (2) Variables on BOTH sides — $5x + 1 = 3x + 9$ : remove three bags from each pan first ( $2x + 1 = 9$ ), a genuinely new move. (3) The distributive opener: $2(x + 3) = 16$ — either share the 2 first ( $2x + 6 = 16$ ) or halve both sides ( $x + 3 = 8$ ); pairs try both and time them.

Each solve gets its symbolic diary written next to the mat work, with the check-by-substitution stamped at the bottom.

Formalize

Formalize the method: simplify each side; collect the variable on one side by adding/subtracting terms from both; then peel constants and coefficients in reverse order of operations:

$5x + 1 = 3x + 9 \;\xrightarrow{-3x}\; 2x + 1 = 9 \;\xrightarrow{-1}\; 2x = 8 \;\xrightarrow{\div 2}\; x = 4$

The two legal-move principles that generate everything: same operation on both sides preserves truth; simplifying one side (combining like terms, distributing) rewrites without changing anything. Every line must remain a TRUE-or-solvable sentence — no orphan expressions.

Practice

Practice: six solves climbing from two-step to variables-both-sides to one distributive; two story-to-equation-to-solution problems; one "find the error" (a sign lost moving a term across).

Exit ticket: solve $4(n - 2) = n + 7$ with every move labelled, then verify. ( $4n - 8 = n + 7 \to 3n = 15 \to n = 5$ ; check: $4(3) = 12 = 5+7$ ✓.)

Exit ticket

Practice: six solves climbing from two-step to variables-both-sides to one distributive; two story-to-equation-to-solution problems; one "find the error" (a sign lost moving a term across).

Exit ticket: solve $4(n - 2) = n + 7$ with every move labelled, then verify. ( $4n - 8 = n + 7 \to 3n = 15 \to n = 5$ ; check: $4(3) = 12 = 5+7$ ✓.)

TIP "Move the term" language causes the lost-sign plague. Enforce "subtract

3x

from both sides" phrasing for two weeks; the sign errors halve.

WORKED EXAMPLES

Example 1 — Variables on both sides: the two phone plans meet

The setup: Plan A costs $25 + 0.10t$ (t texts); Plan B costs $10 + 0.25t$ . When are they equal?

Step 1: Set equal: $25 + 0.10t = 10 + 0.25t$ .

Step 2: Collect variables (subtract $0.10t$ both sides): $25 = 10 + 0.15t$ .

Step 3: Peel the constant: $15 = 0.15t$ → divide: $t = 100$ texts.

Step 4: Verify: A at 100: $25 + 10 = 35$ ; B: $10 + 25 = 35$ ✓.

Step 5: Interpret the crossover like a consumer: under 100 texts, B is cheaper (small base wins); over, A wins (cheap rate wins). Solving found WHERE; the interpretation says WHO cares.

Example 2 — The perimeter equation: geometry hires algebra

The problem: a rectangle's length is 5 cm more than its width; the perimeter is 38 cm. Find both dimensions.

Step 1: One variable for the unknown anchor: let $w$ = width; then length $= w + 5$ .

Step 2: Perimeter equation: $2w + 2(w + 5) = 38$ .

Step 3: Simplify the left side FIRST (distribute, collect): $2w + 2w + 10 = 38 \to 4w + 10 = 38$ .

Step 4: Solve: $4w = 28 \to w = 7$ ; length $= 12$ .

Step 5: Check in the GEOMETRY, not the equation: $2(7) + 2(12) = 14 + 24 = 38$ ✓ and 12 is 5 more than 7 ✓. Two facts claimed, two facts verified — checks should test the story's every promise, not just re-run the algebra.

Example 3 — Error autopsy: the sign that teleported

The submitted solve: $7x - 4 = 2x + 11$ ; line 2: $7x - 2x = 11 - 4$ ; line 3: $5x = 7$ ; line 4: $x = 1.4$ .

Step 1: Line 2 audit — left side: subtracting $2x$ from both sides ✓ legal. Right side: the $-4$ crossed over and became… still $-4$ ? It shows as $11 - 4$ . But moving $-4$ legally means ADDING 4 to both sides → should be $11 + 4$ . The sign teleported unchanged; the balance broke.

Step 2: Repair from line 1: $7x - 4 = 2x + 11 \to 5x - 4 = 11 \to 5x = 15 \to x = 3$ .

Step 3: Verify: $7(3) - 4 = 17$ and $2(3) + 11 = 17$ ✓.

Step 4: The diagnostic for self-checking: an "ugly" answer (1.4) in a friendly-looking problem is a smell, and substitution ( $7(1.4) - 4 = 5.8$ vs $2(1.4) + 11 = 13.8$ — not equal!) is the smoke alarm. Substitution catches EVERY algebra error, including the ones you can't name.

MATERIALS

•Balance mats, bags, chips

•Solve diaries (two-column: moves | reasons)

•Story cards

•Practice set (PDF)

WATCH FOR

!Terms "moved" across the equals sign with signs intact:

2x + 1 = 9 \to 2x = 9 + 1

. Both-sides language prevents the teleport error.

!Distributing to only the first term:

2(x+3) \to 2x + 3

. The area picture (two rows of

x

AND two rows of 3) re-anchors it.

!Dividing before collecting:

5x + 1 = 3x + 9

divided by 5 immediately — legal but chaos. Collect first, peel second.