Mathematics • Year 7 • Unit 2 • Lesson 15
Bracket Equations — Mixed Challenge
Mix it up: six equations of varied bracket forms, find a classic distributive-law mistake, then build your own bracket equation with a specified solution.
1. Mixed problems — expand and solve
Show working: expand first (or divide first, if cleaner), then SADMEP. Always check. 2 marks each
1.1 Solve 3(x + 7) = 27.
1.2 Solve 2(x − 3) = 8.
1.3 Solve 4(2x + 5) = 28.
1.4 Solve (x − 5)⁄3 = 4. (Multiply both sides by 3 first.)
1.5 Solve 5(x − 1) = 2(x + 7). (Brackets on BOTH sides — expand both, then collect x-terms on one side.)
1.6 Solve 3(x + 4) = 2(x + 9).
2. Find the mistake
Another Year 7 student has tried to solve 4(x + 2) = 24 by expanding. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then redo the solve correctly. 3 marks
Student's working — Solve 4(x + 2) = 24:
Line 1: Expand: 4(x + 2) = 4x + 2
Line 2: New equation: 4x + 2 = 24
Line 3: Subtract 2: 4x = 22
Line 4: Divide by 4: x = 5.5
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (which rule did they break?).
(c) Redo the working correctly, including a check in the ORIGINAL equation.
Stuck? The 4 outside must multiply BOTH the x AND the 2, not just the x. So 4(x + 2) = 4x + 8, not 4x + 2.3. Open-ended challenge — design a bracket equation
This question has more than one correct answer. Show one that works and explain. 4 marks
3.1 Design a bracket equation of the form a(x + b) = c with these rules:
(i) a, b and c are POSITIVE whole numbers
(ii) a divides cleanly into c (so "divide by a first" gives a whole number on the RHS)
(iii) the solution is x = 7
(iv) make sure a ≥ 3 and b ≥ 2 (so it's not trivial).
Write down:
(1) the equation,
(2) a real-world story it could represent (one sentence),
(3) the solving working using your preferred strategy (expand-first OR divide-first), naming which one you chose,
(4) the check in the ORIGINAL equation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 3(x + 7) = 27
Divide by 3: x + 7 = 9. Subtract 7: x = 2. Check: 3(2 + 7) = 3(9) = 27 ✓.
1.2 — 2(x − 3) = 8
Divide by 2: x − 3 = 4. Add 3: x = 7. Check: 2(7 − 3) = 2(4) = 8 ✓.
1.3 — 4(2x + 5) = 28
Divide by 4: 2x + 5 = 7. Subtract 5: 2x = 2. Divide by 2: x = 1. Check: 4(2(1) + 5) = 4(7) = 28 ✓.
1.4 — (x − 5)⁄3 = 4
Multiply both sides by 3: x − 5 = 12. Add 5: x = 17. Check: (17 − 5)⁄3 = 12⁄3 = 4 ✓.
1.5 — 5(x − 1) = 2(x + 7)
Expand both sides: 5x − 5 = 2x + 14. Subtract 2x from both sides: 3x − 5 = 14. Add 5: 3x = 19. Divide by 3: x = 19⁄3 ≈ 6.33. Check: LHS = 5(19⁄3 − 1) = 5(16⁄3) = 80⁄3. RHS = 2(19⁄3 + 7) = 2(40⁄3) = 80⁄3 ✓.
1.6 — 3(x + 4) = 2(x + 9)
Expand both sides: 3x + 12 = 2x + 18. Subtract 2x: x + 12 = 18. Subtract 12: x = 6. Check: 3(6 + 4) = 3(10) = 30; 2(6 + 9) = 2(15) = 30 ✓.
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The student multiplied the 4 only by the x and forgot to multiply it by the +2. The distributive law says 4(x + 2) = 4 × x + 4 × 2 = 4x + 8, not 4x + 2.
(c) Correct working: 4(x + 2) = 24. Expand: 4x + 8 = 24. Subtract 8: 4x = 16. Divide by 4: x = 4. Check in original: 4(4 + 2) = 4(6) = 24 ✓.
3 — Open-ended (sample solution)
Equation: 4(x + 3) = 40 (a = 4, b = 3, c = 40).
Story: "Four identical lunchboxes each hold x sandwiches and 3 cookies. Together they contain 40 items in total."
Solving (divide-first): divide both sides by 4 → x + 3 = 10. Subtract 3 → x = 7. (Divide-first was chosen because 40 ÷ 4 = 10 is clean and avoids large expanded numbers.)
Check in original: 4(7 + 3) = 4(10) = 40 ✓.
Marking: 1 for a valid equation matching all rules; 1 for a sensible "groups of" story; 1 for showing the chosen strategy and reason; 1 for the substitution check in the original.