Mathematics • Year 7 • Unit 2 • Lesson 14
Two-Step Equations — Mixed Challenge
Mix it up: six two-step equations of varied form, find a classic SADMEP-order mistake, then design a "fee + per-item" equation from scratch.
1. Mixed problems — use SADMEP
For each: undo the +/− first, then undo the ×/÷. Always check. 2 marks each
1.1 Solve 7x + 3 = 31.
1.2 Solve x⁄4 + 5 = 9.
1.3 Solve 2x − 9 = −5.
1.4 Solve 5x + 12 = 2. (Negative answer.)
1.5 Solve x⁄3 − 7 = −4.
1.6 Solve 3x + 4 = 19 by SADMEP and also by checking that x = 5 satisfies it. Confirm both routes agree.
2. Find the mistake
Another Year 7 student has tried to solve 5x − 3 = 22. 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 5x − 3 = 22:
Line 1: x is multiplied by 5, then 3 is subtracted.
Line 2: Divide both sides by 5 first: x − 3 = 22 ÷ 5
Line 3: x − 3 = 4.4
Line 4: Add 3: x = 7.4
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Redo the working correctly using SADMEP, including a check.
Stuck? Two problems with Line 2: (1) wrong order — should undo −3 FIRST, then divide. (2) When you divide the LHS by 5, you only divide 5x by 5, not the −3, unless you put brackets first.3. Open-ended challenge — design a "fee + per-item" equation
This question has more than one correct answer. Show one that works and explain. 4 marks
3.1 Design a real-world scenario that leads to a two-step equation, with these rules:
(i) the situation must involve a fixed fee AND a per-item cost (or per-hour, per-km, etc.)
(ii) your equation must be of the form ax + b = c, where a, b and c are POSITIVE whole numbers
(iii) the solution must be a positive whole number (no fractions in the answer)
(iv) make at least one of a, b, c bigger than 10 (so it's not trivial).
Write down:
(1) the scenario in one or two sentences,
(2) what each letter stands for,
(3) the equation,
(4) the full SADMEP solving working,
(5) the answer in a sentence.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 7x + 3 = 31
Subtract 3: 7x = 28. Divide by 7: x = 4. Check: 7(4) + 3 = 31 ✓.
1.2 — x⁄4 + 5 = 9
Subtract 5: x⁄4 = 4. Multiply by 4: x = 16. Check: 16⁄4 + 5 = 4 + 5 = 9 ✓.
1.3 — 2x − 9 = −5
Add 9: 2x = 4. Divide by 2: x = 2. Check: 2(2) − 9 = 4 − 9 = −5 ✓.
1.4 — 5x + 12 = 2
Subtract 12: 5x = −10. Divide by 5: x = −2. Check: 5(−2) + 12 = −10 + 12 = 2 ✓.
1.5 — x⁄3 − 7 = −4
Add 7: x⁄3 = 3. Multiply by 3: x = 9. Check: 9⁄3 − 7 = 3 − 7 = −4 ✓.
1.6 — 3x + 4 = 19, both routes
SADMEP: subtract 4 → 3x = 15 → divide by 3 → x = 5. Substitution check: 3(5) + 4 = 15 + 4 = 19 ✓. Both routes agree.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) Two problems: (i) wrong order — they should have undone the −3 FIRST and then divided (SADMEP), and (ii) when you divide the LHS by 5, the 5 only cancels 5x, leaving x and the −3 unchanged inside brackets: (5x − 3)⁄5 ≠ x − 3, it's actually x − 3⁄5. The student treated the equation incorrectly.
(c) Correct working: add 3 to both sides first → 5x − 3 + 3 = 22 + 3 → 5x = 25. Divide by 5: x = 5. Check: 5(5) − 3 = 25 − 3 = 22 ✓.
3 — Open-ended (sample solution)
Scenario: Tash hires a jumping castle for a birthday party. The hire company charges a $20 set-up fee plus $15 per hour. Her total bill was $110.
Variable: h = number of hours the castle was hired.
Equation: 15h + 20 = 110.
SADMEP: subtract 20 → 15h = 90 → divide by 15 → h = 6.
Answer: the jumping castle was hired for 6 hours. Check: 15(6) + 20 = 90 + 20 = 110 ✓.
Marking: 1 for a sensible scenario with a fixed fee + per-item cost; 1 for a correct equation matching the scenario; 1 for full SADMEP working; 1 for a whole-number answer with check.