Mathematics • Year 8 • Unit 2 • Lesson 18
Substitution Method — Mixed Challenge
Combine all four steps — Isolate, Substitute, Solve, Back-substitute — across six mixed systems. Then spot a classic substitution mistake (the minus-sign trap) and build your own system that solves cleanly by substitution.
1. Mixed problems — solve each by substitution
Show ALL steps: isolate, substitute (with brackets!), solve, back-substitute, verify. 3 marks each
1.1 y = 3x and 2x + y = 20.
1.2 y = 2x − 3 and x + y = 9.
1.3 x + y = 8 and 2x − y = 7. (Isolate y in Eq 1.)
1.4 3x + y = 11 and 2x − y = −1. (Isolate y in Eq 1, then watch the sign in Eq 2.)
1.5 x = 2y + 1 and 3x + y = 17. (Sub x straight in.)
1.6 x − 2y = 4 and 2x + 3y = 15. (Isolate x in Eq 1, sub, expand, solve.)
2. Find the mistake
A student tries to solve x + y = 6 and 3x − y = 2 by substitution. Their working is below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then redo correctly. 3 marks
Student's working:
Line 1: From Eq 1: y = 6 − x.
Line 2: Sub into Eq 2: 3x − 6 − x = 2.
Line 3: 2x − 6 = 2 → 2x = 8 → x = 4.
Line 4: y = 6 − 4 = 2.
Line 5: Final: (x, y) = (4, 2).
(a) Which line contains the mistake?
(b) Explain in one or two sentences exactly what went wrong on that line.
(c) Re-do the working from that line onwards. State the correct solution and verify in both original equations.
Stuck? Revisit lesson § "Spot the Trap" — the minus sign distributes to every term inside the brackets. 3x − (6 − x) ≠ 3x − 6 − x.3. Open-ended challenge — design a system
This question has more than one valid answer. 4 marks
3.1 Design your own system of TWO simultaneous equations such that:
(a) the solution is (x, y) = (3, 4),
(b) at least ONE variable in one equation has coefficient 1 or −1 (so substitution is clean), and
(c) the two equations are clearly DIFFERENT (not multiples of each other).
Write your two equations, then SOLVE your own system using substitution to show it works. Finally, verify (3, 4) in both equations.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — y = 3x, 2x + y = 20
Sub: 2x + 3x = 20 → 5x = 20 → x = 4. y = 3(4) = 12. (4, 12). Check: 2(4) + 12 = 20 ✓.
1.2 — y = 2x − 3, x + y = 9
Sub: x + (2x − 3) = 9 → 3x − 3 = 9 → 3x = 12 → x = 4. y = 2(4) − 3 = 5. (4, 5). Check: 4 + 5 = 9 ✓; 5 = 2(4) − 3 ✓.
1.3 — x + y = 8, 2x − y = 7
Isolate y in Eq 1: y = 8 − x. Sub: 2x − (8 − x) = 7 → 2x − 8 + x = 7 → 3x = 15 → x = 5. y = 8 − 5 = 3. (5, 3). Check: 5 + 3 = 8 ✓; 2(5) − 3 = 7 ✓.
1.4 — 3x + y = 11, 2x − y = −1
Isolate y in Eq 1: y = 11 − 3x. Sub: 2x − (11 − 3x) = −1 → 2x − 11 + 3x = −1 → 5x = 10 → x = 2. y = 11 − 3(2) = 5. (2, 5). Check: 3(2) + 5 = 11 ✓; 2(2) − 5 = −1 ✓.
1.5 — x = 2y + 1, 3x + y = 17
Sub: 3(2y + 1) + y = 17 → 6y + 3 + y = 17 → 7y = 14 → y = 2. x = 2(2) + 1 = 5. (5, 2). Check: 3(5) + 2 = 17 ✓.
1.6 — x − 2y = 4, 2x + 3y = 15
Isolate x in Eq 1: x = 4 + 2y. Sub: 2(4 + 2y) + 3y = 15 → 8 + 4y + 3y = 15 → 7y = 7 → y = 1. x = 4 + 2(1) = 6. (6, 1). Check: 6 − 2(1) = 4 ✓; 2(6) + 3(1) = 15 ✓.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) When substituting y = 6 − x into 3x − y, the (6 − x) must be in brackets and the minus sign distributes to BOTH terms: 3x − (6 − x) = 3x − 6 + x. The student wrote 3x − 6 − x instead — they forgot to flip the sign of −x.
(c) Correct: 3x − (6 − x) = 2 → 3x − 6 + x = 2 → 4x − 6 = 2 → 4x = 8 → x = 2. y = 6 − 2 = 4. Solution: (2, 4). Verify: 2 + 4 = 6 ✓; 3(2) − 4 = 2 ✓.
3 — Design a system with solution (3, 4)
Sample student answer.
Eq 1: y = x + 1 (coefficient of y is 1 — clean isolation). Verify (3, 4): 4 = 3 + 1 ✓.
Eq 2: 2x + y = 10. Verify (3, 4): 2(3) + 4 = 10 ✓.
Solve by substitution: Sub y = x + 1 into Eq 2: 2x + (x + 1) = 10 → 3x + 1 = 10 → 3x = 9 → x = 3. Then y = 3 + 1 = 4. Solution: (3, 4) ✓.
Other valid student systems include: y = 2x − 2 and x + y = 7; x = y − 1 and 3x + 2y = 17; etc.
Marking: 1 mark for each equation correctly passing through (3, 4); 1 mark for at least one variable having coefficient 1 or −1; 1 mark for showing substitution arrives back at (3, 4).