Mathematics • Year 8 • Unit 2 • Lesson 19

Elimination Method — Mixed Challenge

Master the full elimination workflow: direct add/subtract, multiply one equation, and multiply both. Six mixed systems, one "find the mistake" focused on the same-sign / opposite-sign rule, and one open-ended challenge: choose substitution OR elimination and justify.

Master · Mixed Challenge

1. Mixed problems — solve each by elimination

Before you operate, write a short note: "Add" or "Subtract" or "Multiply X by ___ first, then add/subtract". Show working. 3 marks each

1.1 x + y = 10 and x − y = 4. (Direct.)

1.2 3x + y = 13 and 3x − 2y = 4. (Direct.)

1.3 2x + y = 7 and x + 3y = 11. (Multiply Eq 2 by 2 OR Eq 1 by 3.)

1.4 2x + 3y = 13 and 5x − 3y = 1. (Opposite +3y and −3y signs.)

1.5 3x + 2y = 17 and 2x + 3y = 13. (Multiply BOTH — Eq 1 by 3, Eq 2 by 2.)

1.6 4x + 3y = 23 and 2x − y = 5. (Multiply Eq 2 by 3 to match y-coefficients with opposite signs.)

Stuck on 1.6? Multiply Eq 2 by 3: 6x − 3y = 15. Now add to Eq 1: 10x = 38 → x = 3.8.

2. Find the mistake

A student tries to solve 3x + 2y = 11 and 3x − 2y = 1 by elimination. 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: y terms are +2y and −2y — same signs.

Line 2: Same signs → SUBTRACT.

Line 3: (3x + 2y) − (3x − 2y) = 11 − 1 → 4y = 10 → y = 2.5.

Line 4: Back-sub into Eq 1: 3x + 5 = 11 → x = 2.

Line 5: Final: (x, y) = (2, 2.5).

(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 with the CORRECT operation. State the correct solution and verify in both originals.

Stuck? +2y and −2y are OPPOSITE signs, not the same. The rule: opposite signs ADD; same signs subtract.

3. Open-ended challenge — choose your method

This question rewards good strategy as well as correct working. 4 marks

3.1 Below are three systems. For EACH system:

(i) State whether you'd choose substitution or elimination and give a one-sentence reason based on the form of the equations (e.g., "y is already isolated" or "coefficients match"). (ii) Solve the system using your chosen method, showing every step.

System A: y = 4x − 1 and 2x + y = 11
System B: 3x + 2y = 16 and 3x − 2y = 4
System C: 2x + 5y = 17 and 3x − 2y = 1

Stuck? Use the table in lesson § "Elimination vs Substitution". Substitution is best when a variable is isolated; elimination is best when coefficients match (or can be made to match easily).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — x + y = 10, x − y = 4

Add (opposite y signs): 2x = 14 → x = 7. Back-sub: 7 + y = 10 → y = 3. (7, 3). Check: 7 − 3 = 4 ✓.

1.2 — 3x + y = 13, 3x − 2y = 4

Subtract (same +3x): (3x + y) − (3x − 2y) = 13 − 4 → 3y = 9 → y = 3. Back-sub: 3x + 3 = 13 → x = 10/3. (10/3, 3). Check Eq 2: 3(10/3) − 2(3) = 10 − 6 = 4 ✓.

1.3 — 2x + y = 7, x + 3y = 11

Multiply Eq 1 by 3: 6x + 3y = 21. Subtract Eq 2: 5x = 10 → x = 2. Back-sub: 2(2) + y = 7 → y = 3. (2, 3). Check Eq 2: 2 + 3(3) = 11 ✓.

1.4 — 2x + 3y = 13, 5x − 3y = 1

Add (opposite y signs): 7x = 14 → x = 2. Back-sub: 2(2) + 3y = 13 → 3y = 9 → y = 3. (2, 3). Check Eq 2: 5(2) − 3(3) = 10 − 9 = 1 ✓.

1.5 — 3x + 2y = 17, 2x + 3y = 13

Multiply Eq 1 by 3: 9x + 6y = 51. Multiply Eq 2 by 2: 4x + 6y = 26. Subtract: 5x = 25 → x = 5. Back-sub into Eq 1: 3(5) + 2y = 17 → 2y = 2 → y = 1. (5, 1). Check Eq 2: 2(5) + 3(1) = 13 ✓.

1.6 — 4x + 3y = 23, 2x − y = 5

Multiply Eq 2 by 3: 6x − 3y = 15. Add Eq 1: 10x = 38 → x = 3.8 (= 19/5). Back-sub into Eq 2: 2(3.8) − y = 5 → y = 7.6 − 5 = 2.6 (= 13/5). (3.8, 2.6). Check Eq 1: 4(3.8) + 3(2.6) = 15.2 + 7.8 = 23 ✓.

2 — Find the mistake

(a) The mistake is on Line 1 (and Line 2 follows from it).
(b) The y-terms are +2y and −2y — those are OPPOSITE signs, not the same. The student misread the signs. The correct rule is: opposite signs ADD.
(c) Add the equations: (3x + 2y) + (3x − 2y) = 11 + 1 → 6x = 12 → x = 2. Back-sub into Eq 1: 3(2) + 2y = 11 → 2y = 5 → y = 2.5. Solution: (2, 2.5). Verify Eq 1: 6 + 5 = 11 ✓; Eq 2: 6 − 5 = 1 ✓. (Coincidentally the student's final pair was correct — they got lucky because their subtraction of (3x − 2y) flipped the sign of 2y, which accidentally cancelled the x-terms anyway. But on a different system this would not save them.)

3 — Choose your method

System A: y = 4x − 1 is already isolated → choose SUBSTITUTION. Sub: 2x + (4x − 1) = 11 → 6x = 12 → x = 2. y = 4(2) − 1 = 7. (2, 7). Check: 2(2) + 7 = 11 ✓.
System B: Both in ax + by = c, coefficients +3x and +3x match → choose ELIMINATION (subtract). Subtract: (3x + 2y) − (3x − 2y) = 16 − 4 → 4y = 12 → y = 3. Back-sub: 3x + 6 = 16 → x = 10/3. (10/3, 3). Check Eq 2: 3(10/3) − 2(3) = 10 − 6 = 4 ✓.
System C: Both in ax + by = c, no variable has coefficient 1 → choose ELIMINATION (multiply both). Multiply Eq 1 by 3 and Eq 2 by 2: 6x + 15y = 51 and 6x − 4y = 2. Subtract: 19y = 49 → y = 49/19. Back-sub into Eq 1: 2x + 5(49/19) = 17 → 2x = 17 − 245/19 = (323 − 245)/19 = 78/19 → x = 39/19. (39/19, 49/19). Check Eq 2: 3(39/19) − 2(49/19) = 117/19 − 98/19 = 19/19 = 1 ✓.

Marking: 1 mark for a correctly justified method choice across all three systems; 1 mark each for the three correct solutions with verification (3 marks). Reasonable alternative method choices (e.g., elimination on System A) accepted if the working is correct.