Mathematics • Year 10 • Unit 2 • Lesson 14
Elimination — Mixed Challenge
Apply elimination to six pairs of varying difficulty, spot a sign-flip error when subtracting, then design your own pair where elimination is clearly the better tool than substitution.
1. Mixed problems — solve by elimination
Show your "add or subtract?" reasoning and any multipliers. Always check both originals. 3 marks each
1.1 3x + y = 13 and 3x − y = 5.
1.2 4x + 5y = 22 and 4x + 2y = 16.
1.3 2x + 3y = 12 and 5x − 3y = 9.
1.4 x + 2y = 8 and 3x − 4y = 4. (Multiply Eq 1 to match either coefficient.)
1.5 4x + 3y = 24 and 5x + 2y = 23. (Multiply to match: LCM of 3 and 2 is 6.)
1.6 A right triangle's two legs satisfy 2a + b = 19 and a + 3b = 22. Use elimination to find a and b, then compute the hypotenuse using c = √(a² + b²).
2. Find the mistake
A Year 10 student tries to solve 4x + 3y = 22 and 4x − 5y = −10 by elimination. Their working is below. One line contains a sign error when subtracting. Find it. 3 marks
Student's working:
Line 1: Same sign on x (both +4) → subtract.
Line 2: (4x + 3y) − (4x − 5y) = 22 − (−10)
Line 3: 0 + 3y − 5y = 22 + 10
Line 4: −2y = 32 → y = −16
Line 5: Back: 4x + 3(−16) = 22 → 4x = 70 → x = 17.5
(a) Which line is wrong?
(b) Explain the sign mistake in one or two sentences.
(c) Redo correctly and verify in both originals.
Stuck? When subtracting, the minus sign flips every term in the second bracket: 3y − (−5y) = 3y + 5y = +8y, not −2y.3. Open-ended challenge — design your own
Many valid answers. 4 marks
3.1 Design a pair of simultaneous equations where (i) elimination is clearly the smarter choice (substitution would force ugly fractions), (ii) you need to multiply both equations before adding or subtracting, and (iii) the solution is (x, y) = (3, −2).
Submit:
(a) Your pair of equations.
(b) The multipliers you would use to eliminate one variable.
(c) Full elimination working that lands on (3, −2).
(d) A check in both originals.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 3x + y = 13; 3x − y = 5
Opposite sign on y → add: 6x = 18 → x = 3, y = 4. (3, 4).
1.2 — 4x + 5y = 22; 4x + 2y = 16
Same sign on x → subtract: 3y = 6 → y = 2, x = 3. (3, 2).
1.3 — 2x + 3y = 12; 5x − 3y = 9
Opposite sign on y → add: 7x = 21 → x = 3, y = 2. (3, 2).
1.4 — x + 2y = 8; 3x − 4y = 4
× Eq 1 by 2: 2x + 4y = 16. Add to Eq 2 (opposite y): 5x = 20 → x = 4, y = 2. (4, 2).
1.5 — 4x + 3y = 24; 5x + 2y = 23
× Eq 1 by 2: 8x + 6y = 48. × Eq 2 by 3: 15x + 6y = 69. Subtract Eq 1 form: 7x = 21 → x = 3, then 4(3) + 3y = 24 → y = 4. (3, 4). Check Eq 2: 5(3) + 2(4) = 15 + 8 = 23 ✓.
1.6 — Right triangle
2a + b = 19; a + 3b = 22. × Eq 1 by 3: 6a + 3b = 57. Subtract Eq 2: 5a = 35 → a = 7, b = 19 − 14 = 5. Hypotenuse: c = √(49 + 25) = √74 ≈ 8.60.
2 — Find the mistake
(a) Line 3.
(b) When subtracting (4x − 5y) from (4x + 3y), every term in the second bracket flips sign: 3y − (−5y) = 3y + 5y = +8y, not 3y − 5y = −2y. The student forgot to flip the sign on the −5y.
(c) Corrected: 0 + 8y = 32 → y = 4. Back: 4x + 3(4) = 22 → 4x = 10 → x = 2.5. Check Eq 2: 4(2.5) − 5(4) = 10 − 20 = −10 ✓.
3 — Open-ended (sample solution)
Sample pair: 2x + 3y = 0 and 5x − 4y = 23. At (3, −2): 2(3) + 3(−2) = 0 ✓ and 5(3) − 4(−2) = 15 + 8 = 23 ✓.
Multipliers: × Eq 1 by 4 and × Eq 2 by 3 to eliminate y.
Working: 8x + 12y = 0 and 15x − 12y = 69. Add (opposite sign on y): 23x = 69 → x = 3. Back: 2(3) + 3y = 0 → 3y = −6 → y = −2 ✓.
Check originals: (3, −2) satisfies both as above ✓.
Marking: 1 for a valid pair with (3, −2) as the unique solution; 1 for choosing multipliers that make elimination clean; 1 for the elimination working; 1 for the substitution check.