Mathematics • Year 8 • Unit 2 • Lesson 19
Simultaneous Equations — Elimination Method
Build fluency with the sign rule: opposite signs ADD, same signs SUBTRACT. One worked example with direct elimination, one guided example where you have to subtract, then eight independent problems graduated from direct add/subtract to "multiply one equation first".
1. I do — fully worked example
Read every line. The sign rule: opposite signs ADD, same signs SUBTRACT. The aim is to make one variable cancel to zero.
Problem. Solve x + y = 7 and x − y = 3 by elimination.
Step 1 — Line the equations up and inspect the y-terms.
Eq 1: x + y = 7
Eq 2: x − y = 3
Reason: y has coefficients +1 and −1 — OPPOSITE signs. Add the equations to cancel y.
Step 2 — Add the two equations (LHS + LHS = RHS + RHS).
(x + y) + (x − y) = 7 + 3
2x = 10
Reason: +y and −y combine to 0, leaving an equation with one variable.
Step 3 — Solve for x.
x = 5
Step 4 — Back-substitute into either original to find y.
Eq 1: 5 + y = 7 → y = 2
Step 5 — VERIFY in the OTHER equation.
Eq 2: 5 − 2 = 3 ✓
Answer: (x, y) = (5, 2).
2. We do — fill in the missing steps
This one needs SUBTRACTION (same-sign coefficients). Fill every blank. 5 marks
Problem. Solve 2x + 3y = 13 and 2x + y = 7 by elimination.
Step 1 — Inspect the x-terms: both are +2x. Same signs → ______________ the equations.
Step 2 — Subtract Eq 2 from Eq 1 (carefully — distribute the minus):
(2x + 3y) − (2x + y) = 13 − ______
2x − 2x + 3y − y = ______
______y = ______
Step 3 — Solve for y:
y = ______
Step 4 — Back-substitute into Eq 2 to find x:
2x + ______ = 7 → 2x = ______ → x = ______
Step 5 — Verify in Eq 1: 2(______) + 3(______) = ______ ✓ ?
Answer: (x, y) = (______, ______).
3. You do — independent practice
Show every step. 3.1–3.3 are foundation (direct add or subtract). 3.4–3.6 are standard (read the sign rule, then operate). 3.7–3.8 are extension (multiply one equation first).
Foundation — direct add or subtract
3.1 x + y = 9 and x − y = 3. Add to eliminate y. 2 marks
3.2 2x + y = 11 and 2x − y = 5. Add to eliminate y. 2 marks
3.3 3x + 2y = 14 and 3x − y = 8. Subtract to eliminate x. 2 marks
Standard — choose add or subtract
3.4 x + 2y = 8 and x + y = 5. (Same +x sign — subtract.) 3 marks
3.5 4x − 3y = 6 and 2x + 3y = 12. (Opposite y signs — add.) 3 marks
3.6 5x + 3y = 22 and 2x + 3y = 13. (Same +3y — subtract.) 3 marks
Extension — multiply one equation first
3.7 3x + 2y = 12 and x − y = 1. Multiply Eq 2 by 2 so the y-coefficients become +2y and −2y, then add. 3 marks
3.8 x + 4y = 14 and 3x − 2y = 0. Multiply Eq 2 by 2 so the y-coefficients become +4y and −4y, then add. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do 2x + 3y = 13 and 2x + y = 7
Step 1: SUBTRACT.
Step 2: (2x + 3y) − (2x + y) = 13 − 7 → 2x − 2x + 3y − y = 6 → 2y = 6.
Step 3: y = 3.
Step 4: 2x + 3 = 7 → 2x = 4 → x = 2.
Step 5: 2(2) + 3(3) = 4 + 9 = 13 ✓.
Answer: (x, y) = (2, 3).
3.1 — x + y = 9, x − y = 3
Add: 2x = 12 → x = 6. Back-sub: y = 9 − 6 = 3. (6, 3). Check: 6 − 3 = 3 ✓.
3.2 — 2x + y = 11, 2x − y = 5
Add: 4x = 16 → x = 4. Back-sub: y = 11 − 2(4) = 3. (4, 3). Check: 2(4) − 3 = 5 ✓.
3.3 — 3x + 2y = 14, 3x − y = 8
Subtract: (3x + 2y) − (3x − y) = 14 − 8 → 3y = 6 → y = 2. Back-sub: 3x − 2 = 8 → x = 10/3. (10/3, 2). Check Eq 1: 3(10/3) + 2(2) = 10 + 4 = 14 ✓.
3.4 — x + 2y = 8, x + y = 5
Subtract: y = 3. Back-sub: x + 3 = 5 → x = 2. (2, 3). Check: 2 + 2(3) = 8 ✓.
3.5 — 4x − 3y = 6, 2x + 3y = 12
Add (opposite y signs): 6x = 18 → x = 3. Back-sub: 2(3) + 3y = 12 → 3y = 6 → y = 2. (3, 2). Check Eq 1: 4(3) − 3(2) = 6 ✓.
3.6 — 5x + 3y = 22, 2x + 3y = 13
Subtract (same +3y): 3x = 9 → x = 3. Back-sub: 2(3) + 3y = 13 → 3y = 7 → y = 7/3. (3, 7/3). Check Eq 1: 5(3) + 3(7/3) = 15 + 7 = 22 ✓.
3.7 — 3x + 2y = 12, x − y = 1 (×2)
Multiply Eq 2 by 2: 2x − 2y = 2. Add to Eq 1: 5x = 14 → x = 14/5. Back-sub into Eq 2: 14/5 − y = 1 → y = 14/5 − 1 = 9/5. (14/5, 9/5). Check Eq 1: 3(14/5) + 2(9/5) = 42/5 + 18/5 = 60/5 = 12 ✓.
3.8 — x + 4y = 14, 3x − 2y = 0 (×2)
Multiply Eq 2 by 2: 6x − 4y = 0. Add to Eq 1: 7x = 14 → x = 2. Back-sub into Eq 2 (original): 3(2) − 2y = 0 → 2y = 6 → y = 3. (2, 3). Check Eq 1: 2 + 4(3) = 14 ✓.