Mathematics • Year 10 • Unit 2 • Lesson 13
Simultaneous Equations (Substitution) — Skill Drill
Build fluency with the substitution method from Lesson 13: pick the equation with an isolated variable, substitute the expression into the other equation, solve, then back-substitute and verify the (x, y) pair in both originals.
1. I do — fully worked example
Substitution where one variable is already isolated.
Problem. Solve simultaneously: y = 3x − 2 and 2x + 5y = 24.
Step 1 — Spot the isolated variable.
Equation 1 already has y on its own: y = 3x − 2.
Reason: Substitution is fastest when one variable is already alone.
Step 2 — Substitute the expression for y into Equation 2.
2x + 5(3x − 2) = 24
Step 3 — Expand, collect, solve for x.
2x + 15x − 10 = 24 → 17x = 34 → x = 2
Step 4 — Back-substitute to find y.
y = 3(2) − 2 = 4
Step 5 — Check in BOTH originals.
y: 3(2) − 2 = 4 ✓ 2x + 5y: 2(2) + 5(4) = 4 + 20 = 24 ✓
Answer: x = 2, y = 4.
2. We do — fill in the missing steps
Substitution with rearrangement first. 5 marks
Problem. Solve: 2x + 3y = 12 and y = x − 1.
Step 1 — Spot the isolated variable: ____ is isolated in equation 2.
Step 2 — Substitute y = x − 1 into equation 1:
2x + 3( ____ ) = 12
Step 3 — Expand and collect like terms:
2x + 3x − ____ = 12 → ____ x = ____ → x = ____
Step 4 — Back-substitute into y = x − 1:
y = ____ − 1 = ____
Step 5 — Check in both originals: 2x + 3y = 2( ____ ) + 3( ____ ) = ____ ✓.
3. You do — independent practice
For each pair, give the solution as an ordered pair (x, y) and show your check.
Foundation — y already isolated
3.1 y = 2x and x + y = 9. 1 mark
3.2 y = x + 3 and 2x + y = 15. 1 mark
3.3 x = 3y and 2x + y = 14. 1 mark
3.4 y = 4x − 1 and 3x + y = 13. 1 mark
Standard — rearrange first
3.5 x + y = 10 and 2x − y = 5. (Make y the subject of equation 1 first.) 2 marks
3.6 2x − y = 4 and 3x + 2y = 13. (Make y the subject of equation 1.) 2 marks
Extension — non-integer solution + word setup
3.7 y = 2x + 1 and 3x + 4y = 18. State your answer as exact fractions. 3 marks
3.8 A movie ticket costs $a for adults and $c for children. Two adults and three children pay $58, while one adult and two children pay $33. Set up two equations, isolate one variable, and use substitution to find a and c. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (2x + 3y = 12; y = x − 1)
Step 1: y. Step 2: 2x + 3( x − 1 ) = 12. Step 3: 2x + 3x − 3 = 12 → 5x = 15 → x = 3. Step 4: y = 3 − 1 = 2. Step 5: 2(3) + 3(2) = 12 ✓.
3.1 — y = 2x; x + y = 9
x + 2x = 9 → x = 3, y = 6. (3, 6).
3.2 — y = x + 3; 2x + y = 15
2x + x + 3 = 15 → 3x = 12 → x = 4, y = 7. (4, 7).
3.3 — x = 3y; 2x + y = 14
2(3y) + y = 14 → 7y = 14 → y = 2, x = 6. (6, 2).
3.4 — y = 4x − 1; 3x + y = 13
3x + 4x − 1 = 13 → 7x = 14 → x = 2, y = 7. (2, 7).
3.5 — x + y = 10; 2x − y = 5
y = 10 − x. Sub: 2x − (10 − x) = 5 → 3x − 10 = 5 → x = 5, y = 5. (5, 5).
3.6 — 2x − y = 4; 3x + 2y = 13
y = 2x − 4. Sub: 3x + 2(2x − 4) = 13 → 7x − 8 = 13 → 7x = 21 → x = 3, y = 2. (3, 2). Check: 3(3) + 2(2) = 13 ✓.
3.7 — y = 2x + 1; 3x + 4y = 18
3x + 4(2x + 1) = 18 → 11x + 4 = 18 → 11x = 14 → x = 14/11. y = 2(14/11) + 1 = 28/11 + 11/11 = 39/11. (14/11, 39/11).
3.8 — Tickets
2a + 3c = 58 (eq 1); a + 2c = 33 (eq 2) → a = 33 − 2c. Sub into eq 1: 2(33 − 2c) + 3c = 58 → 66 − 4c + 3c = 58 → 66 − c = 58 → c = $8. Then a = 33 − 16 = $17. Check eq 1: 2(17) + 3(8) = 34 + 24 = 58 ✓.