Mathematics • Year 10 • Unit 2 • Lesson 13
Substitution — Mixed Challenge
Apply substitution to six varied pairs (one-variable isolated, both need rearrangement, fractional answers, word setup). Spot a missing back-substitution, then design your own simultaneous pair with a chosen solution.
1. Mixed problems — solve by substitution
Always state the solution as (x, y), and always check both originals. 3 marks each
1.1 y = 3x and x + y = 16.
1.2 x = 2y − 5 and 3x + y = 13.
1.3 2x + y = 11 and 4x − y = 7. (Isolate y in equation 1.)
1.4 y = 4 − x and 5x + 2y = 17. (Watch the brackets.)
1.5 3x + 2y = 8 and x − y = 1. (Make x the subject of eq 2: x = 1 + y.)
1.6 A rectangle has perimeter 30 cm and length is 3 cm more than width. Let L = length, W = width. Write two equations, then use substitution to find L and W.
2. Find the mistake
A Year 10 student solves y = 2x + 1 and 3x + y = 21. Their working is below. One step contains an error — find it. 3 marks
Student's working:
Line 1: Substitute y into eq 2: 3x + 2x + 1 = 21
Line 2: 5x + 1 = 21
Line 3: 5x = 20 → x = 4
Line 4: Final answer: x = 4
(a) What is missing or wrong?
(b) Explain in one or two sentences why it matters.
(c) Complete the solution properly, including a check in both originals.
Stuck? "Solving simultaneously" requires both x and y, and back-substitution into the y = expression is the missing step.3. Open-ended challenge — design your own pair
Many valid answers. 4 marks
3.1 Design a pair of simultaneous equations whose only solution is (x, y) = (4, 3). They must:
(i) Use both x and y in both equations.
(ii) Not be multiples of each other.
(iii) Be set up so substitution is the obvious method (one variable easily isolated).
Submit: (a) your pair, (b) the substitution working that produces (4, 3), and (c) a check in both originals.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — y = 3x; x + y = 16
x + 3x = 16 → 4x = 16 → x = 4, y = 12. (4, 12).
1.2 — x = 2y − 5; 3x + y = 13
3(2y − 5) + y = 13 → 7y − 15 = 13 → 7y = 28 → y = 4, x = 2(4) − 5 = 3. (3, 4).
1.3 — 2x + y = 11; 4x − y = 7
y = 11 − 2x. Sub: 4x − (11 − 2x) = 7 → 6x = 18 → x = 3, y = 11 − 6 = 5. (3, 5).
1.4 — y = 4 − x; 5x + 2y = 17
5x + 2(4 − x) = 17 → 3x + 8 = 17 → 3x = 9 → x = 3, y = 1. (3, 1).
1.5 — 3x + 2y = 8; x − y = 1
x = 1 + y. Sub: 3(1 + y) + 2y = 8 → 5y + 3 = 8 → y = 1, x = 2. (2, 1).
1.6 — Rectangle
2L + 2W = 30 and L = W + 3. Sub: 2(W + 3) + 2W = 30 → 4W + 6 = 30 → 4W = 24 → W = 6 cm, L = 9 cm. Check: 2(9) + 2(6) = 30 ✓.
2 — Find the mistake
(a) The student stopped after finding x and never back-substituted to find y.
(b) "Solving simultaneously" means finding the pair (x, y) that satisfies both originals. Reporting only x is half a solution.
(c) Complete: x = 4. Back-substitute into y = 2x + 1 → y = 2(4) + 1 = 9. Final answer: (4, 9). Check: y = 2(4) + 1 = 9 ✓ and 3(4) + 9 = 21 ✓.
3 — Open-ended (sample solution)
Sample pair: y = x − 1 and 2x + 3y = 17.
Substitution: 2x + 3(x − 1) = 17 → 5x − 3 = 17 → 5x = 20 → x = 4, y = 4 − 1 = 3 ✓.
Check: y = 4 − 1 = 3 ✓ and 2(4) + 3(3) = 8 + 9 = 17 ✓.
Both equations include x and y. Eq 2 is not a multiple of eq 1 (different gradient when rearranged).
Marking: 1 for two valid equations with both variables; 1 for not being multiples of each other; 1 for substitution working that lands on (4, 3); 1 for a successful check in both equations.