20 questions · 56 marks · step-by-step working for every question. Marks are shown in square brackets. Every answer on this sheet has been checked by computer algebra.
Q1 · x = 6, y = 3 [2]
Adding the two equations removes y: 2x = 12, so x = 6
Substitute into x + y = 9: 6 + y = 9, so y = 3
Check in the second equation: 6 − 3 = 3 ✓
Q2 · x = 3, y = 2 [2]
Both equations have a single y, so subtract them: (3x + y) − (2x + y) = 11 − 8
x = 3
Substitute into 3x + y = 11: 9 + y = 11, so y = 2
Q3 · x = 3, y = 2 [2]
The y terms are +3y and −3y, so add: 6x = 18
x = 3
Substitute into 2x − 3y = 0: 6 − 3y = 0, so y = 2
Q4 · x = 4, y = 2 [2]
Both have −2y, so subtract: 2x = 8
x = 4
Substitute into 3x − 2y = 8: 12 − 2y = 8 → 2y = 4, so y = 2
Q5 · x = 4, y = 1 [2]
The y terms are −y and +y, so add: 3x = 12
x = 4
Substitute into x + y = 5: 4 + y = 5, so y = 1
Check in the first equation: 2(4) − 1 = 7 ✓
Q6 · x = 3, y = 2 [2]
Add the two equations: 10x = 30
x = 3
Substitute into 6x + y = 20: 18 + y = 20, so y = 2
Check in the second equation: 12 − 2 = 10 ✓
Q7 · x = 3, y = 2 [3]
Multiply the first by 3 and the second by 2 so the x terms match: 6x + 9y = 36 and 6x + 4y = 26
Subtract: 5y = 10, so y = 2
Substitute into 2x + 3y = 12: 2x + 6 = 12 → x = 3
Q8 · x = 3, y = 4 [3]
Multiply the first by 3 and the second by 4 to match the y terms: 9x + 12y = 75 and 8x + 12y = 72
Subtract: x = 3
Substitute into 3x + 4y = 25: 9 + 4y = 25 → 4y = 16, so y = 4
Q9 · x = 3, y = 4 [3]
Doubling the first equation matches the y terms: 10x + 4y = 46
Subtract the second equation: 7x = 21, so x = 3
Substitute into 5x + 2y = 23: 15 + 2y = 23 → y = 4
Q10 · x = 4, y = 9 [3]
Multiply the first equation by 6 to clear the fractions: 3x + 2y = 30
The second equation gives x = 13 − y; substituting: 3(13 − y) + 2y = 30
39 − 3y + 2y = 30 → y = 9, so x = 4
Check: 42 + 93 = 2 + 3 = 5 ✓
Q11 · x = 5, y = 2 [3]
Multiply the first by 3 and the second by 2 to match the y terms: 9x + 6y = 57 and 8x + 6y = 52
Subtract: x = 5
Substitute into 3x + 2y = 19: 15 + 2y = 19 → 2y = 4, so y = 2
Q12 · x = 4, y = 7 [3]
Multiply the first by 4 and the second by 5 to match the x terms: 20x + 12y = 164 and 20x + 25y = 255
Subtract: 13y = 91, so y = 7
Substitute into 5x + 3y = 41: 5x + 21 = 41 → x = 4
Q13 · x = 3, y = 4 [3]
Substitute y = 3x − 5 into the second equation: 2x + (3x − 5) = 10
5x − 5 = 10 → 5x = 15, so x = 3
y = 3(3) − 5 = 4
Q14 · x = 31, y = 16 [3]
Let the numbers be x and y: x + y = 47 and x − y = 15
Adding: 2x = 62, so x = 31
31 + y = 47, so y = 16
Q15 · a pen costs $1.90, a pencil costs $1.10 [3]
3p + 2q = 7.90 and 2p + 4q = 8.20
Doubling the first equation: 6p + 4q = 15.80
Subtract the second equation: 4p = 7.60, so p = 1.90
2(1.90) + 4q = 8.20 → 4q = 4.40, so q = 1.10
Q16 · the son is 12, the father is 36 [4]
Let the father be f and the son s: f = 3s
In 12 years: f + 12 = 2(s + 12)
Substituting f = 3s: 3s + 12 = 2s + 24 → s = 12
f = 3(12) = 36 — check in 12 years: 48 = 2 × 24 ✓
Q17 · width 7 cm, length 10 cm [3]
2(l + w) = 34 → l + w = 17, and l = w + 3
Substituting: (w + 3) + w = 17 → 2w = 14, so w = 7 cm
l = 7 + 3 = 10 cm — check: perimeter = 2(10 + 7) = 34 cm ✓
Q18 · the daughter is 12, the mother is 48 [3]
Let the mother be m and the daughter d: m = 4d
In 6 years: m + 6 = 3(d + 6)
Substituting m = 4d: 4d + 6 = 3d + 18 → d = 12
m = 4(12) = 48 — check: in 6 years 54 = 3 × 18 ✓
Q19 · 7 twenty-cent coins and 5 fifty-cent coins [3]
Let there be x twenty-cent coins and y fifty-cent coins: x + y = 12
Value in cents: 20x + 50y = 390. Divide by 10: 2x + 5y = 39
Substituting x = 12 − y: 2(12 − y) + 5y = 39 → 24 + 3y = 39 → y = 5
x = 12 − 5 = 7 — check: 7(20) + 5(50) = 390 cents ✓
Q20 · an adult ticket costs $12, a child ticket costs $7 [4]
3a + 4c = 64 and 2a + 5c = 59
Multiply the first by 2 and the second by 3 to match the a terms: 6a + 8c = 128 and 6a + 15c = 177
Subtract: 7c = 49, so c = 7
Substitute into 3a + 4c = 64: 3a + 28 = 64 → a = 12
Check in the second equation: 2(12) + 5(7) = 59 ✓
