Solutions — Simultaneous Equations
-
Q1 — Substitution (simpler)
(a) y = x + 2 and 3x + y = 10
Substitute y = x + 2 into the second equation:
3x + (x + 2) = 10 → 4x + 2 = 10 → 4x = 8 → x = 2
y = 2 + 2 = 4
Solution: x = 2, y = 4
Check: 3(2) + 4 = 10 ✓
(b) y = 3 − 2x and x + 2y = 6
Substitute y = 3 − 2x:
x + 2(3 − 2x) = 6 → x + 6 − 4x = 6 → −3x = 0 → x = 0
y = 3 − 2(0) = 3
Solution: x = 0, y = 3
Check: 0 + 2(3) = 6 ✓
-
Q2 — Substitution (rearranging needed)
(a) 2x − y = 5 and y = x − 1
Substitute y = x − 1 into the first equation:
2x − (x − 1) = 5 → 2x − x + 1 = 5 → x + 1 = 5 → x = 4
y = 4 − 1 = 3
Solution: x = 4, y = 3
Check in Eq (1): 2(4) − 3 = 5 ✓
(b) x = 4y + 1 and 2x − 3y = 8
Substitute x = 4y + 1 into the second equation:
2(4y + 1) − 3y = 8 → 8y + 2 − 3y = 8 → 5y = 6 → y = 6/5
x = 4(6/5) + 1 = 24/5 + 5/5 = 29/5
Solution: x = 29/5 = 5.8, y = 6/5 = 1.2
Check: 2(5.8) − 3(1.2) = 11.6 − 3.6 = 8 ✓
-
Q3 — Elimination (subtract to remove a variable)
(a) 3x + 2y = 12 and x + 2y = 8
Both equations have +2y. Subtract Eq (2) from Eq (1) to eliminate y:
(3x + 2y) − (x + 2y) = 12 − 8 → 2x = 4 → x = 2
Substitute into Eq (2): 2 + 2y = 8 → 2y = 6 → y = 3
Solution: x = 2, y = 3
Check in Eq (1): 3(2) + 2(3) = 6 + 6 = 12 ✓
(b) 5x − 3y = 11 and 5x + y = 15
Both have 5x. Subtract Eq (1) from Eq (2) to eliminate x:
(5x + y) − (5x − 3y) = 15 − 11 → 4y = 4 → y = 1
Substitute into Eq (2): 5x + 1 = 15 → 5x = 14 → x = 14/5 = 2.8
Solution: x = 2.8, y = 1
Check in Eq (1): 5(2.8) − 3(1) = 14 − 3 = 11 ✓
-
Q4 — Elimination (add to remove a variable)
(a) 2x + 3y = 13 and 4x − 3y = 11
+3y and −3y cancel when added. Add Eq (1) and Eq (2):
6x = 24 → x = 4
Substitute into Eq (1): 2(4) + 3y = 13 → 3y = 5 → y = 5/3
Solution: x = 4, y = 5/3 ≈ 1.67
Check in Eq (2): 4(4) − 3(5/3) = 16 − 5 = 11 ✓
(b) 3x + 5y = 22 and 6x + 5y = 37
Both have +5y. Subtract Eq (1) from Eq (2):
(6x + 5y) − (3x + 5y) = 37 − 22 → 3x = 15 → x = 5
Substitute into Eq (1): 3(5) + 5y = 22 → 5y = 7 → y = 7/5 = 1.4
Solution: x = 5, y = 1.4
Check in Eq (2): 6(5) + 5(1.4) = 30 + 7 = 37 ✓
-
Q5 — Choose your method
(a) y = 2x and 4x + 3y = 50
Best method: Substitution (y is already isolated).
Substitute y = 2x: 4x + 3(2x) = 50 → 4x + 6x = 50 → 10x = 50 → x = 5
y = 2(5) = 10
Solution: x = 5, y = 10
Check: 4(5) + 3(10) = 20 + 30 = 50 ✓
(b) 2x + 3y = 18 and 3x + 2y = 17
Best method: Elimination (both in standard form; scale to eliminate one variable).
Multiply Eq (1) by 3: 6x + 9y = 54
Multiply Eq (2) by 2: 6x + 4y = 34
Subtract: 5y = 20 → y = 4
Substitute into Eq (1): 2x + 12 = 18 → 2x = 6 → x = 3
Solution: x = 3, y = 4
Check in Eq (2): 3(3) + 2(4) = 9 + 8 = 17 ✓
-
Q6 — Special cases
(a) 2x + 4y = 10 and x + 2y = 5
Divide the first equation by 2: x + 2y = 5. This is identical to the second equation.
The two equations describe the same line. Every point on the line satisfies both equations.
Infinitely many solutions (dependent system).
(b) 3x − y = 7 and 3x − y = 2
Both equations have the same left-hand side (3x − y) but different right-hand sides (7 ≠ 2).
Subtracting: 0 = 5 — a contradiction. The lines are parallel (same gradient m = 3, different y-intercepts).
No solution (inconsistent system).
-
Q7 — Word problem: two numbers
Let the two numbers be x and y.
Equation 1 (sum): x + y = 47
Equation 2 (difference): x − y = 13
Add both equations: 2x = 60 → x = 30
Substitute: 30 + y = 47 → y = 17
The two numbers are 30 and 17.
Check: 30 + 17 = 47 ✓ 30 − 17 = 13 ✓
-
Q8 — Word problem: café drinks
Let c = coffees sold, t = teas sold.
Equation 1 (total drinks): c + t = 45
Equation 2 (total revenue): 4.50c + 2.80t = 165.50
From Eq (1): t = 45 − c. Substitute into Eq (2):
4.50c + 2.80(45 − c) = 165.50
4.50c + 126 − 2.80c = 165.50
1.70c = 39.50 → c = 39.50 / 1.70 = 23.24...
Hmm, this should be a whole number. Let’s recheck: 1.70c = 165.50 − 126 = 39.50. c = 39.50/1.70 = 23.23... This is not a whole number.
Checking: if c = 23: revenue = 4.50(23) + 2.80(22) = 103.50 + 61.60 = $165.10 (close, not exact).
If c = 25: t = 20. Revenue = 4.50(25) + 2.80(20) = 112.50 + 56 = $168.50. Not matching.
Working more carefully with exact values: 1.70c = 39.50 → c = 395/17 which is not a whole number. The numbers are most consistent at c = 23, t = 22 with revenue ~$165.10, suggesting a minor rounding in the problem. For exam purposes: c ≈ 23 coffees, t ≈ 22 teas. (Some versions of this problem use $165.10 as the total to give exact integer answers.)
Formal answer using exact arithmetic: c = 395/17 ≈ 23.2, t = 45 − 395/17 = (765 − 395)/17 = 370/17 ≈ 21.8. The nearest whole-number solution is 23 coffees and 22 teas.
-
Q9 — Trains meeting
(a) Position equations (distance from Station 1 in km, t in hours):
Train A starts at Station 1 (position 0) and moves toward Station 2: d_A = 60t
Train B starts at Station 2 (position 180 km) and moves toward Station 1: d_B = 180 − 80t
(b) They meet when d_A = d_B:
60t = 180 − 80t
140t = 180
t = 180/140 = 9/7 ≈ 1.286 hours ≈ 1 hour 17 minutes after departure
(c) Distance from Station 1: d_A = 60 × (9/7) = 540/7 ≈ 77.1 km from Station 1
Check: d_B = 180 − 80(9/7) = 180 − 720/7 = (1260 − 720)/7 = 540/7 ✓
-
Q10 — Production problem
(a) Let x = units of Product X, y = units of Product Y.
Labour equation: 3x + y = 24 (total labour hours = 24)
Machine equation: 2x + 4y = 32 (total machine hours = 32)
(b) From labour equation (1): y = 24 − 3x. Substitute into machine equation (2):
2x + 4(24 − 3x) = 32
2x + 96 − 12x = 32
−10x = −64 → x = 6.4
y = 24 − 3(6.4) = 24 − 19.2 = 4.8
Produce 6.4 units of X and 4.8 units of Y per day.
Check labour: 3(6.4) + 4.8 = 19.2 + 4.8 = 24 ✓
Check machine: 2(6.4) + 4(4.8) = 12.8 + 19.2 = 32 ✓
(c) Daily profit = 80x + 120y = 80(6.4) + 120(4.8) = 512 + 576 = $1 088