Mathematics • Year 10 • Unit 2 • Lesson 13
Simultaneous Equations in the Real World
Set up and solve simultaneous equations from word problems — pens and pencils, phone plan break-even, mixtures, two-rate trip and a coin jar. Each one builds the two equations, then uses substitution to find the pair.
1. Word problems
For each: (i) define your variables, (ii) write the two equations, (iii) substitute and solve, (iv) state the answer in context.
1.1 — Stationery order. A shop sells pens at $1.20 each and pencils at $0.80 each. A school buys 150 items in total and spends $152.
(a) Let p = pens, c = pencils. Write two equations.
(b) Isolate c in the first equation, substitute into the second, and solve.
(c) State how many pens and pencils the school bought. 4 marks
1.2 — Phone plan break-even. Plan A: $30/month + $0.20 per call. Plan B: $15/month + $0.50 per call. Let n = calls per month, C = monthly cost.
(a) Write a cost equation for each plan.
(b) Set the two C-values equal and solve for the break-even n.
(c) For 80 calls/month, which plan is cheaper, and by how much? 4 marks
1.3 — Apples and oranges. Apples cost $2 each, oranges cost $1.50 each. Marco buys 10 pieces of fruit for $17. How many of each did he buy?
Define your variables, write the two equations, isolate one variable, and substitute. 3 marks
1.4 — Two-rate trip. Mai drives at 60 km/h on the highway and 40 km/h in the city. Her 3 hour trip covers 160 km. Let h = hours on the highway, c = hours in the city.
(a) Write a time equation (h + c = 3) and a distance equation (60h + 40c = 160).
(b) Use substitution to find h and c, then state the distance driven in each. 4 marks
1.5 — Coin jar. A jar holds 20-cent and 50-cent coins. There are 30 coins worth $9.60 in total. Let t = number of 20-cent coins, f = number of 50-cent coins.
(a) Write two equations (one for count, one for value in cents).
(b) Solve by substitution. 3 marks
2. Explain your thinking
Communication, not just numbers. 4 marks
2.1 A classmate is given x + y = 7 and 2x + 2y = 14 to solve simultaneously. Using substitution, they substitute y = 7 − x into 2x + 2y = 14 and end up with 14 = 14, which seems to be the trivial true statement. Using the words unique solution and multiple of, explain (i) what this result actually tells us about the two equations, (ii) why there is no single (x, y) pair solving them, and (iii) give two different (x, y) pairs that both satisfy x + y = 7 to demonstrate.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Stationery
(a) p + c = 150 (count); 1.20p + 0.80c = 152 (cost). (b) c = 150 − p. Sub: 1.20p + 0.80(150 − p) = 152 → 0.40p + 120 = 152 → 0.40p = 32 → p = 80. c = 70. (c) 80 pens, 70 pencils. Check: 80 + 70 = 150 and 80(1.20) + 70(0.80) = 96 + 56 = 152 ✓.
1.2 — Phone plans
(a) C_A = 30 + 0.20n; C_B = 15 + 0.50n. (b) 30 + 0.20n = 15 + 0.50n → 15 = 0.30n → n = 50 calls. (c) For 80: C_A = 30 + 16 = $46; C_B = 15 + 40 = $55. Plan A is cheaper by $9.
1.3 — Apples and oranges
a + o = 10 (count); 2a + 1.50o = 17 (cost). o = 10 − a. Sub: 2a + 1.50(10 − a) = 17 → 0.50a + 15 = 17 → 0.50a = 2 → a = 4 apples, o = 6 oranges.
1.4 — Two-rate trip
h + c = 3; 60h + 40c = 160. c = 3 − h. Sub: 60h + 40(3 − h) = 160 → 20h + 120 = 160 → 20h = 40 → h = 2 h, c = 1 h. Highway distance = 60(2) = 120 km; city distance = 40(1) = 40 km. Total 160 km ✓.
1.5 — Coin jar
t + f = 30; 20t + 50f = 960 (cents). t = 30 − f. Sub: 20(30 − f) + 50f = 960 → 600 + 30f = 960 → 30f = 360 → f = 12 fifty-cent coins, t = 18 twenty-cent coins. Check: 18(20) + 12(50) = 360 + 600 = 960 ✓.
2.1 — Explain (sample response)
(i) The result 14 = 14 is a true statement that does not depend on x or y — so the two equations are not really two different conditions, they are the same condition twice. The second equation 2x + 2y = 14 is a multiple of the first (it equals 2 × (x + y = 7)). (ii) Because both equations describe the same line, every (x, y) pair on that line satisfies both — there is no unique solution, there are infinitely many. (iii) Two valid pairs: (3, 4) gives 3 + 4 = 7 ✓, and (10, −3) gives 10 + (−3) = 7 ✓. Both also satisfy 2x + 2y = 14.
Marking: 1 for identifying that the equations are equivalent; 1 for "multiple of" explanation; 1 for explaining "no unique solution" / infinitely many solutions; 1 for two demonstrative pairs.