Mathematics • Year 8 • Unit 2 • Lesson 17

Cafés, Shops & Two-Unknown Stories

Build simultaneous equations from cafés, school dances, shopping, and parking — five word problems where you DEFINE variables, SET UP two equations, and verify a given solution.

Apply · Real-World Maths

1. Word problems

For each problem, (i) define the two variables clearly (including units), (ii) write the two equations, and (iii) do whatever the question asks (verify, classify, or read off the solution).

1.1 — Café order. 2 coffees and 3 muffins cost $13. 3 coffees and 1 muffin cost $11.

(a) Let c = price of a coffee, m = price of a muffin (both in dollars). Write the two equations.
(b) The barista claims the solution is c = $2.86 and m = $2.43 (to 2 dp). Verify in BOTH equations — does the claim work? 3 marks

Stuck? Replace c and m in each equation with the claimed numbers, compute LHS, compare to RHS. Allow tiny rounding error.

1.2 — School dance tickets. Mr Lee sold 100 tickets to the Year 8 dance. Students paid $5 each; adults paid $10 each. Total ticket sales were $620.

(a) Let s = number of student tickets, a = number of adult tickets. Write the two equations relating s and a.
(b) Check that (s = 76, a = 24) satisfies both equations. 3 marks

Stuck? Equation 1 counts tickets (s + a = 100). Equation 2 counts dollars (5s + 10a = 620).

1.3 — Graphical solution. A friend graphs the lines y = x + 1 and y = −x + 5 on the same axes. From the graph the lines cross at the point (2, 3).

(a) State the solution of the simultaneous equations.
(b) Verify (2, 3) by substituting into BOTH equations.
(c) Explain in one sentence why "the intersection point is the solution". 3 marks

Stuck? Every point on a line satisfies that line's equation. The intersection sits on BOTH lines, so it satisfies BOTH equations.

1.4 — Pencils and rulers. The school shop says: "4 pencils and 1 ruler cost $5. 8 pencils and 2 rulers cost $10."

(a) Let p = price of a pencil, r = price of a ruler. Write the two equations.
(b) Show that the second equation is just twice the first.
(c) State how many solutions this system has and what this means about being able to find p and r uniquely. 3 marks

Stuck? Two equations that are multiples of each other describe the same line — infinite solutions. You cannot pin down a unique price.

1.5 — Parking comparison. Lot A charges $4 entry plus $2 per hour. Lot B charges $7 entry plus $2 per hour. Let h = number of hours parked and C = cost in dollars.

(a) Write an equation C = ... for each lot.
(b) Without solving, state how many hours h would make the two lots cost the same — and explain in one sentence why. 2 marks

Stuck? Same gradient (both $2/hr) but different intercepts ($4 vs $7) → parallel lines → no solution. Lot A is ALWAYS $3 cheaper.

2. Explain your thinking

This question is about communication. Use full sentences. 4 marks

2.1 A classmate says: "If I write down ONE equation like x + y = 5, I can pick any pair I like — (1, 4), (2, 3), (0, 5). So I don't need a second equation. The first one already gives me the answer."

In your own words, explain: (i) why one equation alone is not enough to find unique values of x and y, (ii) what role the second equation plays, and (iii) what we mean by saying the solution is the INTERSECTION POINT of two lines. Use the phrase "satisfies both" somewhere in your answer.

Stuck? Revisit lesson § "What Are Simultaneous Equations?". One equation with two unknowns has INFINITELY many solutions — a whole line of them.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Café order

(a) Eq 1: 2c + 3m = 13. Eq 2: 3c + m = 11.
(b) Verify c = 2.86, m = 2.43. Eq 1: 2(2.86) + 3(2.43) = 5.72 + 7.29 = 13.01 ≈ 13 ✓ (tiny rounding). Eq 2: 3(2.86) + 2.43 = 8.58 + 2.43 = 11.01 ≈ 11 ✓. The barista's claim is correct to 2 dp.

1.2 — School dance tickets

(a) Eq 1: s + a = 100 (total tickets). Eq 2: 5s + 10a = 620 (total dollars).
(b) Check (76, 24): Eq 1: 76 + 24 = 100 ✓. Eq 2: 5(76) + 10(24) = 380 + 240 = 620 ✓. Both pass — solution confirmed.

1.3 — Graphical solution

(a) Solution: (2, 3) — i.e., x = 2 and y = 3.
(b) Eq 1: y = x + 1 → 3 = 2 + 1 ✓. Eq 2: y = −x + 5 → 3 = −2 + 5 = 3 ✓.
(c) Every point on each line satisfies that line's equation, so the intersection — the one point on BOTH lines — satisfies BOTH equations at once. That makes it the system's only solution.

1.4 — Pencils and rulers

(a) Eq 1: 4p + r = 5. Eq 2: 8p + 2r = 10.
(b) Multiply Eq 1 by 2: 8p + 2r = 10 — identical to Eq 2. They are the same line.
(c) Infinite solutions. Any (p, r) on the line 4p + r = 5 works. The shop hasn't given enough independent information to pin down unique prices — you'd need a different second equation (e.g., "1 pencil and 3 rulers cost $7").

1.5 — Parking comparison

(a) Lot A: C = 2h + 4. Lot B: C = 2h + 7.
(b) No value of h makes them equal. Both lines have the same gradient (2) but different y-intercepts, so they're parallel. Lot A is always exactly $3 cheaper, no matter how long you park.

2.1 — Explain your thinking (sample response)

The classmate is right that x + y = 5 has many solutions — pairs like (1, 4), (2, 3), (0, 5) ALL satisfy it. In fact, EVERY point on the line y = 5 − x satisfies it, so there are infinitely many solutions. To narrow that down to a single unique pair we need a SECOND equation involving the same two unknowns. The second equation also has infinitely many solutions on its own, but the ONLY pair (x, y) that satisfies BOTH equations is where the two lines cross — the intersection point. That intersection is the solution to the simultaneous equations: the unique pair that satisfies both at the same time. Two equations, two unknowns, one shared answer.

Marking: 1 mark for noting one equation has many / infinite solutions; 1 mark for stating the second equation narrows it to a unique pair; 1 mark for linking solution = intersection point of two lines; 1 mark for using the phrase "satisfies both" correctly.