Mathematics • Year 8 • Unit 2 • Lesson 19

Elimination in Real Life — Tickets, Snacks & Numbers

Use elimination on five real-world problems where both equations come out in ax + by = c form: movie tickets, sport scores, mixtures, chocolate boxes and number puzzles. Set up, line up, and let one variable disappear.

Apply · Real-World Maths

1. Word problems

For each: (i) define variables (with units), (ii) write the two equations, (iii) solve by ELIMINATION (add, subtract, or multiply first), (iv) verify.

1.1 — Movie tickets. A cinema sold adult and child tickets. 2 adults + 3 children cost $72. 1 adult + 4 children cost $58.

(a) Let a = adult price, c = child price (in dollars). Write the two equations.
(b) Eliminate a: multiply the second equation by 2, then subtract from the first to find c. Then back-substitute for a. 3 marks

Stuck? Eq 1: 2a + 3c = 72. Eq 2: a + 4c = 58. Multiplying Eq 2 by 2 gives 2a + 8c = 116. Now you can subtract from Eq 1.

1.2 — Sport scores. A basketball team scored a total of 84 points from x successful 2-point shots and y successful 3-point shots. Altogether they made 36 shots.

(a) Write two equations relating x and y (one for shot count, one for points).
(b) Use elimination to find x and y. (Hint: multiply the count equation by 2 so the x-coefficients match.) 3 marks

Stuck? Count: x + y = 36. Points: 2x + 3y = 84. Multiply count by 2 → 2x + 2y = 72. Subtract.

1.3 — Chocolate boxes. A shop sells small and large chocolate boxes. 5 small + 2 large cost $40. 3 small + 2 large cost $32.

(a) Let s = small box price, l = large box price. Write the two equations.
(b) Notice the +2l terms — same sign! Subtract to eliminate l, find s, then back-substitute. 3 marks

Stuck? Subtraction is the cleanest move when one variable already has matching coefficients.

1.4 — Number puzzle. The sum of two numbers is 50 and their difference is 8.

(a) Let x and y be the numbers (with x > y). Write the two equations.
(b) Add the equations to eliminate y, find x, then back-substitute for y. 2 marks

Stuck? Sum: x + y = 50. Difference: x − y = 8. Add: 2x = 58.

1.5 — School fundraiser. A bake sale sold cookies and cupcakes. 4 cookies + 5 cupcakes cost $19. 3 cookies + 2 cupcakes cost $10.

(a) Let c = price of a cookie, k = price of a cupcake. Write the two equations.
(b) Neither variable has matching coefficients yet. Multiply Eq 1 by 3 and Eq 2 by 4 to make the cookie coefficients both 12, then subtract. Show every step. 3 marks

Stuck? After multiplying: 12c + 15k = 57 and 12c + 8k = 40. Subtract to eliminate c: 7k = 17.

2. Explain your thinking

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

2.1 A classmate is asked to solve the system 2x + 3y = 16 and 3x + 3y = 21. They say: "Both equations have +3y, so I should ADD them to cancel y."

In your own words: (i) why is the classmate's choice WRONG (what would adding give for the y-term)? (ii) state the correct operation and the SIGN RULE that tells you so, (iii) carry out the correct operation and solve the system, and (iv) verify your answer in both original equations. Use the phrase "same signs subtract" somewhere.

Stuck? Revisit lesson § "Elimination by Addition and Subtraction". Adding +3y and +3y gives +6y — nothing cancels.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Movie tickets

(a) Eq 1: 2a + 3c = 72. Eq 2: a + 4c = 58.
(b) Multiply Eq 2 by 2: 2a + 8c = 116. Subtract Eq 1: (2a + 8c) − (2a + 3c) = 116 − 72 → 5c = 44 → c = $8.80. Back-sub into Eq 2: a + 4(8.80) = 58 → a = 58 − 35.20 = $22.80. Check Eq 1: 2(22.80) + 3(8.80) = 45.60 + 26.40 = 72 ✓.

1.2 — Sport scores

(a) Count: x + y = 36. Points: 2x + 3y = 84.
(b) Multiply count by 2: 2x + 2y = 72. Subtract from points: (2x + 3y) − (2x + 2y) = 84 − 72 → y = 12 (three-pointers). Back-sub: x + 12 = 36 → x = 24 (two-pointers). Check: 2(24) + 3(12) = 48 + 36 = 84 ✓.

1.3 — Chocolate boxes

(a) Eq 1: 5s + 2l = 40. Eq 2: 3s + 2l = 32.
(b) Same +2l → subtract: 2s = 8 → s = $4. Back-sub into Eq 2: 3(4) + 2l = 32 → 2l = 20 → l = $10. Check Eq 1: 5(4) + 2(10) = 20 + 20 = 40 ✓.

1.4 — Number puzzle

(a) Eq 1: x + y = 50. Eq 2: x − y = 8.
(b) Add: 2x = 58 → x = 29. Back-sub: 29 + y = 50 → y = 21. Numbers: 29 and 21. Check: 29 − 21 = 8 ✓.

1.5 — School fundraiser

(a) Eq 1: 4c + 5k = 19. Eq 2: 3c + 2k = 10.
(b) Multiply Eq 1 by 3: 12c + 15k = 57. Multiply Eq 2 by 4: 12c + 8k = 40. Subtract: 7k = 17 → k = 17/7 ≈ $2.43. Back-sub into Eq 2: 3c + 2(17/7) = 10 → 3c = 10 − 34/7 = 36/7 → c = 12/7 ≈ $1.71. Check Eq 1: 4(12/7) + 5(17/7) = 48/7 + 85/7 = 133/7 = 19 ✓.

2.1 — Explain your thinking (sample response)

The classmate's choice is wrong because both equations have +3y — same signs — and adding them would give +3y + 3y = +6y, which is NOT zero, so y is not eliminated. The sign rule is "same signs subtract, opposite signs add". Same signs subtract — so we must SUBTRACT one equation from the other. Subtracting Eq 1 from Eq 2: (3x + 3y) − (2x + 3y) = 21 − 16 → x = 5. Back-sub into Eq 1: 2(5) + 3y = 16 → 3y = 6 → y = 2. Solution: (5, 2). Verify Eq 1: 2(5) + 3(2) = 16 ✓; Eq 2: 3(5) + 3(2) = 15 + 6 = 21 ✓.

Marking: 1 mark for noting +3y + 3y = +6y (so adding doesn't cancel); 1 mark for stating the correct sign rule ("same signs subtract"); 1 mark for the correct solution (5, 2); 1 mark for full verification in both equations.