Mathematics • Year 10 • Unit 2 • Lesson 14
Elimination in the Real World
Use elimination on real word setups — admission tickets, two-week part-time work, mixing two solutions, and a movie download bundle. Then justify when elimination beats substitution.
1. Word problems
For each: define your variables, write two equations, choose which variable to eliminate and how, solve, and state the answer in context.
1.1 — Concert tickets. An adult ticket costs $a and a student ticket costs $s. A family buys 3 adult + 4 student tickets for $146. Another group buys 5 adult + 4 student tickets for $214.
(a) Write the two equations.
(b) Use elimination (the student-ticket coefficients already match) to find a and s. 4 marks
1.2 — Two-week pay. Marco earns $w per weekday hour and $W per weekend hour. In week 1 he works 10 weekday hours and 4 weekend hours for $300. In week 2 he works 8 weekday hours and 6 weekend hours for $300.
(a) Write the two equations.
(b) Eliminate either variable. Show your multiplier choice.
(c) Find w and W. 4 marks
1.3 — Mixing solutions. A 20% sugar solution is mixed with a 50% sugar solution to make 10 L of a 35% solution. Let x = litres of 20% solution and y = litres of 50% solution.
(a) Volume equation: x + y = 10. Sugar equation: 0.20x + 0.50y = 0.35(10).
(b) Multiply the volume equation by 0.20 (or 0.50), subtract, and solve. 4 marks
1.4 — Bundle download deal. A movie costs $m and an album costs $a. Bundle A (3 movies + 2 albums) costs $44. Bundle B (1 movie + 4 albums) costs $38.
(a) Write the two equations.
(b) Multiply Bundle B by 3 to match the movie coefficient, then subtract.
(c) Find m and a. 3 marks
1.5 — Boat in a river. A boat travels at b km/h in still water and the river flows at r km/h. Downstream (with the current) the boat covers 36 km in 2 hours. Upstream (against the current) it covers 24 km in 2 hours.
(a) Set up equations: b + r = 18 (downstream speed) and b − r = 12 (upstream speed).
(b) Solve by elimination (add, then subtract). 3 marks
2. Explain your thinking
Communication, not just numbers. 4 marks
2.1 A classmate is given 3x + 5y = 19 and 2x − 7y = −9 to solve. They choose substitution: rearrange equation 1 to x = (19 − 5y)/3, then substitute into equation 2 and end up with messy fractions. Using the words elimination, LCM, and multiply, explain (i) why elimination would have been the faster method here, (ii) what specific multipliers you would use to eliminate x or y, and (iii) which choice creates the smaller arithmetic.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Concert tickets
3a + 4s = 146; 5a + 4s = 214. Subtract: 2a = 68 → a = $34. Back: 3(34) + 4s = 146 → 4s = 44 → s = $11. Check: 5(34) + 4(11) = 170 + 44 = 214 ✓.
1.2 — Two-week pay
10w + 4W = 300; 8w + 6W = 300. Multiply Eq 1 by 3, Eq 2 by 2: 30w + 12W = 900 and 16w + 12W = 600. Subtract: 14w = 300 → w = $150/7 ≈ $21.43. Back: 10(150/7) + 4W = 300 → 4W = 2100/7 − 1500/7 = 600/7 → W = 150/7 ≈ $21.43. (Both equal — try-revealing answer: every-hour rate is the same; the system was set up to be consistent.)
1.3 — Mixing solutions
x + y = 10; 0.20x + 0.50y = 3.5. × volume eq by 0.20: 0.20x + 0.20y = 2. Subtract from sugar eq: 0.30y = 1.5 → y = 5 L, x = 5 L. Check: 0.20(5) + 0.50(5) = 1 + 2.5 = 3.5 ✓.
1.4 — Bundles
3m + 2a = 44; m + 4a = 38. × Eq 2 by 3: 3m + 12a = 114. Subtract Eq 1: 10a = 70 → a = $7. Back: m + 4(7) = 38 → m = $10. Check Eq 1: 3(10) + 2(7) = 30 + 14 = 44 ✓.
1.5 — Boat
b + r = 18; b − r = 12. Add: 2b = 30 → b = 15 km/h (still water speed). Subtract: 2r = 6 → r = 3 km/h (current speed).
2.1 — Explain (sample response)
(i) Elimination would have been faster because neither equation has a variable with coefficient 1, so substitution forces messy fractions like (19 − 5y)/3. Elimination only needs you to multiply each equation by a small integer to match coefficients. (ii) To eliminate x: LCM of 3 and 2 is 6. Multiply Eq 1 by 2 and Eq 2 by 3 → 6x + 10y = 38 and 6x − 21y = −27. Subtract → 31y = 65 → y = 65/31. To eliminate y: LCM of 5 and 7 is 35. Multiply Eq 1 by 7 and Eq 2 by 5 → 21x + 35y = 133 and 10x − 35y = −45. Add → 31x = 88 → x = 88/31. (iii) The y-elimination uses smaller multipliers (2 and 3) and gives cleaner whole-number subtraction; the x-elimination uses larger numbers (7 and 5). Either gives the same answer, but the y-elimination keeps the arithmetic smaller.
Marking: 1 for naming elimination as faster + reason; 1 for using LCM correctly; 1 for showing specific multipliers; 1 for identifying which is smaller arithmetic.