Mathematics • Year 8 • Unit 2 • Lesson 18

Two Variables, Two Stories — Solve by Substitution

Use substitution to crack real-life problems: ages, ride costs, perimeters, mixtures, and money sums. Define your variables clearly, write your two equations, then use substitution to find a unique pair.

Apply · Real-World Maths

1. Word problems

For each problem: (i) define your two variables (with units), (ii) write the two equations, (iii) solve by SUBSTITUTION, (iv) check.

1.1 — Ages. Sam is 4 years older than Lin. The sum of their ages is 22 years.

(a) Let S = Sam's age, L = Lin's age (both in years). Write two equations.
(b) Use substitution to find S and L. Show every step.    3 marks

Stuck? "Sam is 4 older" → S = L + 4. "Sum is 22" → S + L = 22. The first equation already has S isolated.

1.2 — Theme park. A theme park charges $x for one ride and $y for one show. 3 rides + 2 shows cost $26. 2 rides + 1 show cost $15.

(a) Write the two equations (in x and y).
(b) Isolate y in the second equation, substitute into the first, and solve to find x and y.    3 marks

Stuck? From 2x + y = 15: y = 15 − 2x. Sub into 3x + 2y = 26.

1.3 — Rectangle. A rectangle has length L cm and width W cm. The length is 5 cm longer than the width, and the perimeter is 34 cm.

(a) Write two equations in L and W (recall: perimeter = 2L + 2W).
(b) Use substitution to find the length and width.    3 marks

Stuck? L = W + 5 is already isolated. Sub into 2L + 2W = 34.

1.4 — Coins. A piggy bank has only $1 and $2 coins. There are 12 coins in total, worth $19.

(a) Let a = number of $1 coins and b = number of $2 coins. Write two equations.
(b) Use substitution (isolate a from the first equation) to find a and b.    3 marks

Stuck? Eq 1: a + b = 12 (count). Eq 2: 1a + 2b = 19 (value). Isolate a: a = 12 − b.

1.5 — Two numbers. Twice the first number plus the second number is 17. The first number is 1 less than the second.

(a) Let x = first number, y = second number. Write the two equations.
(b) Solve by substitution.    2 marks

Stuck? "Twice x plus y is 17" → 2x + y = 17. "x is 1 less than y" → x = y − 1 (already isolated).

2. Explain your thinking

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

2.1 A classmate is asked to solve the system 3x + 2y = 19 and x + y = 8 by substitution. They start by trying to isolate x in the first equation, getting x = (19 − 2y) / 3, which gives them fractions and a long expansion.

In your own words: (i) why does isolating x in 3x + 2y = 19 lead to fractions? (ii) which variable in which equation should they isolate FIRST to keep the working clean, and why? (iii) using your suggestion, solve the system and verify your answer. Use the phrase "coefficient of 1 or −1" somewhere.

Stuck? Revisit lesson § "Spot the Trap". Scan both equations for any variable whose coefficient is 1 or −1 — that's your isolation target.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Ages

(a) Eq 1: S = L + 4. Eq 2: S + L = 22.
(b) Sub: (L + 4) + L = 22 → 2L + 4 = 22 → 2L = 18 → L = 9. Then S = 9 + 4 = 13. Sam = 13, Lin = 9. Check: 13 + 9 = 22 ✓.

1.2 — Theme park

(a) Eq 1: 3x + 2y = 26. Eq 2: 2x + y = 15.
(b) Isolate y in Eq 2: y = 15 − 2x. Sub into Eq 1: 3x + 2(15 − 2x) = 26 → 3x + 30 − 4x = 26 → −x = −4 → x = 4. y = 15 − 2(4) = 7. Ride = $4, show = $7. Check Eq 1: 3(4) + 2(7) = 12 + 14 = 26 ✓.

1.3 — Rectangle

(a) Eq 1: L = W + 5. Eq 2: 2L + 2W = 34.
(b) Sub: 2(W + 5) + 2W = 34 → 2W + 10 + 2W = 34 → 4W = 24 → W = 6. L = 6 + 5 = 11. Length = 11 cm, width = 6 cm. Check perimeter: 2(11) + 2(6) = 22 + 12 = 34 ✓.

1.4 — Coins

(a) Eq 1: a + b = 12 (count). Eq 2: a + 2b = 19 (value in dollars).
(b) Isolate a in Eq 1: a = 12 − b. Sub into Eq 2: (12 − b) + 2b = 19 → 12 + b = 19 → b = 7. a = 12 − 7 = 5. 5 × $1 coins and 7 × $2 coins. Check: 5 + 7 = 12 ✓ and 5(1) + 7(2) = 19 ✓.

1.5 — Two numbers

(a) Eq 1: 2x + y = 17. Eq 2: x = y − 1.
(b) Sub: 2(y − 1) + y = 17 → 2y − 2 + y = 17 → 3y = 19 → y = 19/3 ≈ 6.33. x = y − 1 = 16/3 ≈ 5.33. (x, y) = (16/3, 19/3). Check: 2(16/3) + 19/3 = 32/3 + 19/3 = 51/3 = 17 ✓; 16/3 = 19/3 − 1 = 16/3 ✓.

2.1 — Explain your thinking (sample response)

Isolating x in 3x + 2y = 19 gives x = (19 − 2y) / 3, which puts a denominator of 3 onto every later step — that's the source of the fractions and the messy expansion. To keep the working clean you should scan for a variable with a coefficient of 1 or −1. In this system, the second equation x + y = 8 has BOTH x and y with coefficient 1, so isolating either is fraction-free. The simplest move is x = 8 − y. Substituting into the first equation: 3(8 − y) + 2y = 19 → 24 − 3y + 2y = 19 → 24 − y = 19 → y = 5. Then x = 8 − 5 = 3. Verify: Eq 1: 3(3) + 2(5) = 9 + 10 = 19 ✓; Eq 2: 3 + 5 = 8 ✓. Solution: (3, 5).

Marking: 1 mark for explaining why a coefficient of 3 causes fractions; 1 mark for naming the variable/equation with coefficient 1 or −1; 1 mark for the correct solution (3, 5); 1 mark for showing verification in both original equations.