Pairs of linear equations

Learning outcome

Solve simultaneous linear equations and distinguish systems with one, no or infinitely many ordered-pair solutions.

Concepts and assumptions

Unknowns are real unless otherwise restricted. Each equation has form ax + by = c, with a and b not both zero. Solutions satisfy both equations.

Substitution replaces a variable using an equivalent expression from one equation. Any common solution obeys that replacement, reducing the second equation to one variable.

Elimination multiplies equations by nonzero constants, then adds or subtracts them to remove a variable. Retaining an original equation makes this reversible: find one unknown, then recover the other.

Non-proportional coefficient pairs give one solution. If the entire second equation is a nonzero multiple of the first, both impose the same condition: infinitely many pairs work. If only the left sides have that relationship, but the constants do not, there is no solution.

Avoid coefficient-ratio tests that divide by zero; compare whole equations instead. Check both originals. For prices, assume positive unit prices, identical prices for identical items, and no additional charges.

Worked examples

Example 1 — Elimination. Solve 2x + 3y = 19 and 3x − 2y = 9.

Multiply the first by 2: 4x + 6y = 38. Multiply the second by 3: 9x − 6y = 27. Add: 13x = 65, so x = 5. Substitute: 10 + 3y = 19, giving y = 3. Thus (x, y) = (5, 3). Checks: 10 + 9 = 19 and 15 − 6 = 9.

Example 2 — Prices. One notebook and two pens cost ₹38; three notebooks and one pen cost ₹79. Find unit prices.

Let prices be n and p: n + 2p = 38; 3n + p = 79. First, n = 38 − 2p. Substitute: 3(38 − 2p) + p = 79. Thus 114 − 5p = 79, giving p = 7. Then n = 38 − 14 = 24. Notebook: ₹24; pen: ₹7. Checks: 24 + 14 = 38; 72 + 7 = 79.

Example 3 — Dependent or inconsistent. Start with x + 2y = 7.

Paired with 2x + 4y = 15, doubling the first gives the same left side equal to 14. Subtraction produces 0 = 1: no solution.

Paired with 2x + 4y = 14, the second merely doubles the first. All pairs (7 − 2t, t), for real t, satisfy both: infinitely many solutions.

Common mistakes

Solving equations independently; multiplying only some terms; checking only one equation; confusing identical conditions with contradictory conditions.

Practice questions

1. Solve x + y = 17 and x − y = 5.
2. Three notebooks and two pens cost ₹72; two notebooks and five pens cost ₹70. Find unit prices.
3. Classify 3x − 2y = 4 and 6x − 4y = 11.
4. Classify 2x − y = 4 and 6x − 3y = 12; describe all solutions.

Worked solutions

1. Add: 2x = 22, so x = 11. Then y = 17 − 11 = 6. Solution: (11, 6).
2. Equations: 3n + 2p = 72; 2n + 5p = 70. Multiplying by 5 and 2 gives 15n + 10p = 360; 4n + 10p = 140. Subtract: 11n = 220, so n = 20. Then 60 + 2p = 72 gives p = 6. Prices: ₹20 and ₹6.
3. Twice the first gives 6x − 4y = 8, contradicting 11. Therefore no solution.
4. The second is three times the first. Set x = t; then y = 2t − 4. All pairs (t, 2t − 4), t real, satisfy both: infinitely many solutions.