Lesson 3: Linear Equations in Two Variables

One equation in two unknowns has infinitely many solutions — x + y = 5 is satisfied by (1, 4), (2, 3), (0, 5), and endlessly more. But two equations together usually pin down exactly one pair. NCERT Class X studies pairs of linear equations in two variables, the geometric meaning of their solutions, and the algebraic methods: substitution, elimination, and cross-multiplication.

The three geometric possibilities

Each equation ax + by + c = 0 is a straight line. For a pair of lines there are exactly three outcomes:

Memorise the test as a table — board papers ask it directly, often with a parameter k hidden in the coefficients.

Worked example 1 (basic — substitution). Solve x + y = 5 and 2x + 3y = 13. From the first, x = 5 − y. Substitute: 2(5 − y) + 3y = 13, so 10 + y = 13, giving y = 3 and x = 2. Solution: (2, 3). Check: 2 + 3 = 5 ✓; 4 + 9 = 13 ✓.

Worked example 2 (intermediate — elimination). Solve 3x + 4y = 10 and 2x − 2y = 2. Multiply the second equation by 2: 4x − 4y = 4. Add to the first: 7x = 14, so x = 2. Then 6 + 4y = 10 gives y = 1. Solution: (2, 1). Elimination is fastest when a coefficient can be matched with one multiplication — always scan for that before reaching for heavier machinery.

Worked example 3 (NCERT exemplar style). For what value(s) of k does the pair kx + 3y = k − 3 and 12x + ky = k have *no* solution? No solution needs k/12 = 3/k ≠ (k − 3)/k. From k/12 = 3/k: k² = 36, k = ±6.

• k = 6: ratios are 1/2, 1/2, and (6−3)/6 = 1/2 — all equal, so the lines are *coincident* (infinitely many solutions). Reject.
• k = −6: ratios are −1/2, −1/2, and (−9)/(−6) = 3/2 — the first two equal, the third different, so the lines are parallel.

Answer: k = −6 only. The trap — reporting both ±6 — is precisely what this question tests.

Special forms: reduce first

Some pairs are not linear at first sight but become linear after substitution. For 2/x + 3/y = 13 and 5/x − 4/y = −2, set u = 1/x, v = 1/y and solve the linear pair in u and v. NCERT includes such problems; the skill being tested is recognising the hidden linear structure.

Competency application: forming the equations is the real skill

Most marks in applications are earned in the *translation*, not the solving.

Two-digit number. The sum of the digits of a two-digit number is 9, and the number formed by reversing the digits is 27 more than the original. Let the tens digit be x and units digit y. Then x + y = 9 and (10y + x) = (10x + y) + 27, which simplifies to y − x = 3. Adding the two equations: 2y = 12, y = 6, x = 3. The number is 36. Check: 6 + 3 = 9 ✓; 63 − 36 = 27 ✓.

Fixed charge plus rate. A taxi service charges a fixed ₹50 plus ₹14 per kilometre. The fare for x km is y = 50 + 14x — one equation of the pair; a second condition (say, two trips costing known totals) creates a solvable system. For a 12 km ride the fare is 50 + 168 = ₹218.

Common mistakes

• Reading the coincident case as "no solution". Coincident lines have *infinitely many* solutions; *parallel* lines have none. Keep the ratio table in front of you.
• Dropping the third ratio. k/12 = 3/k alone does not distinguish parallel from coincident — the c-ratio decides, as Worked example 3 shows.
• Arithmetic slips in elimination signs. When you multiply an equation by a negative, every term flips sign, including the constant.
• Stopping at x. Most application questions ask for the number, the ages, or the fare — both values, interpreted in words, with a check.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. The pair 2x + 3y = 7 and 4x + 6y = 10 has: (a) one solution (b) no solution (c) infinitely many solutions (d) two solutions. Answer: (b) no solution. Ratios: 2/4 = 3/6 = 1/2 ≠ 7/10, so the lines are parallel.

Q2. If x + y = 14 and x − y = 4, then x equals: (a) 5 (b) 7 (c) 9 (d) 10. Answer: (c) 9. Adding gives 2x = 18, so x = 9 (and y = 5).

Section B — Short answer (3 marks) Q3. Solve by elimination: 3x + 4y = 10 and 2x − 2y = 2. Marking scheme: 1 mark for multiplying the second equation by 2 (or an equivalent step) to get 4x − 4y = 4; 1 mark for adding to obtain 7x = 14, x = 2; 1 mark for back-substituting to get y = 1 and stating the solution (2, 1).

Section C — Case-based long answer (5 marks) Q4. A stationery shop sells pens and notebooks. On Monday, 3 pens and 4 notebooks cost ₹250. On Tuesday, 5 pens and 2 notebooks cost ₹230. (i) Form the pair of equations, taking the pen's price as ₹x and a notebook's as ₹y. (1 mark) (ii) Solve the pair and state both prices. (3 marks) (iii) What would 2 pens and 5 notebooks cost? (1 mark) Marking scheme: (i) 3x + 4y = 250 and 5x + 2y = 230 (1 mark). (ii) Doubling the second equation gives 10x + 4y = 460 (1 mark); subtracting the first equation leaves 7x = 210, so x = 30 (1 mark); substituting back, 90 + 4y = 250, so y = 40 (1 mark). Pen ₹30, notebook ₹40. (iii) 2(30) + 5(40) = 60 + 200 = ₹260 (1 mark).

Going deeper: the cross-multiplication formula

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 (constants on the left, signs included), elimination once and for all gives: x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁). Apply it to 2x + 3y − 13 = 0 and x + y − 5 = 0: x/(3×(−5) − 1×(−13)) = x/(−15 + 13) = x/(−2); the third ratio: 1/(2×1 − 1×3) = 1/(−1). So x = (−2)/(−1) = 2, and y/(c₁a₂ − c₂a₁) = y/((−13)(1) − (−5)(2)) = y/(−3) gives y = (−3)/(−1) = 3 ✓, matching Lesson 3's substitution answer. The formula is fast but sign-fragile: NCERT's convention needs the equations written with all terms on one side. If your signs are shaky, elimination is the safer route — CBSE awards the answer, not the method.

Going deeper: an ages problem, fully translated

The sum of the ages of a mother and daughter is 44 years. Four years ago, the mother was three times as old as the daughter. Find their present ages. Let the daughter be x now; the mother is 44 − x. Four years ago: daughter x − 4, mother 40 − x. Condition: 40 − x = 3(x − 4) = 3x − 12, so 52 = 4x and x = 13; the mother is 31. Check the past: 9 and 27 ✓ (27 = 3 × 9), and the present sum: 13 + 31 = 44 ✓. Note how "four years ago" subtracts from *both* ages — subtracting from only one is the standard error in this genre.

Going deeper: a system hiding in prices

2 chairs and 3 tables cost ₹4,600; 3 chairs and 5 tables cost ₹7,400. Find each price. Multiply the first equation by 3 and the second by 2: 6x + 9y = 13,800 and 6x + 10y = 14,800. Subtract: y = ₹1,000 (a table), then 2x = 4,600 − 3,000 = 1,600, x = ₹800 (a chair). Choosing multipliers to eliminate x here (LCM of 2 and 3) rather than y was arbitrary — either works; pick whichever pair of coefficients has the smaller LCM to keep numbers small.

One more model: age problems

A father is 30 and his son is 6, so the father is currently five times as old. After how many years will the father be exactly twice as old as the son? Let the time be x years: 30 + x = 2(6 + x), so 30 + x = 12 + 2x and x = 18. Check: father 48, son 24, and 48 = 2 × 24 ✓. Age problems are Lesson 3 in disguise — both ages shift by the *same* x, which is the single fact the equation encodes. The classic error is adding x to one age only; ages move together through time, and the algebra insists on it.

Final drills

Drill 1 — the NCERT pencils-and-pens classic. 5 pencils and 7 pens together cost ₹50; 7 pencils and 5 pens cost ₹46. Find each price. Let a pencil cost ₹x and a pen ₹y: 5x + 7y = 50 and 7x + 5y = 46. *Add* the equations: 12x + 12y = 96, so x + y = 8. *Subtract*: 2y − 2x = 4, so y − x = 2. Solving the tiny pair: y = 5, x = 3. Check: 15 + 35 = 50 ✓; 21 + 25 = 46 ✓. When coefficients are swapped between the two equations, add-and-subtract beats elimination — the symmetry is the hint.

Drill 2 — consistency by inspection. Do the lines 2x − y = 2 and 4x − 2y = 5 intersect? Ratios: 2/4 = (−1)/(−2) = 1/2, but 2/5 ≠ 1/2 — the lines are parallel, no solution. Note the c values must be compared with both equations written in the same form (constants moved to one side, e.g. 2x − y − 2 = 0 and 4x − 2y − 5 = 0, giving c₁/c₂ = 2/5); sign conventions in the ratio test are where this drill is lost, not in the idea.