Pair of Linear Equations in Two Variables: Two Clues, One Answer
Two straight lines, drawn on the same graph, can only relate to each other in three ways — and each way tells you exactly how many solutions their equations share. This companion connects the graphs to the algebra, and gives you tools, flashcards and a CBSE-style quiz to test yourself.
xya₁x+b₁y+c₁=0a₂x+b₂y+c₂=0
🎡3.1 Two clues, two unknowns — the fair problem
Akhila goes to a village fair. She wants rides on the Giant Wheel and games of Hoopla. She played Hoopla half as many times as she rode the Giant Wheel. Each ride costs ₹3, each Hoopla game costs ₹4, and she spent ₹20 in total. How many rides did she take, and how many games did she play?
Letting x be the number of rides and y be the number of Hoopla games, this story translates into two equations:
y = (1/2)x ...(1)
3x + 4y = 20 ...(2)
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Like a detective with two clues. One equation alone leaves too many possibilities open — lots of (x, y) pairs satisfy y = x/2 by itself. But combine it with a second, independent clue (3x + 4y = 20), and suddenly the possibilities collapse down to a single answer. This chapter is all about how two linear equations, taken together, pin down (or fail to pin down) exact values for two unknowns.
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This isn't just a textbook trick. Any time you're juggling two unknown quantities and have two independent pieces of information about them — two prices, two rates, two ratios — you can almost always turn the situation into a pair of linear equations like this one, and solve for both unknowns at once.
📈3.2 Solving by drawing: the graphical method
Every linear equation in two variables draws a straight line when graphed. So a pair of linear equations draws two lines — and where those two lines stand relative to each other tells you everything about the solutions.
Three key terms. Consistent — the pair has at least one solution. Inconsistent — the pair has no solution at all. Dependent — the pair has infinitely many solutions (the lines are identical). A dependent pair is always automatically consistent too, since "infinitely many solutions" certainly includes "at least one".
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Think of train tracks. Intersecting lines are like two tracks crossing at a station — exactly one shared point, one unique solution. Parallel lines are like two separate tracks running side by side forever — they never share a point, no solution. Coincident lines are like the same track drawn twice, on top of itself — every point is shared, infinitely many solutions.
Intersecting — unique solution
Parallel — no solution
Coincident — infinitely many solutions
🔢Spotting the case from the coefficients
You don't actually have to draw anything to know which case you're in. Write both equations in the general form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, then compare the ratios of coefficients:
Pair of lines
a₁/a₂
b₁/b₂
c₁/c₂
Ratio comparison
Graph
Meaning
x − 2y = 0
3x + 4y − 20 = 0
1/3
−2/4
0/−20
a₁/a₂ ≠ b₁/b₂
Intersecting
Exactly one solution
2x + 3y − 9 = 0
4x + 6y − 18 = 0
2/4
3/6
−9/−18
a₁/a₂ = b₁/b₂ = c₁/c₂
Coincident
Infinitely many solutions
x + 2y − 4 = 0
2x + 4y − 12 = 0
1/2
2/4
−4/−12
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Parallel
No solution
The three rules. (i) If a₁/a₂ ≠ b₁/b₂ → lines intersect → unique solution (consistent).
(ii) If a₁/a₂ = b₁/b₂ = c₁/c₂ → lines coincide → infinitely many solutions (dependent, consistent).
(iii) If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → lines are parallel → no solution (inconsistent).
Worked example: x + 3y = 6 and 2x − 3y = 12
x
0
6
y = (6 − x)/3
2
0
x
0
3
y = (2x − 12)/3
−4
−2
Plotting both lines, they cross at the point (6, 0). So the pair of equations is consistent, with unique solution x = 6, y = 0.
Worked example: Champa's pants and skirts
Champa says: "The number of skirts is two less than twice the number of pants. Also, the number of skirts is four less than four times the number of pants." Letting x = pants and y = skirts:
y = 2x − 2
y = 4x − 4
These two lines intersect at (1, 0) — so Champa bought 1 pant and 0 skirts. It's a perfectly valid answer even though "zero skirts" might seem like an odd thing to solve for.
⚠️Where the graphical method struggles. Some solutions land on "ugly" coordinates — things like (√3, 2√7) or (49/29, 19/29). Try plotting those precisely by eye on graph paper and you'll quickly make mistakes. That's exactly the gap the algebraic methods in the next section are built to close.
Since graphs can't always be read precisely, we need exact algebraic techniques. There are two standard ones: substitution and elimination. Both always give exact answers — no eyeballing required.
🔁Substitution method
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Like swapping in a description for a name. If you know "x is just 3 minus twice y", you can swap that description in wherever x appears in the other equation — turning a two-variable puzzle into a one-variable one you already know how to solve.
STEP 3Substitute back: x = 3 − 2(19/29) = 49/29. Solution: x = 49/29, y = 19/29.
Real-world example: Aftab and his daughter's ages
"Seven years ago I was seven times as old as you. Three years from now I'll be three times as old as you." Letting s = Aftab's age and t = his daughter's age:
SET UPs − 7 = 7(t − 7) → s − 7t + 42 = 0. And s + 3 = 3(t + 3) → s − 3t = 6.
SUBSTITUTEFrom the second equation, s = 3t + 6. Plug into the first: (3t + 6) − 7t + 42 = 0 → 4t = 48 → t = 12.
BACK-SUBSTITUTEs = 3(12) + 6 = 42. So Aftab is 42 and his daughter is 12.
🔀Two special cases to recognise. All variables cancel and you get a TRUE statement (like 18 = 18) — this means the two equations were actually equivalent all along, so there are infinitely many solutions (e.g. 2x+3y=9 and 4x+6y=18, from a pencils-and-erasers problem). All variables cancel and you get a FALSE statement (like −4 = 0) — the equations flatly contradict each other, so there's no solution (e.g. the "two rail tracks" x+2y−4=0 and 2x+4y−12=0, which — sensibly — never cross).
➖Elimination method
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Like cancelling a shared ingredient. If two recipes both call for 12 grams of salt, you can subtract one from the other and the salt cancels out entirely, leaving a simpler comparison of just the remaining ingredients. Elimination scales two equations until one variable's coefficient matches exactly, then adds or subtracts to make that variable vanish.
Real-world example: two incomes in a 9:7 ratio
Two people's incomes are in ratio 9:7, expenditures in ratio 4:3, and each saves ₹2000/month. Writing incomes as 9x, 7x and expenditures as 4y, 3y:
SET UP9x − 4y = 2000 and 7x − 3y = 2000.
STEP 1Multiply the first equation by 3 and the second by 4, to match the y-coefficients: 27x − 12y = 6000 and 28x − 12y = 8000.
STEP 3–4Substitute back: 9(2000) − 4y = 2000 → y = 4000. Incomes are 9(2000)=₹18,000 and 7(2000)=₹14,000.
Worked example: 2x + 3y = 8 and 4x + 6y = 7 (no solution)
STEP 1Multiply the first equation by 2: 4x + 6y = 16.
STEP 2Subtract the second equation (4x + 6y = 7): 0 = 9 — a false statement.
CONCLUDEThe pair has no solution; the equations are inconsistent (the lines are parallel — notice the coefficients of x and y match exactly between the two equations, but the constants don't).
Real-world example: the two-digit number puzzle
A two-digit number plus its digit-reversal equals 66; the digits differ by 2. Letting x = tens digit, y = units digit:
SET UP(10x + y) + (10y + x) = 66 → 11(x + y) = 66 → x + y = 6. The digits differ by 2, so either x − y = 2 or y − x = 2 (we don't know in advance which digit is bigger).
CASE 1x + y = 6 and x − y = 2 → x = 4, y = 2 → the number 42.
CASE 2x + y = 6 and y − x = 2 → x = 2, y = 4 → the number 24.
CONCLUDEThere are two valid answers: 42 and 24 — both satisfy every condition in the problem.
💡Which method should you pick?. Substitution tends to be quickest when one equation already has a variable with coefficient 1 (or close to it) — less rearranging needed. Elimination shines when the equations look symmetric or share similar coefficients that are easy to match by multiplying. Either method always gets you to the same correct answer — it's purely a matter of convenience.
📋3.4 Summary
🎨 A pair of linear equations in two variables can be solved by the graphical method or an algebraic method.
📈 Graphically: the two lines can intersect (unique solution, consistent), be coincident (infinitely many solutions, dependent & consistent), or be parallel (no solution, inconsistent).
🧩 Algebraically, the two standard techniques are the substitution method and the elimination method.
🔢 For a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0: a₁/a₂≠b₁/b₂ → consistent (unique solution); a₁/a₂=b₁/b₂≠c₁/c₂ → inconsistent; a₁/a₂=b₁/b₂=c₁/c₂ → dependent & consistent (infinite solutions).
🌍 Many real situations that don't look linear at first (ages, ratios, digit puzzles) can be translated into a pair of linear equations and solved the same way.