Pair of Linear Equations in Two Variables
Why This Matters
So many everyday puzzles have two unknowns tangled together. How many ₹50 and ₹100 notes make up ₹2000 in 25 notes? If 5 pencils and 7 pens cost ₹50 but 7 pencils and 5 pens cost ₹46, what’s the price of each? You can’t pin down either unknown from a single fact — you need two conditions, and a way to untangle them.
That’s exactly what a pair of linear equations is: two facts about two unknowns, written as two straight-line equations. Solving the pair means finding the one pair of values (x, y) that makes both true at once.
This chapter gives you three tools and, just as importantly, the picture behind them. Each equation is a straight line; solving the pair is asking where the two lines meet. Sometimes they cross at one point (one answer), sometimes they run parallel (no answer), sometimes they’re the very same line (endless answers). Once you see that, you can predict the kind of answer before doing any algebra — and then use substitution or elimination to nail down the exact numbers.
The Big Idea
Each linear equation in x and y is a straight line. A solution of the pair is a point (x, y) that lies on both lines — so solving the pair means finding where the two lines meet. Two lines can meet at one point (unique solution), never meet (parallel → no solution), or lie on top of each other (coincident → infinitely many). The ratios of the coefficients tell you which case you’re in before you draw anything.
Let’s Break It Down
What a pair of linear equations is
A linear equation in two variables has the form ax + by + c = 0 (with a and b not both zero) — its graph is a straight line. A pair is just two of them:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
A solution is a pair of values (x, y) satisfying both equations. Take Akhila at a fair: if she had x rides (₹3 each) and played Hoopla y times (₹4 each), spending ₹20, with the games being half the rides, then y = x/2 and 3x + 4y = 20 — two equations, two unknowns.
The graphical method: where do the lines meet?
Plot both lines on the same axes. The point(s) they share are the solution(s). Here’s a worked case — the lines x + 3y = 6 and 2x − 3y = 12:
Because two straight lines can sit in only three ways, a pair of linear equations always falls into one of three cases:
| Lines | Solutions | Name | Ratio test |
|---|---|---|---|
| Intersecting | exactly one | consistent | a₁/a₂ ≠ b₁/b₂ |
| Parallel | none | inconsistent | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ |
| Coincident | infinitely many | dependent (consistent) | a₁/a₂ = b₁/b₂ = c₁/c₂ |
Why the ratio test works (this is worth understanding, not just memorising). Rearrange ax + by + c = 0 into slope form: y = (−a/b)x − c/b, so the slope is −a/b and the y-intercept is −c/b.
- If a₁/a₂ ≠ b₁/b₂, the two slopes −a₁/b₁ and −a₂/b₂ are different → the lines tilt differently → they must cross once (unique solution).
- If a₁/a₂ = b₁/b₂, the slopes are equal → the lines are parallel or the same line. The intercepts decide which: if c₁/c₂ is also equal, the equations are just multiples of each other → same line (coincident); if c₁/c₂ differs, the lines are distinct parallels that never meet (no solution).
Without drawing, decide how many solutions each pair has: (a) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0; (b) 2x − 3y = 8 and 4x − 6y = 9.
- (a) Compare: a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2. All three are equal.
- a₁/a₂ = b₁/b₂ = c₁/c₂ → coincident lines → infinitely many solutions.
- (b) Write as 2x − 3y − 8 = 0 and 4x − 6y − 9 = 0. Then a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −3/−6 = 1/2, c₁/c₂ = −8/−9 = 8/9.
- Here a₁/a₂ = b₁/b₂ (= 1/2) but ≠ c₁/c₂ (= 8/9) → parallel lines → no solution (inconsistent).
The graphical method is great for seeing the answer, but it’s awkward when the meeting point has messy coordinates like (49/29, 19/29). For exact answers we turn to algebra.
Method 1 — Substitution
The idea: from one equation, write one variable in terms of the other, then substitute into the second equation, leaving a single equation in one variable.
Solve 7x − 15y = 2 and x + 2y = 3.
- From the simpler equation x + 2y = 3, make x the subject: x = 3 − 2y.
- Substitute into 7x − 15y = 2: 7(3 − 2y) − 15y = 2 → 21 − 14y − 15y = 2 → 21 − 29y = 2.
- So −29y = −19, giving y = 19/29.
- Put y back into x = 3 − 2y: x = 3 − 2(19/29) = 3 − 38/29 = (87 − 38)/29 = 49/29. So x = 49/29, y = 19/29.
Substitution shines when one variable already has coefficient 1 (so making it the subject is clean).
Method 2 — Elimination
The idea: scale the equations so one variable has the same coefficient in both, then add or subtract to make it vanish.
The incomes of two people are in the ratio 9 : 7 and their expenditures in the ratio 4 : 3. Each saves ₹2000 a month. Find their incomes.
- Let incomes be 9x and 7x, expenditures 4y and 3y. Saving = income − expenditure = 2000, so: 9x − 4y = 2000 … (1) and 7x − 3y = 2000 … (2).
- Make the y-coefficients equal: (1) × 3 gives 27x − 12y = 6000; (2) × 4 gives 28x − 12y = 8000.
- Subtract the first from the second to eliminate y: (28x − 27x) = 8000 − 6000 → x = 2000.
- Put x = 2000 into (1): 9(2000) − 4y = 2000 → 18000 − 4y = 2000 → y = 4000. So incomes are 9x = ₹18,000 and 7x = ₹14,000.
Both algebraic methods also reveal the special cases. If, while solving, every variable cancels and you’re left with a true statement like 18 = 18, the pair has infinitely many solutions (same line). If you’re left with a false statement like 0 = 9 or −4 = 0, the pair has no solution (parallel).
Use elimination on 2x + 3y = 8 and 4x + 6y = 7.
- Make x-coefficients equal: multiply the first by 2 → 4x + 6y = 16; the second is 4x + 6y = 7.
- Subtract: (4x − 4x) + (6y − 6y) = 16 − 7, i.e. 0 = 9.
- 0 = 9 is false, so the lines are parallel and the pair has no solution (inconsistent). (Check: a₁/a₂ = b₁/b₂ = 1/2 but c₁/c₂ = 8/7 — agrees.)
Equations that become linear
Some problems aren’t linear as written but turn linear with a smart substitution. For example 2/x + 3/y = 13 and 5/x − 4/y = −2 become linear if you let u = 1/x and v = 1/y, giving 2u + 3v = 13 and 5u − 4v = −2 — solve for u, v, then flip back to x = 1/u, y = 1/v.
Common Mistakes
A pair of linear equations always has exactly one solution.
Most practice problems are built to have a single neat answer, so the 'no solution' and 'infinitely many' cases feel like exceptions you can ignore.
There are THREE outcomes: one solution (intersecting), none (parallel), or infinitely many (coincident). Always check the coefficient ratios — a pair can easily be inconsistent or dependent.
If both equations have no common solution, but the algebra gives 0 = 0, that means no solution.
Getting '0 = ...' with the variables gone looks like a dead end, so any such line feels like 'no solution'.
It depends on whether the leftover statement is TRUE or FALSE. A true statement (0 = 0, 18 = 18) → infinitely many solutions (same line). A false statement (0 = 9) → no solution (parallel). The truth value is the whole point.
For the ratio test, parallel lines satisfy a₁/a₂ = b₁/b₂ = c₁/c₂.
Parallel and coincident lines both start with a₁/a₂ = b₁/b₂, so it's easy to blur the last ratio.
Parallel (no solution) is a₁/a₂ = b₁/b₂ ≠ c₁/c₂. When c₁/c₂ ALSO matches, the equations are multiples of each other — the SAME line (coincident, infinitely many). The c-ratio is exactly what separates 'never meet' from 'always meet'.
In elimination, when the coefficients are equal you always subtract the equations.
The textbook examples often make like-signed coefficients equal, so subtracting becomes a habit.
Add when the equal coefficients have OPPOSITE signs (e.g. +6y and −6y → add to cancel); subtract when they have the SAME sign (+6y and +6y → subtract to cancel). Pick whichever makes the variable disappear.
Quick Check
Two lines are parallel and distinct. How many solutions does the pair of equations have?
For 3x + 2y = 5 and 2x − 3y = 7, which describes the pair?
While solving a pair by elimination, all variables cancel and you get 18 = 18. What does this mean?
Which substitution turns 2/x + 3/y = 13 into a linear equation?
Practice Problems
Easy
Solve by substitution: x + y = 14 and x − y = 4.
From the second equation, x = 4 + y.
Substitute into the first: (4 + y) + y = 14 → 4 + 2y = 14 → 2y = 10 → y = 5.
Then x = 4 + 5 = 9. So x = 9, y = 5. (Check: 9 + 5 = 14 ✓, 9 − 5 = 4 ✓.)
Without solving, state whether 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0 intersect, are parallel, or coincide.
a₁/a₂ = 5/7 and b₁/b₂ = −4/6 = −2/3.
Since 5/7 ≠ −2/3, the slopes differ, so the lines intersect at one point (a unique solution).
Medium
The coach buys 7 bats and 6 balls for ₹3800; later 3 bats and 5 balls for ₹1750. Find the cost of each bat and ball.
Let a bat cost ₹x and a ball ₹y. Then 7x + 6y = 3800 … (1) and 3x + 5y = 1750 … (2).
Eliminate x: (1) × 3 → 21x + 18y = 11400; (2) × 7 → 21x + 35y = 12250.
Subtract: (35y − 18y) = 12250 − 11400 → 17y = 850 → y = 50.
Put y = 50 into (2): 3x + 5(50) = 1750 → 3x = 1500 → x = 500.
So a bat costs ₹500 and a ball ₹50.
A two-digit number plus the number with its digits reversed is 66, and the two digits differ by 2. Find all such numbers.
Let the tens digit be x and units digit y. The number is 10x + y; reversed it’s 10y + x.
Sum: (10x + y) + (10y + x) = 66 → 11(x + y) = 66 → x + y = 6 … (1).
Digits differ by 2: either x − y = 2 or y − x = 2.
With x − y = 2 and (1): adding gives 2x = 8 → x = 4, y = 2 → the number 42.
With y − x = 2 and (1): adding gives 2y = 8 → y = 4, x = 2 → the number 24.
So there are two such numbers: 42 and 24 (and indeed 42 + 24 = 66 ✓).
Challenge
For what value of k does the pair kx + 3y = k − 3 and 12x + ky = k have no solution?
“No solution” means the lines are parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Here a₁/a₂ = k/12, b₁/b₂ = 3/k, c₁/c₂ = (k − 3)/k.
Set the first two equal: k/12 = 3/k → k² = 36 → k = 6 or k = −6.
Now we need a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so test each:
- k = 6: ratios a₁/a₂ = b₁/b₂ = 6/12 = 1/2, while c₁/c₂ = (6 − 3)/6 = 1/2. All three equal → coincident (infinitely many), not “no solution”. Reject.
- k = −6: a₁/a₂ = b₁/b₂ = −6/12 = −1/2, while c₁/c₂ = (−6 − 3)/(−6) = −9/−6 = 3/2. Here the first two match but ≠ c-ratio → parallel → no solution. ✓
So k = −6.
Summary
You should now be able to explain:
- A pair of linear equations is two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0; a solution is a point on both lines.
- Graphically there are three cases: intersecting (one solution, consistent), parallel (no solution, inconsistent), coincident (infinitely many, dependent/consistent).
- The ratio test: a₁/a₂ ≠ b₁/b₂ → unique; a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → none; a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many. (It works because −a/b is the slope and −c/b the intercept.)
- Substitution: make one variable the subject and plug into the other equation.
- Elimination: equalise one variable’s coefficient, then add/subtract to cancel it.
- During algebra, a leftover true statement → infinitely many solutions; a false one → no solution.
- Some non-linear pairs become linear with a substitution like u = 1/x, v = 1/y.
What’s Next
So far every equation has been degree 1 — straight lines. Next, in Quadratic Equations, the variable gets squared (ax² + bx + c = 0). One equation, one unknown, but now the graph is a parabola, so there can be two solutions, one, or none — and you’ll learn the famous quadratic formula to find them every time.