Pair of Linear Equations in Two Variables
Two linear equations in the same two variables form a 'pair'. This chapter teaches how to solve such a pair by substitution and elimination, and how to tell — just from the coefficients — whether the pair has one solution, no solution, or infinitely many. Geometrically, each equation is a line, and the solution is where the two lines meet.
Learning objectives
- Represent a pair of linear equations graphically as two lines.
- Solve a pair by the substitution method.
- Solve a pair by the elimination method.
- Decide consistency from the ratios of coefficients.
- Translate word problems into a pair of linear equations.
Key concepts
Geometric meaning
Each equation a₁x + b₁y + c₁ = 0 is a straight line. The pair's solution is the point where the lines intersect. Lines may intersect at one point, be parallel, or coincide.
Conditions for consistency
Compare ratios: if a₁/a₂ ≠ b₁/b₂ there is a unique solution (intersecting lines). If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ there is no solution (parallel lines). If a₁/a₂ = b₁/b₂ = c₁/c₂ there are infinitely many solutions (coincident lines).
Substitution method
Express one variable from one equation and substitute it into the other, reducing the pair to a single equation in one variable.
Elimination method
Make the coefficient of one variable equal in both equations (by multiplying), then add or subtract to eliminate that variable.
Important formulas
Unique solution
a₁/a₂ ≠ b₁/b₂
Lines intersect at one point (consistent).
No solution
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Parallel lines (inconsistent).
Infinite solutions
a₁/a₂ = b₁/b₂ = c₁/c₂
Coincident lines (dependent, consistent).
Key definitions
- Consistent pair
- A pair of equations that has at least one solution.
- Inconsistent pair
- A pair that has no solution (parallel lines).
- Dependent pair
- A consistent pair with infinitely many solutions (the lines coincide).
Solved examples
Q1. Solve x + y = 14 and x − y = 4 by elimination.
Solution: Add the two equations: 2x = 18, so x = 9. Substitute into x + y = 14: y = 5. Solution: (9, 5).
Q2. Solve x + 2y = −1 and 2x − 3y = 12 by substitution.
Solution: From the first, x = −1 − 2y. Substitute: 2(−1 − 2y) − 3y = 12 ⇒ −2 − 4y − 3y = 12 ⇒ −7y = 14 ⇒ y = −2. Then x = −1 − 2(−2) = 3. Solution: (3, −2).
Q3. Check the consistency of 3x + 2y = 5 and 6x + 4y = 10.
Solution: Write as 3x + 2y − 5 = 0 and 6x + 4y − 10 = 0. a₁/a₂ = 3/6 = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = −5/−10 = 1/2. All equal, so there are infinitely many solutions — consistent and dependent.
Common mistakes to avoid
- Comparing c₁/c₂ without writing both equations in the form ax + by + c = 0 first.
- Sign errors while substituting a negative expression.
- Eliminating before making the chosen variable's coefficients equal.
- Calling a dependent pair (infinite solutions) inconsistent — it is actually consistent.
Pair of Linear Equations in Two Variables — MCQ Quiz
10 questions with instant feedback. Use number keys 1–4 to answer.
The pair x + 2y = 4 and 2x + 4y = 12 has:
Practice questions
Short answer
Is the pair 2x + 3y = 7 and 4x + 6y = 14 consistent?
Ratios: 2/4 = 3/6 = 7/14 = 1/2, so the lines coincide — consistent with infinitely many solutions.
Solve 2x = y and x + y = 6.
Substitute y = 2x: x + 2x = 6 ⇒ x = 2, y = 4.
Name the two algebraic methods of solving a pair.
Substitution and elimination.
Long answer
Solve 3x − 5y = −2 and 2x + y = 4 by elimination.
Multiply the second by 5: 10x + 5y = 20. Add to the first: 13x = 18 ⇒ x = 18/13. Hmm; instead multiply second by 5 and add: 3x − 5y = −2 and 10x + 5y = 20 give 13x = 18, x = 18/13, then y = 4 − 2(18/13) = (52 − 36)/13 = 16/13. Solution: (18/13, 16/13).
The sum of two numbers is 50 and their difference is 10. Form equations and find the numbers.
Let the numbers be x and y. x + y = 50 and x − y = 10. Adding: 2x = 60 ⇒ x = 30, y = 20.
HOTS (Higher Order Thinking)
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it goes 40 km upstream and 55 km downstream. (Set up only.) Why is a pair of linear equations useful here?
Let boat speed be x and stream speed y. Upstream speed = x − y, downstream = x + y. The two time conditions give two equations in 1/(x−y) and 1/(x+y), which form a solvable pair of linear equations in those reciprocals.
For what value of k will the pair x + 2y = 3 and 5x + ky + 7 = 0 have no solution?
Write 5x + ky + 7 = 0 and x + 2y − 3 = 0. No solution ⇒ 1/5 = 2/k ≠ −3/7 ⇒ k = 10 (and 2/10 ≠ −3/7 holds).
Quick revision
Revision notes
- Each equation = a line; solution = intersection point.
- Unique: a₁/a₂ ≠ b₁/b₂.
- None: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
- Infinite: a₁/a₂ = b₁/b₂ = c₁/c₂.
- Methods: substitution and elimination.
Key takeaways
- Always rewrite as ax + by + c = 0 before comparing ratios.
- Pick elimination when coefficients line up easily, substitution otherwise.
- Word problems: name the variables, write two equations, then solve.
Frequently asked questions
How do I know if a pair of equations has a solution?
Compare the ratios a₁/a₂, b₁/b₂ and c₁/c₂. If the first two are unequal there is a unique solution; equal first two but different third means no solution; all three equal means infinitely many.
Which method is better, substitution or elimination?
Use elimination when a variable's coefficients are easy to match, and substitution when one variable is already isolated or simple to isolate.
What does 'consistent' mean?
A consistent pair has at least one solution; an inconsistent pair has none.