Linear equations in two variables are a guaranteed topic in every CBSE and ICSE Class 9-10 exam. They’re predictable, the marks are defined, and the three methods — substitution, elimination, and graphical — each have their uses. Let’s sort all three out systematically.
What Is a Linear Equation in Two Variables?
A linear equation in two variables is an equation of the form:
ax + by + c = 0, where a and b are not both zero
Each such equation represents a straight line on a coordinate plane. When you have two such equations together (called a system or pair of linear equations), you’re finding the point(s) where both lines intersect.
Examples:
- 2x + 3y = 7
- x − y = 1
- 4x − 2y + 8 = 0
A solution to the system is a pair (x, y) that satisfies both equations simultaneously. That’s why they’re also called simultaneous equations.
Types of Solutions — The Consistency Conditions
Before you solve, you should understand what kind of solution to expect. Compare the coefficients of the two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
Unique solution (consistent): a₁/a₂ ≠ b₁/b₂
No solution (inconsistent): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Infinite solutions (dependent): a₁/a₂ = b₁/b₂ = c₁/c₂
Geometrically:
- Unique solution → Two lines intersect at one point
- No solution → Two lines are parallel (same slope, different intercept)
- Infinite solutions → Two lines are the same (coincident)
The “for what value of k” type question always uses these ratio conditions. If the question asks for a unique solution, set a₁/a₂ ≠ b₁/b₂. If it asks for no solution, set a₁/a₂ = b₁/b₂ ≠ c₁/c₂. This is a standard 3-mark question in CBSE.
Method 1: Substitution Method
When to use: When one equation already has a single variable isolated, or when one coefficient is 1 (making isolation easy).
The logic: Express one variable in terms of the other, substitute into the second equation, and solve. You’ve reduced two equations in two unknowns to one equation in one unknown.
Worked Example: Solve 2x + 3y = 7 and x − y = 1
Step 1: From the simpler equation, isolate one variable. x − y = 1 → x = y + 1
Step 2: Substitute into the first equation. 2(y + 1) + 3y = 7 2y + 2 + 3y = 7 5y = 5 y = 1
Step 3: Back-substitute to find x. x = y + 1 = 1 + 1 = 2
Step 4: Verify in both original equations. 2(2) + 3(1) = 7 ✓ and 2 − 1 = 1 ✓
Solution: x = 2, y = 1
Always do step 4. A small arithmetic error in step 2 will give you a wrong answer, and verification catches it. In board exams, you don’t lose marks for writing the check — you save them.
Method 2: Elimination Method
When to use: When coefficients of one variable are the same or easily made equal by multiplication. Often faster than substitution for “messy” equations.
The logic: Make the coefficient of one variable equal in both equations, then add or subtract to eliminate that variable. One equation, one unknown — easy solve.
Worked Example: Solve 3x + 2y = 12 and 4x − 3y = 10
Step 1: Choose a variable to eliminate. Let’s eliminate y. LCM of 2 and 3 is 6. Multiply equation 1 by 3: 9x + 6y = 36 Multiply equation 2 by 2: 8x − 6y = 20
Step 2: Add both equations (signs are opposite, so the y terms cancel). 17x = 56 x = 56/17… hmm, let me try eliminating x instead.
LCM of 3 and 4 is 12. Multiply equation 1 by 4: 12x + 8y = 48 Multiply equation 2 by 3: 12x − 9y = 30
Step 3: Subtract equation 2 from equation 1: (12x + 8y) − (12x − 9y) = 48 − 30 17y = 18 y = 18/17
Step 4: Substitute back: 3x + 2(18/17) = 12 → 3x = 12 − 36/17 = 168/17 → x = 56/17
Solution: x = 56/17, y = 18/17
The answer doesn’t always come out as neat integers. If you get fractions, check your work once but don’t panic — fractions are valid solutions.
Method 3: Graphical Method
When to use: When the question specifically asks for a graphical solution, or when you want a visual confirmation. For board exams, this method gets its own dedicated question type.
The logic: Plot each equation as a straight line on the coordinate plane. The point of intersection is the solution.
Steps:
- For each equation, find at least 2 points (usually x = 0 and y = 0 work well).
- Plot both lines on the same graph using a scale.
- Read off the intersection point.
Example: Solve x + y = 5 and x − y = 1 graphically.
For x + y = 5: When x = 0, y = 5. When y = 0, x = 5. Points: (0, 5) and (5, 0). For x − y = 1: When x = 0, y = −1. When y = 0, x = 1. Points: (0, −1) and (1, 0).
Plot both lines. They intersect at (3, 2).
Verify: 3 + 2 = 5 ✓ and 3 − 2 = 1 ✓
In CBSE board exams, the graphical method question (3-4 marks) expects you to draw neat axes with labelled scales, plot at least 2 points per line, and clearly mark the intersection. Use a ruler. Messy graphs lose presentation marks.
What the graph looks like for each case:
- Unique solution: Two lines crossing at one point
- No solution: Two parallel lines (no crossing)
- Infinite solutions: Both equations produce the same line (overlapping)
Cross-Multiplication Method (Bonus)
This is a faster algebraic method, useful when substitution and elimination are both slow. Given a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁)
This looks intimidating but follows a cyclic pattern. Write the coefficients in a specific arrangement and the numerators follow automatically.
Word Problems — How to Set Them Up
Word problems are where students drop marks — not because the algebra is hard, but because setting up the equations takes thought.
Standard types in CBSE Class 10:
Age problems: “A is twice as old as B. 5 years ago, the sum of their ages was 25.” → Let A = 2B, then (A − 5) + (B − 5) = 25.
Speed-distance-time: “A train covers distance d₁ at speed v₁ and d₂ at speed v₂. Total time = T.” → d₁/v₁ + d₂/v₂ = T.
Fraction problems: “When numerator increases by 1, fraction becomes 1/2. When denominator increases by 1, fraction becomes 1/3.” → Two equations in numerator n and denominator d.
Money problems: “2 apples and 3 oranges cost ₹70. 4 apples and 5 oranges cost ₹130.” → Straightforward setup, then solve.
When reading a word problem, identify the two unknowns first and assign variables. Write down each condition as a separate equation. Don’t try to do everything in your head — write it out.
Solved Examples: Easy to Hard
Easy: Solve by substitution — x + 2y = 6 and x − y = 0
From eq 2: x = y. Substitute into eq 1: y + 2y = 6 → 3y = 6 → y = 2, x = 2. Solution: (2, 2)
Medium: A says to B, “I am twice your age.” If the sum of their ages is 60, find their ages.
Let B’s age = y, A’s age = x. x = 2y and x + y = 60. Substitute: 2y + y = 60 → y = 20, x = 40. B is 20 years old, A is 40 years old.
Hard: A man rows 24 km downstream in 4 hours and 18 km upstream in 6 hours. Find the speed of the boat in still water and the current.
Let boat speed = x km/h, current speed = y km/h. Downstream speed = x + y = 24/4 = 6 Upstream speed = x − y = 18/6 = 3
Adding: 2x = 9 → x = 4.5 Subtracting: 2y = 3 → y = 1.5
Boat speed: 4.5 km/h, Current: 1.5 km/h
Real-World Examples
Example 1: Buying Parle-G and Dairy Milk at a Kirana Store
Ravi walks into his neighbourhood kirana store in Pune with ₹50. He picks up some Parle-G biscuit packets at ₹5 each and Dairy Milk bars at ₹10 each, spending exactly ₹50 on 7 items total. How many of each did he buy? We instantly get two equations: and , where and are the counts of each item. Solving by substitution gives , — four Parle-G and three Dairy Milk bars.
Connect to the syllabus: This is the classic “two unknowns, two conditions” setup that the substitution method is designed for — exactly the pattern CBSE Class 10 favours in 3-mark word problems.
Example 2: Mumbai Local — Fast Train vs. Slow Train Timing
Priya commutes from Thane to Churchgate. The fast train covers the route in minutes and the slow one in minutes. One Monday she takes the fast train and arrives 15 minutes early; on Tuesday the slow train makes her 10 minutes late. Setting up two linear equations in the unknowns — scheduled travel time and the actual journey times — lets us pinpoint both values exactly. The key insight is that each constraint (early/late by a fixed amount) gives one linear equation, and two such constraints uniquely determine two unknowns.
Connect to the syllabus: This is the “time and work / distance” variant of linear equations; ICSE 10 often phrases it as and , solved by the elimination method.
Example 3: Wagh Bakri Tea — Mixing Two Grades
A tea blender at Wagh Bakri mixes a premium Darjeeling grade (₹800/kg) with an everyday Assam grade (₹400/kg) to produce 10 kg of a blend priced at ₹560/kg. Let kg be Darjeeling and kg be Assam. The quantity condition gives , and the cost condition gives . Elimination knocks out one variable in one step, and we find kg and kg — a result any quality-control team would verify on the floor.
Connect to the syllabus: Mixture problems are a high-weightage application in CBSE Class 10 Chapter 3; the elimination method shines here because the coefficient of one variable () is already set to 1.
Common Mistakes
Mistake 1: Not substituting back correctly After finding y, students substitute into the wrong equation or make a sign error. Always substitute into the simpler of the two original equations.
Mistake 2: Forgetting to change sign when subtracting equations When using elimination and you subtract one equation from another, every term of the subtracted equation changes sign. Write it out explicitly — don’t do it in your head.
Mistake 3: Wrong consistency ratio check The ratios use a₁/a₂, b₁/b₂, c₁/c₂ from the form ax + by + c = 0. If your equations are in ax + by = c form, the c values have opposite signs. Rearrange to standard form before applying the ratios.
Mistake 4: Not verifying the solution A simple arithmetic error can give wrong x and y. A 10-second verification in both equations catches this.
Mistake 5: Reading the graph incorrectly If the intersection point is between gridlines, estimate carefully. A common error is reading (3, 2) as (2, 3) — swap x and y. Remember: first coordinate is horizontal (x), second is vertical (y).
Practice Questions
Q1. Solve by substitution: 3x − y = 7 and x + 2y = 6
From eq 1: y = 3x − 7. Substitute: x + 2(3x − 7) = 6 → x + 6x − 14 = 6 → 7x = 20 → x = 20/7. y = 3(20/7) − 7 = 60/7 − 49/7 = 11/7. x = 20/7, y = 11/7
Q2. Solve by elimination: 5x + 2y = 16 and 3x + 5y = 21
Multiply eq 1 by 5: 25x + 10y = 80. Multiply eq 2 by 2: 6x + 10y = 42. Subtract: 19x = 38 → x = 2. Then: 10 + 2y = 16 → y = 3. x = 2, y = 3
Q3. For what value of k does the system 2x + 3y = 4 and (k)x + 6y = 8 have infinitely many solutions?
For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂ 2/k = 3/6 = 4/8 → 2/k = 1/2 → k = 4
Q4. The sum of two numbers is 60 and their difference is 20. Find the numbers.
x + y = 60 and x − y = 20. Adding: 2x = 80 → x = 40. Then y = 20. The numbers are 40 and 20.
Q5. Solve graphically: x + y = 4 and 2x − y = 2.
For x + y = 4: Points (0, 4) and (4, 0). For 2x − y = 2: Points (0, −2) and (1, 0). Intersection: Adding equations: 3x = 6 → x = 2, y = 2. Plot confirms (2, 2).
Q6. 5 pencils and 7 pens cost ₹50. 7 pencils and 5 pens cost ₹46. Find the cost of each.
5p + 7q = 50 and 7p + 5q = 46. (p = pencil, q = pen) Multiply eq 1 by 7 and eq 2 by 5: 35p + 49q = 350 and 35p + 25q = 230. Subtract: 24q = 120 → q = 5. Then: 5p + 35 = 50 → p = 3. Pencil = ₹3, Pen = ₹5
Q7. Check whether the system 3x + 2y = 5 and 6x + 4y = 8 is consistent.
a₁/a₂ = 3/6 = 1/2. b₁/b₂ = 2/4 = 1/2. c₁/c₂ = 5/8. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system has no solution (inconsistent). The lines are parallel.
Q8. A fraction becomes 9/11 when 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Let fraction = x/y. (x + 2)/(y + 2) = 9/11 → 11x + 22 = 9y + 18 → 11x − 9y = −4 … (1) (x + 3)/(y + 3) = 5/6 → 6x + 18 = 5y + 15 → 6x − 5y = −3 … (2) From (2): multiply by 9: 54x − 45y = −27. From (1): multiply by 5: 55x − 45y = −20. Subtract: x = 7. Then: 77 − 9y = −4 → 9y = 81 → y = 9. The fraction is 7/9.
Frequently Asked Questions
Which method is best for CBSE boards? Substitution and elimination both score the same marks. Use substitution when one variable is easy to isolate. Use elimination when both equations have large coefficients. Graphical is only when specifically asked.
Can a pair of equations have exactly two solutions? No. A pair of linear equations can have either zero, one, or infinitely many solutions. Two solutions is not possible for linear (degree 1) equations.
Do I need to know cross-multiplication for CBSE Class 10? It’s in the NCERT syllabus, so yes. But it’s tested less frequently than substitution and elimination. Know it, but prioritise the other two methods.
How do I form equations from word problems quickly? Identify the unknowns (usually two, since it’s a pair). Read each condition and translate it into an equation. Practice 5-10 word problems of each type and you’ll recognise the patterns automatically.
What if I get fractions while solving — have I made an error? Not necessarily. Some problems have fractional solutions. If you get fractions, verify in both original equations. If both check out, the fraction is correct.
Is this topic in Class 9 or Class 10? Linear equations in one variable is Class 9. Pair of linear equations in two variables is the Class 10 chapter. This guide covers the Class 10 topic.