Class 10 Maths Notes

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Pair of Linear Equations Class 10 Maths Notes

A pair of linear equations in two variables is either two intersecting lines, two parallel lines, or one line counted twice, and the entire chapter is about deciding which picture you are looking at and then finding the common point. The algebraic test with a₁/a₂, b₁/b₂ and c₁/c₂ tells you the picture, and substitution and elimination give you the point.

Class:10Subject:MathematicsUnit:IICovers:CBSE 2024-25

7 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

What are the algebraic conditions for a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be consistent or inconsistent?

If a₁/a₂ ≠ b₁/b₂ the pair is inconsistent: the two lines are parallel and distinct, so there is no solution. If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ the pair is consistent with infinitely many solutions: the equations represent the same line. If a₁/a₂ = b₁/b₂ = c₁/c₂ the pair is consistent with a unique solution: the lines coincide. The third case can only arise when the equations are not really two different equations.

The Form of a Pair

Definition

Pair of linear equations in two variables

Two equations of the second degree in two variables, each of the form ax + by + c = 0, taken together. The word linear means that x and y appear only to the first power and never multiplied together, so x² , xy and y² are all excluded. The standard form is a₁x + b₁y + c₁ = 0 together with a₂x + b₂y + c₂ = 0.

The standard form. Every linear equation in two variables can be put into it.

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0

3x + 2y − 7 = 0 is already in standard form: a = 3, b = 2, c = −7.
2x − 3y = 8 is put into standard form by moving everything to the left: 2x − 3y − 8 = 0, so c = −8.
Notice that c carries the sign of whatever was moved across. Write the equation with everything on one side before you read off the ratios.
A pair is consistent when the two equations have at least one common solution, and inconsistent when they have none.

Two methods that are no longer in the course

The cross-multiplication method, where you equate a₁/a₂ = b₁/b₂ = c₁/c₂ directly without saying what it means, has been removed, and so have the problems whose equations are reducible to linear equations. Everything below works with the three cases stated, with the graphical method, and with substitution and elimination on word problems that give two genuine linear equations.

The Graphical Method

Each equation of the pair is the equation of a straight line, because it is of the form ax + by + c = 0. Plotting the first line and then the second turns the algebra question into a picture question, and the position of the two lines tells you immediately how many solutions there are. This is the method the chapter opens with, and it is worth doing on paper even when you intend to solve algebraically.

Solve each equation for y so that it reads y = mx + c, which is the form you can plot from two points.
For x + 2y = 7 we get y = −x/2 + 7/2, so it passes through (0, 7/2) and (2, 0).
For 2x + y = 5 we get y = −2x + 5, so it passes through (0, 5) and (5/2, 0).
The two lines meet at (1, 3), and the single common point is the unique solution of the pair.
The x-intercepts are where y = 0, and the y-intercepts are where x = 0; these give the two easiest points on each line.

What the picture tells you

Two lines crossing once means one solution. Two lines that never touch mean no solution. Two lines lying exactly on top of each other mean every point on that line is a solution, so there are infinitely many. Count the intersections and you have counted the solutions.

A pair that plots badly is still easy algebraically

Lines with steep slopes and awkward intercepts are miserable to draw accurately on squared paper, and a drawing one millimetre out can turn one solution into no solution. In the examination, use the graph for the three or four mark consistency question and use substitution or elimination for everything that asks for an exact answer.

The Three Consistency Cases

There are exactly three pictures a pair can make, and each has an algebraic signature. Work out the three ratios a₁/a₂, b₁/b₂ and c₁/c₂, compare them, and the picture is decided without drawing anything.

Unique solution, the lines cross at one point. Here a₁/a₂ ≠ b₁/b₂. Example: x + 2y = 7 and 2x + y = 5, for which a₁/a₂ = 1/2 and b₁/b₂ = 2.
No solution, the lines are parallel and distinct, so they never touch. Here a₁/a₂ = b₁/b₂ but c₁/c₂ is different. Example: 2x + 3y = 9 and 4x + 6y = 8, where a₁/a₂ = 1/2, b₁/b₂ = 1/2 and c₁/c₂ = 9/8.
Infinitely many solutions, the equations represent the same line drawn twice. Here a₁/a₂ = b₁/b₂ = c₁/c₂. Example: 2x + 3y = 9 and 4x + 6y = 18, where all three ratios equal 1/2.

The most inverted item in the whole unit

If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ then there is NO solution, because the slopes are equal so the lines are parallel, and the intercepts differ so the lines are distinct. If all three ratios are equal then the equations are the same equation written differently, so there are INFINITELY MANY solutions. Students routinely swap these two. Copy the table into your notebook in the correct order and read from it, never from memory.

Check the ratios only as far as they need to be checked

You do not always need all three. If a₁/a₂ and b₁/b₂ are already unequal, the answer is a unique solution and c₁/c₂ is irrelevant. Only when the first two agree do you have to look at the third.

Algebraic Conditions for the Number of Solutions

Inconsistent? No — consistent, with exactly one solution. The lines intersect.

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \Rightarrow unique solution

Inconsistent pair: the lines are parallel and distinct.

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}} \Rightarrow no solution

Consistent pair whose lines coincide.

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} \Rightarrow infinitely many solutions

Write both equations in the standard form ax + by + c = 0 before taking any ratio, so the signs of a, b and c are fixed and comparable.
Never compare b₁/b₂ with b₂/b₁. The three ratios all have their terms in the same order: numerator 1 over numerator 2.
Once the ratios have told you how many solutions there are, switch to substitution or elimination to find the actual point. The ratios alone never give the value of x or y.
If a₂ or b₂ is zero, the ratio that needs it is undefined, so state that fact and use the slope comparison instead: equal slopes mean parallel, different slopes mean intersecting.

The usual five-mark shape

The common form of the long question is: given the pair, find whether it is consistent or inconsistent, and if it is consistent, find the solution. That is two marks for the verdict and three for the solution, which is why the verdict is never left until the end.

Solution by Substitution

The idea is to use one equation to remove a variable, leaving a single equation in a single unknown, which you then solve and substitute back. Pick the equation that gives the cleanest expression for one variable — the one with the coefficient ±1 is always the best choice — and write every step on its own line.

Solve x − y = 1 for y: y = x − 1.
Substitute that into the second equation: 2x + (x − 1) = 8.
Simplify: 3x − 1 = 8, so 3x = 9 and x = 3.
Substitute back into y = x − 1: y = 3 − 1 = 2.
The solution is (x, y) = (3, 2). Check: 3 − 2 = 1 and 2 × 3 + 2 = 8. Both equations hold.

The pair x − y = 1 and 2x + y = 8 solved by substitution, with the check.

x - y = 1 \Rightarrow y = x - 1; 2 x + y = 8 \Rightarrow 2 x + (x - 1) = 8 \Rightarrow x = 3, y = 2

The two usual slips

Forgetting to substitute back for the other variable, and so reporting only x, and losing the brackets when you substitute, as in writing 2x + x − 1 = 8 instead of 2x + (x − 1) = 8. Both are avoidable by writing the substitution step as its own line before simplifying.

Solution by Elimination

The idea is the mirror image: multiply one or both equations by a number chosen so that the coefficients of the same variable become equal, then add or subtract the two equations so that variable cancels and one unknown disappears. Adding is for equal coefficients with opposite signs waiting to cancel, subtracting is for equal coefficients with the same sign.

Start from x + 2y = 7 and 2x + y = 5.
Multiply the first equation by 2: 2x + 4y = 14.
Write the second equation as it stands: 2x + y = 5.
Subtract the second from the first: 3y = 14 − 5 = 9, so y = 3.
Substitute into the first equation: x + 6 = 7, so x = 1.
The solution is (x, y) = (1, 3). Check: 1 + 6 = 7 and 2 + 3 = 5.

The pair solved by elimination: equalise the x coefficients, then subtract.

x + 2 y = 72 x + y = 5} \times 2 on the first \Rightarrow 2 x + 4 y = 14; 14 - 5 \Rightarrow 3 y = 9 \Rightarrow y = 3, x = 1

Equalising a coefficient in one step

If the x coefficients are 1 and 2, multiply the equation with coefficient 1. If they are 2 and 3, multiply the first by 3 and the second by 2. If they are 3 and 5, multiply the first by 5 and the second by 3. Multiply by the other equation's coefficient and the coefficients always come out equal.

The Same Pair Solved Two Ways

Both methods must reach the same answer, and checking that they do is the best possible revision. Take the pair x + 2y = 7 and 2x + y = 5, whose solution is (1, 3) and whose ratios a₁/a₂ = 1/2 against b₁/b₂ = 2 already announce a unique solution.

By substitution: from the first equation x = 7 − 2y. Put this into 2x + y = 5 to get 2(7 − 2y) + y = 5, then 14 − 4y + y = 5, then 14 − 3y = 5, then 3y = 9, then y = 3, and finally x = 7 − 6 = 1.
By elimination: multiply the first equation by 2 to get 2x + 4y = 14. Subtract the second equation, 2x + y = 5, giving 3y = 14 − 5 = 9, so y = 3, and then x + 6 = 7 gives x = 1.
Both give x = 1 and y = 3, and substituting back satisfies both equations: 1 + 2 × 3 = 7 and 2 × 1 + 3 = 5.
Substitution is quicker when one coefficient is already ±1. Elimination is safer when there are no unit coefficients, because it removes the variable without needing the clean expression first.

Which one to write

The paper accepts either, and the wording of the question decides. If it says by substitution, you must show the expression for one variable taken from one equation and then put into the other. If it says by elimination, you must show a multiplication step and then an addition or subtraction step. If it is silent, either is fine, so write the one you can do without hesitating.

How the Questions Are Asked

This is the second section of Unit II Algebra and it is short, so the questions are short too. Nearly all of them fall into four shapes: solve by substitution, solve by elimination, decide the nature of the solution from the ratios, or turn a simple word problem into two linear equations.

Solve the pair x + 2y = 7 and 2x + y = 5 by substitution. (2 marks, Section B.)
Solve the pair 3x + 2y = 16 and x − y = 2 by elimination. (3 marks, Section C.)
Find whether the pair 2x + 3y = 9 and 4x + 6y = 18 is consistent or inconsistent, without solving. (2 marks.)
Show that the pair x + y = 5 and 2x + 2y = 10 has infinitely many solutions. (3 marks.)
A shopkeeper sells an article at ₹40 more than the cost price. The selling price of 12 articles equals the cost price of 15 articles. Find the cost price. (5 marks.)

How to set up a word problem

Take the cost price as x and the selling price as x + 40. The condition gives 12(x + 40) = 15x, so 12x + 480 = 15x, so 3x = 480 and x = 160. The cost price is ₹160 and the selling price ₹200, and the check is 12 × 200 = 2400 = 15 × 160. Introduce one variable per unknown, translate each condition into an equation, solve, and then state the answer in words with its unit.

Paper reminders

The Class 10 paper is 80 marks in 3 hours with 38 questions. Q1-18 are one-mark MCQs and Q19-20 are one-mark assertion-reason, Q21-25 carry 2 marks, Q26-31 carry 3 marks, Q32-35 carry 5 marks and Q36-38 are 4-mark case studies, with internal choice in two questions each of Sections B, C and D. Unit II Algebra carries 20 of the 80 marks, and the other 20 marks are internal assessment, 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Standard form of a pair

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0

Put every equation into this form before taking any ratio, so the signs are comparable.

Unique solution

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

Consistent with exactly one solution — the two lines intersect at a single point.

No solution

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

Inconsistent: parallel and distinct lines that never meet. This is the case students swap.

Infinitely many solutions

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

Consistent: the two equations are the same line written in two different ways.

Substitution step

y = x - 1; 2 x + (x - 1) = 8 \Rightarrow 3 x = 9 \Rightarrow x = 3, y = 2

Use one equation to express one variable, then put it into the other.

Elimination step

2 x + 4 y = 14 - (2 x + y = 5) \Rightarrow 3 y = 9 \Rightarrow y = 3, x = 1

Equalise one pair of coefficients, then add or subtract so that variable cancels.

Equalising coefficients

a_{1}, a_{2} \Rightarrow multiply the first by a_{2} and the second by a_{1}

Multiplying each equation by the other equation's coefficient always makes them equal.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

If a₁/a₂ ≠ b₁/b₂ the pair is consistent with a unique solution, because the two lines intersect once.
If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ the pair is inconsistent: the lines are parallel and distinct, so there is NO solution.
If a₁/a₂ = b₁/b₂ = c₁/c₂ the equations are the same line, so there are INFINITELY MANY solutions. This swap is the single most inverted item in the chapter.
The ratios never give the value of x or y. Use them for the verdict, then switch to substitution or elimination for the actual solution.
Put every equation into ax + by + c = 0 before taking ratios, and never compare b₁/b₂ with b₂/b₁ — the terms stay in the same order.
Elimination rule: multiply the equation with coefficient 1 by the coefficient of the equation with coefficient 2, and the coefficients come out equal.
Do not use the cross-multiplication method and do not attempt problems whose equations are reducible to linear equations. Neither is in the 2024-25 course.
In a word problem, take one unknown per quantity, translate each condition into an equation, solve the pair, and then restate the answer with its unit.
Unit II Algebra carries 20 of the 80 written marks; the remaining 20 marks are internal assessment, 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical.

FAQ

Frequently asked questions

How do you decide whether a pair of linear equations is consistent or inconsistent?

Compare the three ratios a₁/a₂, b₁/b₂ and c₁/c₂. If a₁/a₂ is not equal to b₁/b₂, the two lines intersect once and the pair is consistent with a unique solution. If a₁/a₂ = b₁/b₂ but c₁/c₂ is different, the lines are parallel and distinct, so the pair is inconsistent and has no solution. If all three ratios are equal, the equations are the same line and there are infinitely many solutions.

What is the difference between the substitution and elimination methods?

Both reduce two equations in two unknowns to one equation in one unknown. Substitution uses one equation to express one variable in terms of the other, then puts that expression into the second equation. Elimination multiplies one or both equations so that the coefficients of the same variable become equal, then adds or subtracts so that variable cancels. Substitution is quickest when a coefficient is already ±1; elimination is more reliable when none is.

What does it mean geometrically when a pair has infinitely many solutions?

It means the two equations represent the same straight line. A pair such as 2x + 3y = 9 and 4x + 6y = 18 has a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2, so the second equation is exactly twice the first and the two lines lie on top of each other. Every point on that line satisfies both equations, which is why there are infinitely many solutions.

How do you know a pair has no solution without drawing the graph?

Take the ratios. If a₁/a₂ = b₁/b₂ but c₁/c₂ is different, the lines have the same slope and different intercepts, so they are parallel and distinct and never meet. For example, 2x + 3y = 9 and 4x + 6y = 8 give 1/2, 1/2 and 9/8, so the pair is inconsistent and has no solution.

How is a simple word problem turned into a pair of linear equations?

Take one unknown for each quantity, for instance the cost price x and the selling price x + 40. Translate each stated condition into an equation. A shopkeeper whose selling price is ₹40 above the cost price and whose selling price of 12 articles equals the cost price of 15 articles gives 12(x + 40) = 15x, so 3x = 480 and x = 160. The cost price is ₹160 and the check is 12 × 200 = 15 × 160.

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Class 10 Maths Notes Polynomials Class 10 Maths Notes Quadratic Equations Class 10 Maths Notes Arithmetic Progressions Class 10 Maths Notes All Class 11 & 12 Notes

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