When should I use substitution instead of elimination?

Use substitution when one variable can be isolated quickly. Use elimination when coefficients can be matched easily by multiplying the equations.

How do I know whether a pair has no solution?

If a1/a2 = b1/b2 but c1/c2 is different, the two lines are parallel and the system has no solution.

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Home /Notes /Class 10 /Maths /Pair of Linear Equations in Two Variables

Chapter 3 | Class 10 | CBSE / NCERT

Pair of Linear Equations
in Two Variables

A carefully corrected Class 10 chapter guide covering graphical interpretation, algebraic methods, consistency conditions, reducible forms, and exam-style word problems.

Chapter Roadmap

01Linear Equation in Two Variables 02Pair of Linear Equations 03Graphical Meaning and Three Cases 04Graphical Method of Solving 05Substitution Method 06Elimination Method 07Cross-Multiplication Method 08Equations Reducible to Linear Form 09Consistency Conditions and Finding k 10Word Problems: Numbers and Digits 11Word Problems: Ages and Fractions 12Word Problems: Cost, Distance and Speed

Linear Equation in Two Variables

One equation, two variables, and infinitely many points on a straight line

Not Linear in Two Variables

$x^2+y=5$ because the power of $x$ is 2
$xy=12$ because the variables are multiplied
$\frac{1}{x}+y=4$ because the variable appears in the denominator

A linear equation in two variables is an equation in

x

and

y

where each variable has power 1.

Every solution is an ordered pair

(x,y)

, and all such solutions lie on a straight line in the coordinate plane.

ax+by+c=0

Here a and b are not both zero.

Practice

Write any two solutions of

2x+y=8

.

Example answers:

(0,8)

and

(4,0)

Pair of Linear Equations in Two Variables

A common solution must satisfy both equations together

A pair of linear equations in two variables is written as two equations in the same variables

x

and

y

A solution of the pair is a point that satisfies both equations simultaneously.

a_1x+b_1y+c_1=0

a_2x+b_2y+c_2=0

The ordered pair

(x,y)

must satisfy both equations.

Worked Example

Verify a solution

Check whether

(2,1)

is a solution of

3x-2y=4

and

2x+y=5

Show solution

For the first equation: $3(2)-2(1)=6-2=4$ , so it is satisfied.

For the second equation: $2(2)+1=5$ , so it is also satisfied.

Therefore $(2,1)$ is a solution of the pair.

Practice

(3,2)

a solution of

2x+y=8

and

x-y=1

?

Answer: Yes, because

2(3)+2=8

and

3-2=1

Graphical Meaning and the Three Cases

Two lines can intersect, stay parallel, or coincide

Consistency

A system is consistent if it has at least one solution. It is inconsistent if it has no solution.

Each equation represents a straight line. So a pair of linear equations represents two lines drawn on the same coordinate plane.

The nature of the solution depends on how these two lines are placed.

\frac{a_1}{a_2}\ne\frac{b_1}{b_2}

Intersecting lines, so there is one unique solution.

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}

Parallel lines, so there is no solution.

\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

Coincident lines, so there are infinitely many solutions.

Intersecting Lines

One point of intersection means one unique solution.

Parallel Lines

Same slope but different intercepts means no solution.

Coincident Lines

The same line repeated means infinitely many solutions.

Graphical Method of Solving

The intersection point of the two lines gives the solution

To solve graphically, find two convenient points on each line, plot them, and draw both lines on the same graph.

If the lines intersect, the coordinates of the intersection point are the solution.

x+y=3 \Rightarrow y=3-x

3x-2y=4 \Rightarrow y=\frac{3x-4}{2}

Graph of x + y = 3 and 3x - 2y = 4

The two lines meet at

(2,1)

, so that ordered pair is the solution.

Worked Example

Solve graphically

Solve

x+y=3

and

3x-2y=4

Show solution

For $x+y=3$ , points like $(1,2)$ and $(2,1)$ lie on the line.

For $3x-2y=4$ , points like $(0,-2)$ and $(2,1)$ lie on the line.

Both lines meet at $(2,1)$ . Hence the solution is $x=2$ and $y=1$ .

Practice

For

2x-y=1

and

x+2y=8

, the graph meets at

(2,3)

. Verify by substitution.

Check:

2(2)-3=1

and

2+2(3)=8

Substitution Method

Express one variable from one equation and substitute into the other

This method is best when one equation can be rewritten easily as

x=...

y=...

After substitution, the pair reduces to one equation in one variable.

x-y=-1 \Rightarrow y=x+1

Worked Example

Solve by substitution

Solve

3x-5y=-1

and

x-y=-1

Show solution

From $x-y=-1$ , we get $y=x+1$ .

Substitute into the first equation: $3x-5(x+1)=-1$ .

$3x-5x-5=-1 \Rightarrow -2x=4 \Rightarrow x=-2$ .

Then $y=x+1=-1$ .

So the solution is $(-2,-1)$ .

Practice

Solve

x+2y=-1

and

2x-3y=12

.

Answer:

x=3

and

y=-2

Elimination Method

Make one coefficient equal and eliminate that variable

In elimination, we multiply the equations so that the coefficient of one variable becomes the same in both equations.

Then we add or subtract the equations to remove that variable completely.

Worked Example

Solve by elimination

Solve

3x+2y=11

and

2x+3y=4

Show solution

Multiply the first equation by 3: $9x+6y=33$ .

Multiply the second equation by 2: $4x+6y=8$ .

Subtract: $5x=25$ , so $x=5$ .

Substitute into $3x+2y=11$ : $15+2y=11$ , so $2y=-4$ and $y=-2$ .

Hence the solution is $(5,-2)$ .

Practice

Solve

2x-3y=13

and

3x-2y=12

.

Answer:

x=2

and

y=-3

Cross-Multiplication Method

A direct formula method when the equations are in standard form

This method works smoothly when the equations are written in standard form with zero on the right side.

It gives direct expressions for both

x

and

y

\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

Equivalent board-friendly form of the cross-multiplication relation.

x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\quad y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}

Worked Example

Solve by cross-multiplication

Solve

2x+3y=17

and

3x-2y=6

Show solution

Write them as $2x+3y-17=0$ and $3x-2y-6=0$ .

So $a_1=2$ , $b_1=3$ , $c_1=-17$ and $a_2=3$ , $b_2=-2$ , $c_2=-6$ .

$a_1b_2-a_2b_1=2(-2)-3(3)=-13$ .

$b_1c_2-b_2c_1=3(-6)-(-2)(-17)=-52$ , so $x=\frac{-52}{-13}=4$ .

$c_1a_2-c_2a_1=(-17)(3)-(-6)(2)=-39$ , so $y=\frac{-39}{-13}=3$ .

Hence the solution is $(4,3)$ .

Practice

Solve

x+2y+1=0

and

2x-3y-12=0

.

Answer:

x=3

and

y=-2

Equations Reducible to Linear Form

Use substitution like $u=1/x$ to turn the system into a linear one

Some equations do not look linear at first, but become linear after a suitable substitution.

The most common pattern is to let

u=\frac{1}{x}

and

v=\frac{1}{y}

u=\frac{1}{x},\quad v=\frac{1}{y}

Worked Example

Reduce and solve

Solve

\frac{2}{x}+\frac{3}{y}=13

and

\frac{5}{x}-\frac{4}{y}=-2

Show solution

Let $u=\frac{1}{x}$ and $v=\frac{1}{y}$ .

Then the equations become $2u+3v=13$ and $5u-4v=-2$ .

Solving this pair gives $u=2$ and $v=3$ .

So $x=\frac{1}{u}=\frac{1}{2}$ and $y=\frac{1}{v}=\frac{1}{3}$ .

Practice

Let

u=\frac{1}{x+y}

and

v=\frac{1}{x-y}

when a question contains those two expressions repeatedly. Solve for

u,v

first, then come back to

x,y

Consistency Conditions and Finding k

Use coefficient ratios to predict the nature of the system quickly

Many board questions ask for the value of

k

that makes a pair have a unique solution, no solution, or infinitely many solutions.

These questions are solved by comparing the ratios of coefficients carefully.

\text{Unique solution} \Rightarrow \frac{a_1}{a_2}\ne\frac{b_1}{b_2}

\text{No solution} \Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}

\text{Infinitely many solutions} \Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

Worked Example

Find k for a unique solution

For what value of

k

does the pair

kx+2y=5

and

3x+y=1

have a unique solution?

Show solution

Write the equations as $kx+2y-5=0$ and $3x+y-1=0$ .

For a unique solution, we need $\frac{k}{3}\ne\frac{2}{1}$ .

So $k\ne6$ .

Hence the system has a unique solution for every real value of $k$ except $6$ .

Practice

For

x+2y=5

and

3x+ky=15

, find the value of

k

for infinitely many solutions.

Answer:

k=6

Word Problems: Numbers and Digits

Translate the language into equations before solving

In digit problems, if the tens digit is

x

and the units digit is

y

, then the number is

10x+y

The reversed number is

10y+x

. This one idea unlocks most Class 10 digit questions.

\text{Two-digit number}=10x+y

\text{Reversed number}=10y+x

Worked Example

Digit-reversal problem

The sum of the digits of a two-digit number is 8. If the number obtained by reversing the digits exceeds the given number by 18, find the number.

Show solution

Let the tens digit be $x$ and the units digit be $y$ .

Then $x+y=8$ .

The reversed number exceeds the original by 18, so $(10y+x)-(10x+y)=18$ .

This gives $9y-9x=18$ , so $y-x=2$ .

Now solve $x+y=8$ and $y-x=2$ to get $y=5$ and $x=3$ .

Hence the number is $35$ .

Practice

If the units digit is twice the tens digit and adding 27 reverses the digits, find the number.

Answer:

36

Word Problems: Ages and Fractions

Track present, past, and future values carefully

Age problems become easy when you clearly mark present ages first and then adjust for past or future years.

In fraction problems, let the numerator be

x

and the denominator be

y

, then form equations from the given changes.

\text{After }n\text{ years: }x+n,\ y+n

\text{Fraction}=\frac{x}{y}

Worked Example

Father and son

Five years hence, a father's age will be three times his son's age. Five years ago, he was seven times as old as his son. Find their present ages.

Show solution

Let present ages be $x$ and $y$ .

Five years hence: $x+5=3(y+5)$ , so $x-3y=10$ .

Five years ago: $x-5=7(y-5)$ , so $x-7y=-30$ .

Subtracting gives $4y=40$ , so $y=10$ .

Then $x-3(10)=10$ , so $x=40$ .

Present ages are 40 years and 10 years.

Practice

A fraction becomes

\frac{4}{5}

when 1 is added to numerator and denominator, and becomes

\frac{1}{2}

when 5 is subtracted from both. Find the fraction.

Answer:

\frac{7}{9}

Word Problems: Cost, Distance and Speed

Model real-life situations with two equations and solve steadily

Cost problems usually take the form: fixed price of one item times quantity plus fixed price of another item times quantity.

For boats and streams, upstream speed is

u-v

and downstream speed is

u+v

, where

u

is the speed in still water and

v

is the speed of the stream.

\text{Distance}=\text{Speed}\times\text{Time}

\text{Upstream speed}=u-v,\quad \text{Downstream speed}=u+v

Worked Example

Taxi fare model

For a journey of 10 km, a taxi charges Rs. 75. For 15 km, it charges Rs. 110. Find the fixed charge and the rate per kilometre.

Show solution

Let the fixed charge be $x$ and the rate per kilometre be $y$ .

Then $x+10y=75$ and $x+15y=110$ .

Subtracting gives $5y=35$ , so $y=7$ .

Now $x+70=75$ , so $x=5$ .

The fixed charge is Rs. 5 and the rate is Rs. 7 per km.

Practice

A boat covers 32 km upstream and 36 km downstream in 7 hours, and 40 km upstream with 48 km downstream in 9 hours. Find the speed of the boat and the speed of the stream.

Answer: Boat speed

=10

km/h, stream speed

=2

km/h.

Quick Summary

Concept	Key Idea
Linear equation	Written in the form $ax+by+c=0$ .
Pair of equations	A solution must satisfy both equations together.
Intersecting lines	One unique solution.
Parallel lines	No solution.
Coincident lines	Infinitely many solutions.
Substitution	Rewrite one variable and substitute.
Elimination	Make coefficients equal and remove one variable.
Cross-multiplication	Use coefficient formulas directly after standard form.
Reducible form	Use substitutions like $u=1/x$ and $v=1/y$ .
Digits	Two-digit number $=10x+y$ .
Ages	Past means subtract; future means add.
Distance and speed	$D=ST$ , upstream $=u-v$ , downstream $=u+v$ .

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Formula Reference

ax+by+c=0

\frac{a_1}{a_2}\ne\frac{b_1}{b_2}

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}

x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}

y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}

\text{Distance}=\text{Speed}\times\text{Time}

\text{Upstream}=u-v,\ \text{Downstream}=u+v

Pair of Linear Equations in Two Variables: revision sheet

Close the notes and explain each idea in your own words. Then check the formulas and attempt the examples before opening their solutions.

Recall checklist

What a Pair of Linear Equations Means
Graphical Method — Intersecting, Parallel, Coincident
Substitution Method
Elimination Method
Consistency Conditions and Reducible Equations
Word Problems — Ages, Digits, Speeds
Common Mistakes and Board Answer Template

For each topic: state the rule, explain when it applies, give an example and identify one mistake to avoid. Return to any topic you cannot explain without looking.

Formula recap (1)

Consistency Conditions and Reducible Equations

\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \text{unique solution}; \quad \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \text{no solution}

Consistency conditions. All three ratios equal → infinitely many solutions.

Explain your method

Write the given information, choose a method, show the steps, then check units and whether the answer makes sense.

Solve x + y = 3 and 3x - 2y = 4.

Check worked solution 1

From x + y = 3, write y = 3 - x. Substitute into 3x - 2y = 4: 3x - 2(3 - x) = 4, so 5x = 10 and x = 2. Then y = 1.

For what value of k does x + 2y = 5 and 3x + ky = 15 have infinitely many solutions?

Check worked solution 2

Write in standard form and compare ratios. For infinitely many solutions, 1/3 = 2/k = (-5)/(-15) = 1/3, so k = 6.

Turn mistakes into revision

Record the question, the step that went wrong and the corrected reason. Try it again tomorrow without the solution, then revisit it after a week. A correct guess still needs an explanation.

Practise this chapter

Pair of Linear Equationsin Two Variables

Linear Equation in Two Variables

Pair of Linear Equations in Two Variables

Verify a solution

Graphical Meaning and the Three Cases

Intersecting Lines

Parallel Lines

Coincident Lines

Graphical Method of Solving

Graph of x + y = 3 and 3x - 2y = 4

Solve graphically

Substitution Method

Solve by substitution

Elimination Method

Solve by elimination

Cross-Multiplication Method

Solve by cross-multiplication

Equations Reducible to Linear Form

Reduce and solve

Consistency Conditions and Finding k

Find k for a unique solution

Word Problems: Numbers and Digits

Digit-reversal problem

Word Problems: Ages and Fractions

Father and son

Word Problems: Cost, Distance and Speed

Taxi fare model

Quick Summary

Pair of Linear Equations in Two Variables: revision sheet

Consistency Conditions and Reducible Equations

Explain your method

Turn mistakes into revision

Pair of Linear Equations
in Two Variables