Pair of Linear Equations in Two Variables: Class 10 Maths · Chapter 3

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

60 min
1 simulation
30 practice questions
14 MCQs

What you will learn in Pair of Linear Equations in Two Variables

Write a pair of linear equations in two variables to model an everyday situation.
Solve a pair of linear equations by drawing both lines and reading their common point.
Classify a pair as consistent, inconsistent or dependent from its graph and by comparing a1/a2, b1/b2 and c1/c2.
Solve a pair by the substitution method, and read a variable-free true or false statement as infinitely many or no solutions.
Solve a pair by the elimination method, choosing multipliers that match one variable's coefficients.
Solve word problems on prices, ages, digits, fractions and fees, and check each answer against the original conditions.

Original companion text aligned to NCERT Mathematics Class 10, Chapter 3, pp. 24–37. Not affiliated with or endorsed by NCERT.

3.1 Introduction

Every July, during Van Mahotsav week, your school's eco-club runs a plant stall. Neem saplings sell for ₹15 each and tulsi pots for ₹10 each. At closing time the club has sold 30 plants and the cash box holds ₹380. Nobody kept a tally. How many neem saplings were sold?

You could start by guessing. Suppose 10 neem saplings were sold. Then 20 tulsi pots were sold too, and the cash would be $150 + 200 = 350$ rupees, which is ₹30 short. Swapping one tulsi pot for one neem sapling keeps the total at 30 plants but adds ₹5. So six swaps close the gap: 16 neem and 14 tulsi. Fig. 3.1 shows the guesses side by side.

3/3Sixteen neem and fourteen tulsi bring in exactly ₹380.

Fig. 3.1Guess, check and adjust until the cash matches.

A bar chart of rupees collected, with a dashed target line at 380. Step 1 shows one bar for 10 neem and 20 tulsi reaching 350, marked 30 rupees short. Step 2 adds a bar for 20 neem and 10 tulsi reaching 400, marked 20 rupees over. Step 3 adds a green bar for 16 neem and 14 tulsi that reaches the target line exactly, marked spot on.

Guessing worked because the numbers were small and friendly. It would be slow if the stall had sold 347 plants, and hopeless with prices such as ₹12.50 and ₹8.75. Algebra gives a method that always works. Let $x$ be the number of neem saplings and $y$ the number of tulsi pots. The two facts about the stall become two equations:

\begin{aligned} x + y & = 30 \\ 15 x + 10 y & = 380 \end{aligned}

Each equation on its own has many solutions. For example, $x = 10$ , $y = 20$ fits the first but not the second. We want the values that fit both at once. Two linear equations in the same two unknowns, taken together, form a pair of linear equations in two variables (दो चरों वाले रैखिक समीकरण युग्म).

Recall from Class IX that a linear equation in two variables can be written as $a x + b y + c = 0$ . Here $a$ , $b$ and $c$ are real numbers, and $a$ and $b$ are not both zero. It has infinitely many solutions, and when we plot them they all lie on one straight line. That line is the equation's graph: every point on it is a solution, and every solution is a point on it.

Definition

Pair of linear equations in two variables

Two linear equations in the same two variables, such as $x$ and $y$ , considered together. A solution of the pair is a pair of values that satisfies both equations at the same time.

Example $x + y = 30$ and $15 x + 10 y = 380$ have the solution $x = 16$ , $y = 14$ .

Any pair of linear equations in $x$ and $y$ can be written in the general form below. The six numbers are its coefficients; the small 1 or 2 tells you which equation a number belongs to.

General form of a pair

\begin{aligned} a_{1} x + b_{1} y + c_{1} & = 0 \\ a_{2} x + b_{2} y + c_{2} & = 0 \end{aligned}

(3.1)

where

$a_{1}, b_{1}, c_{1}$: real numbers, with $a_{1}$ and $b_{1}$ not both zero
$a_{2}, b_{2}, c_{2}$: real numbers, with $a_{2}$ and $b_{2}$ not both zero

For the eco-club pair, $a_{1} = 1$ , $b_{1} = 1$ , $c_{1} = - 30$ and $a_{2} = 15$ , $b_{2} = 10$ , $c_{2} = - 380$ . Watch the signs: moving 30 to the left side turns it into $- 30$ . In this chapter you will meet three ways to find the solution: drawing graphs (Section 3.2), substitution (Section 3.3.1) and elimination (Section 3.3.2).

3.2 Graphical Method of Solution of a Pair of Linear Equations

Each equation of a pair has a straight line as its graph, so a pair gives two lines in the same plane. A solution of the pair must lie on both lines. So the solutions of the pair are exactly the points that the two lines share.

Fig. 3.2 draws the eco-club equations. The solid line collects every way to sell 30 plants. The dashed line collects every way to take ₹380. Only one point lies on both lines, $(16, 14)$ , which is the answer we found by guessing.

3/3Only $(16, 14)$ lies on both lines: 16 neem saplings and 14 tulsi pots.

Fig. 3.2The solution of a pair is the point its two lines share.

A grid with neem saplings x across, from 0 to 32, and tulsi pots y up, from 0 to 40. Step 1 draws the solid blue line x + y = 30 with dots at (10, 20) and (20, 10), described as 30 plants. Step 2 adds the dashed orange line 15x + 10y = 380 with dots at (4, 32) and (20, 8), described as 380 rupees in all. Step 3 rings the crossing point (16, 14) and notes that both facts are true there.

SimulationFind the common solution

A coordinate grid from minus 2 to 8 on both axes. Fixed points A (0, 4), B (4, 2) and C (6, 1) are solutions of x + 2y = 8, and fixed points P (0, minus 1), Q (1, 1) and R (3, 5) are solutions of 2x minus y = 1. Point S starts at (6, 6) and can be dragged to whole-number positions while its coordinates are shown, so you can test where it lies in line with each set of points.

Fig. 3.3

A

B

C

solve

x + 2 y = 8

and

P

Q

R

solve

2 x - y = 1

. Drag

S

Try this

Drag $S$ so that it lines up with $A$ , $B$ and $C$ . Do its coordinates satisfy $x + 2 y = 8$ ?
Find the one place where $S$ lines up with both $A, B, C$ and $P, Q, R$ . Test it in both equations.
Put $S$ at $(4, 7)$ . Which equation does it satisfy, and which row of points is it in line with?

Two lines drawn in the same plane can behave in only three ways, shown in Fig. 3.4. Each way tells us how many solutions the pair has.

They intersect at one point. The pair has exactly one solution, called a unique solution.
They are parallel. The lines never meet, so the pair has no solution.
They coincide. One line lies exactly on the other. Every point of it is common, so the pair has infinitely many solutions.

1/4Every pair of lines falls into one of these three cases.

Fig. 3.4Two lines in a plane meet once, never, or everywhere.

Three small graphs side by side, each with a solid line 1 and a dashed line 2. The first shows step by step: in the opening view all three panels are shown together. The intersecting step highlights the left panel, where the lines cross at one ringed point: one solution, consistent. The parallel step highlights the middle panel, where the lines run side by side without meeting: no solution, inconsistent. The coincident step highlights the right panel, where the dashed line lies on top of the solid one: infinitely many solutions, dependent.

These cases give the pairs their names. A pair is consistent (संगत) if it has at least one solution, and inconsistent (असंगत) if it has none. When one equation is simply a multiple of the other, both draw the same line, and the pair is called dependent (आश्रित). Since it has solutions (infinitely many), a dependent pair also counts as consistent.

Definition

Consistent pair

A pair of linear equations that has at least one solution: exactly one (intersecting lines) or infinitely many (coincident lines).

Definition

Inconsistent pair

A pair of linear equations that has no solution. Its graphs are parallel lines.

Definition

Dependent pair

A pair in which one equation is a multiple of the other, so both draw the same line and share infinitely many solutions. Having solutions, it counts as consistent.

Predicting the case from the coefficients

Drawing two lines takes time. Can we tell the case just from the numbers? Write both equations in the general form (3.1) and compare three ratios: $\frac{a_{1}}{a_{2}}$ , $\frac{b_{1}}{b_{2}}$ and $\frac{c_{1}}{c_{2}}$ . Table 3.1 does this for three pairs whose graphs we already know:

Pair A: $2 x - y - 1 = 0$ and $x + 2 y - 8 = 0$ , which meet at $(2, 3)$
Pair B: $3 x - 2 y + 4 = 0$ and $6 x - 4 y + 8 = 0$
Pair C: $x - 3 y + 3 = 0$ and $2 x - 6 y - 6 = 0$

Table 3.1Comparing coefficient ratios for three pairs

	$\frac{a_{1}}{a_{2}}$	$\frac{b_{1}}{b_{2}}$	$\frac{c_{1}}{c_{2}}$	Lines	Solutions
A	$2$	$- \frac{1}{2}$	$\frac{1}{8}$	cross	one
B	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	same line	infinitely many
C	$\frac{1}{2}$	$\frac{1}{2}$	$- \frac{1}{2}$	parallel	none

Read down the columns. When the lines cross, the first two ratios differ. When the lines are parallel or coincide, the first two ratios are equal, and the third ratio settles it: equal for coincident lines, different for parallel ones.

Compare

The ratio test

Intersecting lines

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
exactly one solution
consistent

Coincident lines

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
infinitely many solutions
dependent (consistent)

Parallel lines

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
no solution
inconsistent

PropertyWhy the ratio test works

For a pair in the general form (3.1), the three conditions of the ratio test tell exactly when the lines intersect, coincide or are parallel. Each also works in reverse: intersecting lines always have $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , and so on.

Suppose $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$ , and call this common value $k$ . Then $a_{1} = k a_{2}$ and $b_{1} = k b_{2}$ , so the first equation reads $k (a_{2} x + b_{2} y) + c_{1} = 0$ .

If also $c_{1} = k c_{2}$ , the first equation is just $k$ times the second. They have the same solutions, so they draw the same line.

If $c_{1} \neq k c_{2}$ , a common point would need $a_{2} x + b_{2} y = - c_{2}$ and also $k (a_{2} x + b_{2} y) = - c_{1}$ . That forces $k c_{2} = c_{1}$ , which is false. So there is no common point: the lines are parallel.

If $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , then $a_{1} b_{2} - a_{2} b_{1} \neq 0$ . In that case the elimination method of Section 3.3.2 always gives exactly one value of $x$ and one of $y$ , so the lines cross once.

If a coefficient of the second equation is 0, a ratio with 0 underneath cannot be formed. Then compare $a_{1} b_{2}$ with $a_{2} b_{1}$ (the ratios cross-multiplied), or simply draw the lines. Try all three cases in the lab below.

InteractiveTwo-line lab

An interactive graph from minus 8 to 8 on both axes showing two lines, a solid blue line 1 and a dashed orange line 2, each written in the form a x plus b y plus c equals 0. Plus and minus buttons change each of the six whole-number coefficients from minus 9 to 9. A table shows the ratios a1 over a2, b1 over b2 and c1 over c2 as reduced fractions, and a status box says whether the lines intersect, are parallel or coincide, whether the pair is consistent, inconsistent or dependent, and gives the exact solution when there is one, which is also ringed on the graph. Buttons load a meeting, a parallel and a same-line example. A Make it mode fixes line 1 and asks you to change line 2 so that the lines become parallel, coincide, or meet at a marked point, with hints and a score.

Fig. 3.5Change the six coefficients and watch the lines, the ratios

\frac{a_{1}}{a_{2}}

\frac{b_{1}}{b_{2}}

\frac{c_{1}}{c_{2}}

and the number of solutions.

Solving a pair graphically

To solve a pair by graphs, find two or three solutions of each equation, plot them, rule both lines and read off the common point. Putting $x = 0$ or $y = 0$ often gives easy points. A third point on each line is a cheap check against slips.

Example 3.1NCERT Example 1 · p. 26

Is the pair consistent? Solve it by graph

Use graphs to decide whether $2 x + 3 y = 12$ and $x - y = 1$ have a common solution, and find it if they do.

Try it yourself first. Then reveal the solution one step at a time.

4/4The lines meet at $C (3, 2)$ , the unique solution.

Fig. 3.6Two points fix each line; the crossing point solves the pair.NCERT Fig. 3.1 · p. 27

A grid from minus 1 to 7 across and minus 2 to 5 up. Step 1 plots A (0, 4) and B (6, 0) from the first equation. Step 2 rules the solid blue line through A and B, labelled 2x + 3y = 12. Step 3 plots P (1, 0) and Q (5, 4) and rules the dashed orange line x minus y = 1. Step 4 rings the crossing point C (3, 2) and notes that the lines meet there.

Example 3.2NCERT Example 2 · p. 27

Spotting a coincident pair

How many solutions does the pair $6 x - 9 y + 3 = 0$ and $\frac{x}{2} - \frac{3 y}{4} + \frac{1}{4} = 0$ have? Decide by clearing the fractions, then confirm with a few points.

Try it yourself first. Then reveal the solution one step at a time.

Example 3.3NCERT Example 3 · p. 27

Which bicycle shop is cheaper?

In Manali, Shop P rents bicycles for ₹60 plus ₹20 an hour, and Shop Q for ₹100 plus ₹10 an hour. Write the cost of $x$ hours at each shop as an equation, and use graphs to find when the two costs are equal.

Try it yourself first. Then reveal the solution one step at a time.

2/2The lines meet at $(4, 140)$ : both shops charge ₹140 for 4 hours.

Fig. 3.7Shop P starts cheaper but climbs faster.NCERT Fig. 3.2 · p. 28

Hours x run across from 0 to 8 and the cost y in rupees runs up from 0 to 240. Step 1 shows the solid blue line for Shop P, starting at 60 and climbing 20 per hour, and the dashed orange line for Shop Q, starting at 100 and climbing 10 per hour. Step 2 rings their crossing point (4, 140), with dashed guide lines down to 4 hours and across to 140 rupees, and notes the same cost at 4 hours.

The graph shows more than the meeting point. Left of $x = 4$ , Shop P's line is lower, so P is cheaper for short rides. Right of it, Q is cheaper. Graphs are good at showing comparisons like this at a glance.

Quick check

1 of 4

Match each pair of lines with its number of solutions.

Tap an item on the left, then its match on the right.

Show answer

Answer

Intersecting : exactly one
Parallel : none
Coincident : infinitely many

Solutions are the common points. Crossing lines share one point, parallel lines share none, and coincident lines share all their points.

3.3 Algebraic Methods of Solving a Pair of Linear Equations

Graphs work well when the lines meet at a grid point, but real problems are seldom so kind. Look at the pair $5 x - 3 y = 9$ and $x + 2 y = 6$ in Fig. 3.9. The lines clearly cross, but where exactly? Somewhere near $(2.8, 1.6)$ , and even a magnified view cannot say more.

3/3Algebra gives the exact point $(\frac{36}{13}, \frac{21}{13})$ (Example 3.4).

Fig. 3.9A drawing can only locate a crossing point roughly.

Step 1 shows the lines 5x minus 3y = 9 and x + 2y = 6 crossing between grid lines, with a question mark and the guess about (2.8, 1.6). Step 2 adds a magnified square from x = 2 to 3 and y = 1 to 2 with tenths marked; the crossing point still sits between 2.7 and 2.8 across and between 1.6 and 1.7 up. Step 3 shows the exact answer, 36 over 13 and 21 over 13, in a box.

A pencil line has thickness, graph paper has finite squares, and our eyes are not exact. To find answers like $\frac{36}{13}$ with certainty, we need algebra. The next two subsections give two such methods.

3.3.1 Substitution Method

The idea of the substitution method (प्रतिस्थापन विधि) is simple. Rearrange one equation so that one variable stands alone on one side, such as $x = 6 - 2 y$ . Then replace that variable in the other equation. What is left has only one unknown, and you already know how to solve it.

Example 3.4NCERT Example 4 · p. 30

Solving by substitution

Solve $5 x - 3 y = 9$ (1) and $x + 2 y = 6$ (2) by substitution.

Try it yourself first. Then reveal the solution one step at a time.

Key idea

The substitution method

Express. Rearrange one equation to get one variable alone, like $x = 6 - 2 y$ . A coefficient of 1 or $- 1$ makes this easy.
Substitute. Replace that variable in the other equation by the expression. Only one unknown is left, so solve for it.
Back-substitute. The expression from step 1 now gives the second unknown directly. Finish by checking both original equations.

Sometimes the variable cancels in step 2 and you are left with plain numbers, such as $7 = 7$ or $0 = 4$ . An always-true result like $7 = 7$ means every point of one line is on the other: infinitely many solutions. An impossible result like $0 = 4$ means the lines never meet: no solution. Example 3.6 and Example 3.7 show both.

The name comes from step 2, where an expression takes the place of a variable: it is substituted for it.

Example 3.5NCERT Example 5 · p. 31

Ages of a grandfather and grandson

Dadaji is 63 years older than his grandson Kabir. In 8 years, Dadaji will be four times as old as Kabir. Find their present ages.

Try it yourself first. Then reveal the solution one step at a time.

3/3The gap is still 63 years: $(g + 8) - (k + 8) = g - k$ .

Fig. 3.10Both ages move by the same number of years, so the gap never changes.

A timeline arrow with two marks, now and in 8 years. Rows for Dadaji and Kabir hold their ages. Step 1 shows the present ages g and k with a bracket marking a gap of 63 years and the equation g minus k = 63. Step 2 adds the ages in 8 years, g + 8 and k + 8, and the condition g + 8 = 4 times (k + 8). Step 3 brackets the future ages too and notes that the gap is still 63 years, because both ages grow by the same amount.

On a graph with $k$ across and $g$ up, lines (1) and (2) would meet at $(13, 76)$ . Substitution reaches the same point without any drawing.

Example 3.6NCERT Example 6 · p. 32

The variable vanishes: a true statement

At a bakery, Neha pays ₹90 for 4 cupcakes and 6 cookies. Arjun pays ₹135 for 6 cupcakes and 9 cookies. Can you find the price of one cupcake and of one cookie?

Try it yourself first. Then reveal the solution one step at a time.

Example 3.7NCERT Example 7 · p. 32

The variable vanishes: a false statement

Ravi says that at the school stationery shop, 2 notebooks and 6 pens cost ₹90. Sana says that 1 notebook and 3 pens at the same shop cost ₹50. Can both of them be right?

Try it yourself first. Then reveal the solution one step at a time.

Fig. 3.11 puts the last two examples side by side. A true statement such as $135 = 135$ means the two lines lie on top of each other. A false statement such as $100 = 90$ means they never meet.

1/3A vanished variable leaves either an always-true or a never-true statement.

Fig. 3.11What is left when the variable cancels tells you the case.

Two panels side by side. The left panel is headed 135 = 135, always true, and shows a solid line with a dashed line lying exactly on it: the same line, infinitely many solutions. The right panel is headed 100 = 90, never true, and shows a solid line and a dashed line running parallel: no solution. The opening step shows both panels; the true step highlights the left panel and the false step highlights the right panel.

Quick check

1 of 3

Put the steps of the substitution method in order.

Find $x$ from $y$
Check both equations
Write $x$ in terms of $y$
Substitute into the other equation
Solve for $y$

Show answer

Answer

Write $x$ in terms of $y$
Substitute into the other equation
Solve for $y$
Find $x$ from $y$
Check both equations

Express one variable, substitute it into the other equation, solve, back-substitute, and finally check in both original equations.

Practice · 9 questions

Practice 3.2: questions with answers

1
Short answerEasyNCERT Exercise 3.2, Q1
Solve each pair by the substitution method.
1. $x + y = 11$ and $x - y = 5$
  Show solution
  Answer $x = 8$ , $y = 3$
  From the first, $x = 11 - y$ .
  $11 - y - y = 5$ , so $y = 3$ .
  Then $x = 8$ .
2. $p - q = 6$ and $\frac{p}{4} + \frac{q}{3} = 5$
  Show solution
  Answer $p = 12$ , $q = 6$
  Multiply the second by 12: $3 p + 4 q = 60$ . From the first, $p = q + 6$ .
  $3 (q + 6) + 4 q = 60$ gives $7 q = 42$ , so $q = 6$ .
  Then $p = 12$ .
3. $0.4 x + 0.3 y = 2.5$ and $0.2 x + 0.5 y = 2.3$
  Show solution
  Answer $x = 4$ , $y = 3$
  Multiply both by 10: $4 x + 3 y = 25$ and $2 x + 5 y = 23$ . From the second, $2 x = 23 - 5 y$ .
  $2 (23 - 5 y) + 3 y = 25$ gives $46 - 7 y = 25$ , so $y = 3$ .
  Then $2 x = 23 - 15$ , so $x = 4$ .
4. $2 x - y = 4$ and $6 x - 3 y = 12$
  Show solution
  AnswerInfinitely many solutions.
  From the first, $y = 2 x - 4$ .
  $6 x - 3 (2 x - 4) = 12$ , so $12 = 12$ .
  This is always true: every point of $y = 2 x - 4$ is a solution.
Short answerHardNCERT Exercise 3.2, Q1(v)
Solve $\sqrt{5} x + \sqrt{2} y = 0$ and $\sqrt{2} x - \sqrt{5} y = 0$ by substitution.
Show solution
Answer $x = 0$ , $y = 0$
Substitute $y = - \frac{\sqrt{5}}{\sqrt{2}} x$ into the second equation:
$\sqrt{2} x + \frac{5}{\sqrt{2}} x = 0$ , so $\frac{7}{\sqrt{2}} x = 0$ .
So $x = 0$ , and then $y = 0$ . Both lines pass through the origin and are different lines, so the origin is the only solution.
Short answerMediumNCERT Exercise 3.2, Q2
First solve $3 x + 2 y = 4$ and $5 x - 2 y = 12$ . The solution also lies on the line $y = m x + 5$ : what is $m$ ?
Show solution
Answer $x = 2$ , $y = - 1$ , $m = - 3$
From the first, $2 y = 4 - 3 x$ . Substitute into the second:
$5 x - (4 - 3 x) = 12$ gives $8 x = 16$ , so $x = 2$ .
Then $2 y = 4 - 6$ , so $y = - 1$ . Finally $- 1 = 2 m + 5$ , so $m = - 3$ .
Short answerEasyNCERT Exercise 3.2, Q3(i)
The sum of two numbers is 96, and one of them is five times the other. Find the numbers.
Show solution
Answer80 and 16
Let the numbers be $x$ and $y$ , with $x + y = 96$ and $x = 5 y$ . Then $6 y = 96$ , so $y = 16$ and $x = 80$ .
Short answerEasyNCERT Exercise 3.2, Q3(ii)
Two angles are complementary. The larger one is 12° more than twice the smaller. Find both angles.
Show solution
Answer64° and 26°
Let the angles be $x °$ and $y °$ : $x + y = 90$ and $x = 2 y + 12$ .
Then $3 y + 12 = 90$ , so $y = 26$ and $x = 64$ .
Short answerMediumNCERT Exercise 3.2, Q3(iii)
A sports teacher buys 5 footballs and 8 skipping ropes for ₹2,740. Later she buys 3 footballs and 10 skipping ropes for ₹2,060. Find the price of a football and of a skipping rope.
Show solution
AnswerFootball ₹420, skipping rope ₹80.
Let a football cost ₹ $f$ and a rope ₹ $r$ .
$5 f + 8 r = 2740$ and $3 f + 10 r = 2060$ .
From the second, $3 f = 2060 - 10 r$ . Multiply the first by 3 and substitute:
$5 (2060 - 10 r) + 24 r = 8220$ gives $26 r = 2080$ , so $r = 80$ .
Then $3 f = 2060 - 800 = 1260$ , so $f = 420$ .
Short answerMediumNCERT Exercise 3.2, Q3(iv)
A water-tanker service charges a fixed delivery fee plus a rate for every thousand litres. A delivery of 6 thousand litres costs ₹1,040 and one of 10 thousand litres costs ₹1,600. Find the fee and the rate. What will 15 thousand litres cost?
Show solution
AnswerFee ₹200, rate ₹140 per thousand litres; 15 thousand litres cost ₹2,300.
Let the fee be ₹ $f$ and the rate ₹ $r$ .
$f + 6 r = 1040$ and $f + 10 r = 1600$ .
From the first, $f = 1040 - 6 r$ . Then $1040 + 4 r = 1600$ , so $r = 140$ and $f = 200$ .
For 15 thousand litres: $200 + 2100 = 2300$ rupees.
Short answerMediumNCERT Exercise 3.2, Q3(v)
If 3 is subtracted from both the numerator and the denominator of a fraction, it becomes $\frac{1}{2}$ . If 1 is added to both, it becomes $\frac{2}{3}$ . Find the fraction.
Show solution
Answer $\frac{7}{11}$
First condition: $2 (x - 3) = y - 3$ , so $y = 2 x - 3$ . Second: $3 (x + 1) = 2 (y + 1)$ , so $3 x - 2 y = - 1$ .
$3 x - 2 (2 x - 3) = - 1$ gives $- x + 6 = - 1$ , so $x = 7$ .
So $y = 11$ . Check: $\frac{4}{8} = \frac{1}{2}$ and $\frac{8}{12} = \frac{2}{3}$ .
Short answerMediumNCERT Exercise 3.2, Q3(vi)
Meher's aunt is 26 years older than Meher. Three years ago, the aunt was three times as old as Meher. Find their present ages.
Show solution
AnswerAunt 42 years, Meher 16 years.
Let the aunt be $a$ and Meher $m$ years old.
$a - m = 26$ and $a - 3 = 3 (m - 3)$ .
The first gives $a = m + 26$ . Substitute into the second: $m + 23 = 3 m - 9$ , so $m = 16$ and $a = 42$ .
Check: $42 - 16 = 26$ , and three years ago they were 39 and 13.

3.3.2 Elimination Method

Substitution is quickest when some variable has coefficient 1. When every coefficient is larger, it drags in fractions. The elimination method (विलोपन विधि) avoids them. We multiply the equations by suitable numbers so that one variable has the same coefficient in both. Then subtracting (or adding) the equations wipes that variable out completely, which is what eliminate means.

Example 3.8NCERT Example 8 · p. 34

Marks for right and wrong answers

In a quiz, every right answer earns some marks and every wrong answer loses some. Asha got 14 right and 4 wrong and scored 48. Bilal got 11 right and 5 wrong and scored 34. How many marks does a right answer earn, and how many does a wrong one lose?

Try it yourself first. Then reveal the solution one step at a time.

4/4Putting $r = 4$ into (1) gives $w = 2$ .

Fig. 3.12Match one coefficient, subtract, and that variable disappears.

Rows of equations aligned by term. Step 1 shows equation (1), 14r minus 4w = 48, marked times 5, and equation (2), 11r minus 5w = 34, marked times 4. Step 2 adds equations (3), 70r minus 20w = 240, and (4), 44r minus 20w = 136, with both minus 20w terms highlighted. Step 3 strikes out the two w-terms, rules a line and shows (3) minus (4): 26r = 104, so r = 4, noting that w is gone. Step 4 adds the result of putting r = 4 into (1): w = 2.

We could eliminate $r$ instead, but that needs multipliers 11 and 14 to reach $154 r$ . Choosing the variable whose coefficients have the smaller common multiple keeps the numbers small. Substitution would work too, but it starts with a fraction, $r = \frac{48 + 4 w}{14}$ , which is why elimination is the natural choice here.

Example 3.9NCERT Example 9 · p. 35

Elimination meets a false statement

Use elimination to find all solutions of $3 x - 5 y = 4$ (1) and $9 x - 15 y = 7$ (2).

Try it yourself first. Then reveal the solution one step at a time.

Example 3.10NCERT Example 10 · p. 35

A two-digit number and its reverse

A two-digit number and the number formed by reversing its digits differ by 36. The digits of the number add up to 10. Find all such numbers.

Try it yourself first. Then reveal the solution one step at a time.

3/3The difference is always 9 times the difference of the digits.

Fig. 3.13Reversing the digits swaps their place values.

Two slots labelled tens and ones. Step 1 holds a blue tile x in the tens slot and an orange tile y in the ones slot, with the value 10x + y. Step 2 slides the tiles past each other so that y is in the tens slot and x in the ones slot, with the value 10y + x and the note digits swapped. Step 3 shows the difference (10x + y) minus (10y + x) = 9x minus 9y = 9 times (x minus y), with the example 73 minus 37 = 9 times 4 = 36.

Practice · 6 questions

Practice 3.3: questions with answers

1
Short answerMediumNCERT Exercise 3.3, Q1
Solve each pair by elimination. Then check your answer by substitution.
1. $x + y = 9$ and $3 x - 2 y = 2$
  Show solution
  Answer $x = 4$ , $y = 5$
  Multiply the first by 2 and add it to the second:
  $5 x = 20$ , so $x = 4$ .
  Then $y = 5$ . Check: $12 - 10 = 2$ .
2. $2 x + 5 y = 16$ and $3 x - 2 y = 5$
  Show solution
  Answer $x = 3$ , $y = 2$
  Multiply by 2 and by 5 to get $\pm 10 y$ :
  $4 x + 10 y = 32$ and $15 x - 10 y = 25$ .
  Add: $19 x = 57$ , so $x = 3$ . Then $5 y = 10$ , so $y = 2$ .
3. $4 x - 3 y - 7 = 0$ and $6 x = 5 y + 9$
  Show solution
  Answer $x = 4$ , $y = 3$
  Rewrite as $4 x - 3 y = 7$ and $6 x - 5 y = 9$ . Multiply by 3 and by 2:
  $12 x - 9 y = 21$ and $12 x - 10 y = 18$ .
  Subtract: $y = 3$ . Then $4 x = 16$ , so $x = 4$ .
4. $\frac{x}{3} + \frac{y}{2} = 4$ and $2 x - \frac{y}{4} = 11$
  Show solution
  Answer $x = 6$ , $y = 4$
  Clear the fractions (multiply by 6 and by 4):
  $2 x + 3 y = 24$ and $8 x - y = 44$ .
  Multiply the second by 3 and add: $26 x = 156$ , so $x = 6$ . Then $y = 48 - 44 = 4$ .
Short answerMediumNCERT Exercise 3.3, Q2(i)
If the numerator of a fraction is doubled and the denominator is increased by 1, the fraction becomes 1. If 2 is subtracted from the numerator only, it becomes $\frac{1}{3}$ . Find the fraction.
Show solution
Answer $\frac{5}{9}$
Let the fraction be $\frac{x}{y}$ . Then $2 x = y + 1$ and $3 (x - 2) = y$ .
$2 x - y = 1$ and $3 x - y = 6$ .
Subtract: $x = 5$ , so $y = 9$ . Check: $\frac{10}{10} = 1$ and $\frac{3}{9} = \frac{1}{3}$ .
Short answerMediumNCERT Exercise 3.3, Q2(ii)
Six years ago, Gurpreet was five times as old as her nephew Arman. Twelve years from now, she will be twice as old as he will be. How old are they now?
Show solution
AnswerGurpreet is 36 and Arman is 12.
Six years ago: $g - 6 = 5 (a - 6)$ . In twelve years: $g + 12 = 2 (a + 12)$ .
$g - 5 a = - 24$ and $g - 2 a = 12$ .
Subtract the first from the second: $3 a = 36$ , so $a = 12$ and $g = 36$ . Check: 30 and 6; 48 and 24.
Short answerHardNCERT Exercise 3.3, Q2(iii)
The digits of a two-digit number add up to 6. Seven times the number is equal to four times the number formed by reversing its digits. Find the number.
Show solution
Answer24
Let the digits be $x$ (tens) and $y$ (units), so $x + y = 6$ . The second condition simplifies:
$7 (10 x + y) = 4 (10 y + x)$ gives $66 x = 33 y$ , so $2 x - y = 0$ .
Add this to $x + y = 6$ : $3 x = 6$ , so $x = 2$ and $y = 4$ . The number is 24. Check: $7 \times 24 = 168 = 4 \times 42$ .
Short answerMediumNCERT Exercise 3.3, Q2(iv)
A shopkeeper's drawer holds only ₹20 and ₹50 notes: 40 notes worth ₹1,550 in all. How many notes of each kind are there?
Show solution
Answer15 notes of ₹20 and 25 notes of ₹50.
Let there be $x$ notes of ₹20 and $y$ of ₹50.
$x + y = 40$ and $20 x + 50 y = 1550$ .
Multiply the first by 20 and subtract: $30 y = 750$ , so $y = 25$ and $x = 15$ . Check: $300 + 1250 = 1550$ .
Short answerMediumNCERT Exercise 3.3, Q2(v)
A parking lot charges a fixed fee for the first 2 hours and a further amount for each extra hour. A car parked for 6 hours pays ₹110, and one parked for 4 hours pays ₹80. Find the fixed fee and the charge per extra hour.
Show solution
AnswerFixed fee ₹50; ₹15 per extra hour.
Let the fee be ₹ $f$ and the extra charge ₹ $e$ an hour.
$f + 4 e = 110$ and $f + 2 e = 80$ .
Subtract: $2 e = 30$ , so $e = 15$ and $f = 50$ .

3.4 Summary

Before you solve any pair, decide how. Fig. 3.14 is one way to plan, and the points after it collect the chapter's main ideas.

3/3Whatever the method, put the answer back into both original equations.

Fig. 3.14A plan for any pair of linear equations.

A three-stage flowchart. Step 1 shows the first box, ratio test first, which tells you to expect one solution, none or infinitely many. Step 2 adds three branches: if x or y has coefficient plus or minus 1, use substitution; if the coefficients can be matched, use elimination; if you need a picture or an estimate, use graphs. Step 3 adds the final box: check in both equations.

Recap

Pair of Linear Equations in Two Variables: key points and summary

A pair of linear equations in $x$ and $y$ has the general form (3.1). A solution satisfies both equations at once (Section 3.1).
Each equation is a straight line, so the solutions of the pair are the points common to both lines (Section 3.2).
Intersecting lines: exactly one solution, a consistent pair (Fig. 3.4).
Parallel lines: no solution, an inconsistent pair.
Coincident lines: infinitely many solutions, a dependent pair, which is also consistent.
Ratio test: $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ means one solution; all three ratios equal means infinitely many; only the first two equal means none (the ratio test).
Graphs give exact answers only when the lines meet at a grid point; algebraic methods are always exact (Section 3.3).
Substitution: express one variable from one equation, put it into the other, solve, then back-substitute (Section 3.3.1).
Elimination: multiply to match one variable's coefficients, add or subtract to remove it, solve, then back-substitute (Section 3.3.2).
If a statement with no variable appears, true means infinitely many solutions and false means none (Fig. 3.11).
For word problems, name the unknowns, write one equation per condition, solve, and check the answer against the original words.

Flash cards · 7

Key terms

1 / 7

Board exam practice

These questions use the formats of the CBSE Class 10 board paper: one-mark multiple-choice and assertion–reason items, short answers, a case study and a graph question. Try each one before you open its solution.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 3.1, Q1	Practice 3.1 · Q1
NCERT Exercise 3.1, Q2	Practice 3.1 · Q2
NCERT Exercise 3.1, Q3	Practice 3.1 · Q3
NCERT Exercise 3.1, Q4	Practice 3.1 · Q4
NCERT Exercise 3.1, Q5	Practice 3.1 · Q5
NCERT Exercise 3.1, Q6	Practice 3.1 · Q6
NCERT Exercise 3.1, Q7	Practice 3.1 · Q7
NCERT Exercise 3.2, Q1	Practice 3.2 · Q1
NCERT Exercise 3.2, Q1(v)	Practice 3.2 · Q2
NCERT Exercise 3.2, Q2	Practice 3.2 · Q3
NCERT Exercise 3.2, Q3(i)	Practice 3.2 · Q4
NCERT Exercise 3.2, Q3(ii)	Practice 3.2 · Q5
NCERT Exercise 3.2, Q3(iii)	Practice 3.2 · Q6
NCERT Exercise 3.2, Q3(iv)	Practice 3.2 · Q7
NCERT Exercise 3.2, Q3(v)	Practice 3.2 · Q8
NCERT Exercise 3.2, Q3(vi)	Practice 3.2 · Q9
NCERT Exercise 3.3, Q1	Practice 3.3 · Q1
NCERT Exercise 3.3, Q2(i)	Practice 3.3 · Q2
NCERT Exercise 3.3, Q2(ii)	Practice 3.3 · Q3
NCERT Exercise 3.3, Q2(iii)	Practice 3.3 · Q4
NCERT Exercise 3.3, Q2(iv)	Practice 3.3 · Q5
NCERT Exercise 3.3, Q2(v)	Practice 3.3 · Q6

Quick check: MCQs on Pair of Linear Equations in Two Variables

3.1 Introduction3 questions
Solving a pair graphically4 questions
3.3.1 Substitution Method3 questions
3.3.2 Elimination Method4 questions

Glossary · 11

Key terms (English–Hindi)

coincident linesसंपाती रेखाएँsampati rekhayen: Two lines that lie exactly on top of each other, so every point is common.First seen in the chapter →
consistent pairसंगतsangat: A pair of linear equations that has at least one solution.First seen in the chapter →

dependent pairआश्रितaashrit: A pair in which one equation is a multiple of the other, so both draw the same line and there are infinitely many solutions; it is also consistent.First seen in the chapter →

elimination methodविलोपन विधिvilopan vidhi: Solving a pair by making one variable's coefficients equal, then adding or subtracting the equations to remove it.First seen in the chapter →

inconsistent pairअसंगतasangat: A pair of linear equations that has no solution; its lines are parallel.First seen in the chapter →
intersecting linesप्रतिच्छेदी रेखाएँpratichchhedi rekhayen: Two lines that cross at exactly one point.First seen in the chapter →

linear equation in two variablesदो चरों वाला रैखिक समीकरणdo charon wala rekhik samikaran: An equation that can be written as $a x + b y + c = 0$ , with $a$ , $b$ , $c$ real and $a$ , $b$ not both zero. Its graph is a straight line.First seen in the chapter →

pair of linear equationsरैखिक समीकरण युग्मrekhik samikaran yugm: Two linear equations in the same two variables, considered together.First seen in the chapter →
parallel linesसमांतर रेखाएँsamantar rekhayen: Lines in a plane that never meet, however far they are drawn.First seen in the chapter →

solution of a pairहलhal: Values of $x$ and $y$ that satisfy both equations of the pair at the same time.First seen in the chapter →
substitution methodप्रतिस्थापन विधिpratisthapan vidhi: Solving a pair by writing one variable in terms of the other and putting that expression into the other equation.First seen in the chapter →

Pair of Linear Equations in Two Variables: Class 10 Maths · Chapter 3

What you will learn in Pair of Linear Equations in Two Variables

3.1 Introduction

3.2 Graphical Method of Solution of a Pair of Linear Equations

Predicting the case from the coefficients

Solving a pair graphically

Practice 3.1: questions with answers

3.3 Algebraic Methods of Solving a Pair of Linear Equations

3.3.1 Substitution Method

Practice 3.2: questions with answers

3.3.2 Elimination Method

Practice 3.3: questions with answers

3.4 Summary

Pair of Linear Equations in Two Variables: key points and summary

Board exam practice

Practice questions with answers (NCERT “Board exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Pair of Linear Equations in Two Variables

Key terms (English–Hindi)

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