NCERT Solutions Class 9 Maths Chapter 2 Introduction to Linear Polynomials

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

Chapter 2: Introduction to Linear
Polynomials

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

16 – 40 26 65 English

Solutions, notes, sample papers & more at 86 pages

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 2: Introduction to Linear
Polynomials
A polynomial in one variable is an algebraic expression built from a single letter and its whole-number
powers; the highest power is its degree. This chapter stays with degree 1 — the linear polynomials — and
shows that one idea runs through all of them: equal steps in the input produce equal steps in the output, which is
exactly why their graphs are straight lines.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 16 – 40

SECTIONS QUESTIONS

26 65

MEDIUM

English

Think and Reflect — Page 17
Section 2.1 Introduction

Q1 Can you identify the terms, variables and coefficients of this algebraic expression?

The expression is the total cost of the garden in Example 2, 200l + 160w + 50lw.

PART WHAT IT IS HERE

Terms 200l, 160w, 50lw

Variables l (length in m) and w (width in m)

Coefficients 200 of l, 160 of w, 50 of lw

Constant term none — every term contains a variable

Why it happens: each coefficient is a rate, and reading it off tells you where it came
from. The wire runs along both lengths, so 2 × ₹100 = ₹200 per metre of length; the
wooden fence runs along both widths, so 2 × ₹80 = ₹160 per metre of width. The
seed is charged by area, and the area is l × w, which is why the third term carries the
product lw and not l or w alone.

Page 1 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Tip: The coefficient of a term is everything except the letters. In 50lw the coefficient
is 50, and both l and w are variables of that single term.

Q2 How is it different from the algebraic expression in Example 1?

Example 1 gave 4x + 5y + 3. Example 2 gives 200l + 160w + 50lw. Two differences matter.

The highest power. In 4x + 5y + 3 every variable appears to the power 1 and no term mixes
the two letters. In 200l + 160w + 50lw the term 50lw multiplies the two variables together, so
that term has degree 1 + 1 = 2.
The constant. 4x + 5y + 3 has the constant term 3 (the free pens). The garden expression has
no constant — if the garden shrinks to nothing, the cost is 0.

Why it happens: the difference is not cosmetic, it is about how the quantities
behave. Doubling the garden’s length doubles the wire cost, but doubling both
length and width multiplies the seed cost by four, because area grows with the
product. An expression in which the variables only ever appear singly, to the first
power, describes something that grows in step with its inputs; a product term does
not.

Did you know? A term such as 50lw is called a second-degree term, since the powers
of the variables in it add to 2. Because of it, the garden expression is not linear, even
though the first two terms are.

In-text Questions — Page 17
Section 2.1 Introduction (Example 3)

Q1 A wire of length 20 cm is bent in different ways to form rectangles. For example, we
can have a rectangle with length 7 cm and width 3 cm. We can also have one of
length 5.5 cm and width 4.5 cm. (Think of a few more ways of forming such
rectangles.)

The wire is the perimeter, so for every such rectangle

Page 2 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

2(length + width) = 20

length + width = 10

So any pair of positive lengths adding to 10 works. A few more:

LENGTH (CM) 9 8 6.5 6 5

Width (cm) 1 2 3.5 4 5

Area (cm²) 9 16 22.75 24 25

Why it happens: the wire fixes the perimeter, not the shape. Since 2l + 2w = 20, we
get w = 10 − l, so choosing the length fixes the width as well. Only one number is
free, and it may be any value with 0 < l < 10 — there are infinitely many such
rectangles, not a handful.

Try This: the area is l(10 − l). Work it out for l = 4, 4.5, 5, 5.5, 6 and you will get 24,
24.75, 25, 24.75, 24. The area is largest, 25 cm², when the rectangle is the square of
side 5 cm — and it falls away symmetrically on both sides of that.

Think and Reflect — Page 17
Section 2.1 Introduction (Example 3)

Q1 Can you identify the terms, variables and coefficients of this algebraic expression?

The expression is the area of the bent-wire rectangle, 10x − x².

PART WHAT IT IS HERE

Terms 10x and −x²

Variable x only (the length in cm)

Coefficients 10 of x, −1 of x²

Constant term none (it is 0)

Degree 2 — the highest power of x present

Page 3 of 86

as e
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the area is x(10 − x). Expanding the bracket gives 10x − x², and the

m l a se
minus sign belongs to the term, so the coefficient of x² is −1, not 1. The constant is 0
o
because a.crectangle of zero length has zero area. a g
se m
g l a
aTip: Always read a coefficient with its sign. Writing “the coefficient of x² is 1” here

com
ag
would change the expression completely.
m .
as e
a g l
Can you point out any similarity or difference between the algebraic expressions
m
Q2

co
m.
obtained in Examples 1 and 3?

o m l a se
ANSWER c
. a g
m
se both are algebraic expressions — sums of terms, each term a number multiplying
l a
agletters.
Similarity:

Differences:
m a s
m .co agl
l a se
EXAMPLE 1: 4X + 5Y + 3 EXAMPLE 3: 10X − X²

a g
Number of variables two (x, y) one (x)

co m
Highest power 1 2
m .
m as e
.co a g l
m
Constant term 3 none

a s eUnivariate
agl polynomial? no — two variables yes, of degree 2

se m
com g l a
.
Why it matters: this is exactly the split the chapter is about to make. Expressions
m a
ase
with a single variable and its powers are the ones we call univariate polynomials, and

a gl
only those get a degree. 10x − x² qualifies; 4x + 5y + 3 does not, because it needs two
independent letters.

co m
m .
m l a se
Check it yourself: in Example 1 you can double x and leave y alone. In Example 3
o
m ag and its area, is fixed.
.cthere is only one dial to turn — fix x and the whole rectangle,
l a se
ag
.c
s e m
Exercise Set 2.1 — Pages 18–19om a
em.
c agl
g l as
a

co m
m .
m ase
.co


a g l Page 4 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.1 Introduction

Q1 Find the degrees of the following polynomials: (i) 2x² – 5x + 3 (ii) y³ + 2y – 1 (iii) – 9
(iv) 4z – 3

The degree is the highest power of the variable that actually appears.

POLYNOMIAL POWERS PRESENT DEGREE NAME

(i) 2x² − 5x + 3 2, 1, 0 2 quadratic

(ii) y³ + 2y − 1 3, 1, 0 3 cubic

(iii) −9 0 only 0 constant

(iv) 4z − 3 1, 0 1 linear

Why (iii) has degree 0 and not “no degree”: −9 can be written as −9x⁰, since x⁰ = 1
for every x ≠ 0. The variable is there, carrying the power 0, so the highest power
present is 0. Notice too that in (ii) the y² term is missing; a missing term simply has
coefficient 0 and does not affect the degree, which is decided by the largest power
with a non-zero coefficient.

Tip: Degree is read off the exponents, never off the coefficients. 2x² − 5x + 3 has
degree 2 even though 5 and 3 are bigger numbers than 2.

Q2 Write polynomials of degrees 1, 2 and 3.

Any expression in one variable whose highest power is the required number will do. For
example:

Degree 1 (linear): 3x + 7

Degree 2 (quadratic): x² − 4x + 1

Degree 3 (cubic): 2x³ + x − 5

Page 5 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

What has to be checked: only two things. First, exactly one variable must appear —
3x + 7y is not a univariate polynomial. Second, the leading coefficient must not be
zero: in 2x³ + x − 5 the number multiplying x³ is 2, which is non-zero, so the degree
really is 3. Lower terms may be missing freely — x³ on its own is a perfectly good
cubic.

Try This: write three more of each, using different letters (t, m, z) and at least one
negative coefficient, then swap with a friend and check the degrees.

Q3 What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?

Read the number attached to each power, together with the sign in front of it.

x⁴ − 3x³ + 6x² − 2x + 7

Coefficient of x² = 6

Coefficient of x³ = −3

Why the sign is part of the answer: the polynomial is a sum of terms, so “− 3x³” is
really “+ (−3)x³”. If you drop the minus, substituting a number gives the wrong value:
at x = 1 the true value is 1 − 3 + 6 − 2 + 7 = 9, but with +3 you would get 15.

Tip: the coefficient of x⁴ here is 1 (an unwritten 1), and the constant term 7 is the
coefficient of x⁰.

Q4 What is the coefficient of z in the polynomial 4z³ + 5z² – 11?

There is no z term written, so its coefficient is

4z³ + 5z² − 11 = 4z³ + 5z² + 0·z − 11

Coefficient of z = 0

Page 6 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why 0 and not “none”: a polynomial of degree 3 has room for all four powers z³, z²,
z, z⁰. A power that is not printed is present with coefficient 0 — adding 0·z changes
nothing, so the two ways of writing the polynomial are the same object. Saying the
coefficient is 0 keeps the list of coefficients complete: 4, 5, 0, −11.

Check it yourself: at z = 2, 4(8) + 5(4) − 11 = 32 + 20 − 11 = 41, and adding 0·2 leaves
it at 41.

Q5 What is the constant term of the polynomial 9x³ + 5x² – 8x –10?

The constant term is the one with no x in it.

9x³ + 5x² − 8x − 10

Constant term = −10

Why it is worth naming: the constant term is the value of the polynomial when the
variable is 0. Substituting x = 0 kills every other term: 9(0) + 5(0) − 8(0) − 10 = −10. So
the constant term is the “starting value” of the input–output machine, and later in
this chapter it is exactly the b of y = ax + b — the y-intercept.

Tip: Keep the minus sign with the 10. “10” would be the constant of a different
polynomial.

Think and Reflect — Page 19
Section 2.2 Linear Polynomials (Example 4)

Q1 Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What
will happen to the perimeters if the sides increase by 0.5 cm?

The perimeter of a square of side x is the linear polynomial 4x.

Page 7 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

SIDE X (CM) 1 1.5 2 2.5 3

Perimeter 4x (cm) 4 6 8 10 12

Increase — +2 +2 +2 +2

Each extra 0.5 cm of side adds 2 cm to the perimeter, every single time.

Why it happens: do not just read it off the table — prove it. For any side x,

4(x + 0.5) − 4x = 4x + 2 − 4x = 2

The x cancels, so the increase does not depend on where you started. That
cancellation is possible only because the variable appears to the first power; it is the
algebraic reason a linear polynomial produces equal steps.

Check it yourself: try the same subtraction on a square’s area, x². (x + 0.5)² − x² = x +
0.25, which still contains x — so the area grows by different amounts at different
stages. That is precisely what makes area non-linear.

Think and Reflect — Page 19
Section 2.2 Linear Polynomials (Example 5)

Q1 If a player paid ₹750, how many matches did he play?

The amount for m matches is the linear polynomial 200 + 50m. Set it equal to 750 — that turns
the polynomial into a linear equation.

200 + 50m = 750

50m = 750 − 200 = 550

m = 550 ÷ 50 = 11 matches

Check: 200 + 50 × 11 = 200 + 550 = 750. ✓

Page 8 of 86

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Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Why there is exactly one answer: the joining fee ₹200 is paid once, so of the ₹750

m l a se
only ₹550 was spent on matches, and each match costs the same ₹50. Because the
o
.c never changes, ₹550 can be made up in only oneaway g — 11 matches.
m
cost per match

l a se polynomial takes each output value exactly once, which is why solving ax + b
a=g c with a ≠ 0 gives a single value of the variable.
A linear

o m
e
. c
m(200) from the repeating part (50m) before ag
s
Tip: Always separate the one-time part
a away would wrongly suggest 15 matches.
agl
dividing. Dividing 750 by 50 straight

co m
se m.
o m
Think and Reflect — Page 20
g l a
m .c Polynomials a
se
Section 2.2 Linear

l a
ag Q1 We have learnt that to evaluate the value of an algebraic expression, we substitute
m a s
.co agl
a value of the variable in the given expression. Consider Example 3, where the wire

se m
is bent to form a rectangle. Here, the area of the rectangle, 10x – x², is a function of

l a
x. Can you interpret this as an input-output process? What value does the
g
expression take when x = 6 cm? a

co m
m .
as e
com l
Yes. The machine takes in a length x and returns the area of the rectangle that the 20 cm wire
. a g
em
then makes.

a s
agl
m
input x = 6 → 10 × 6 − 6²

a se
= 60 − 36
. com a g l
m
ase
agl
= 24 cm²

m
Check by the shape: if the length is 6 cm, the width is 10 − 6 = 4 cm, and 6 × 4 = 24 cm². ✓

. co
em
m l as
.co
INPUT X (CM) 2 4 6 8

a g
se m Output 10x − x² (cm²)
l a
16 24 24 16

ag
.c
m
Step — +8 0 −8

m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why this machine is not linear: the steps in the bottom row are not equal — they
are +8, 0, −8. Two different inputs, 4 and 6, even give the same output, 24. A linear
machine can never do that. The reason is the x² term: subtracting consecutive values
leaves an x behind instead of cancelling it, so the step size itself depends on where
you are. That is why 10x − x² is called a quadratic function while 2x + 3 is linear.

Try This: feed in x = 5. You get 25 cm², the largest output of all — the square. Beyond
it the outputs come back down, which is why the graph of this machine is a curve,
not a straight line.

Exercise Set 2.2 — Page 21
Section 2.2 Linear Polynomials

Q1 Find the value of the linear polynomial 5x – 3 if: (i) x = 0 (ii) x = –1 (iii) x = 2

Substitute each value in turn.

(i) 5(0) − 3 = 0 − 3 = −3

(ii) 5(−1) − 3 = −5 − 3 = −8

(iii) 5(2) − 3 = 10 − 3 = 7

Why the answers rise in equal steps: from x = −1 to x = 0 the value goes −8 → −3, a
rise of 5; from x = 0 to x = 1 it would go −3 → 2, again 5. In general 5(x + 1) − 3 − (5x −
3) = 5, the coefficient of x. Each one-unit step in the input changes the output by
exactly the coefficient — this is the number we will soon call the slope.

Tip: Put the substituted value in brackets, especially when it is negative. Writing 5−1
instead of 5(−1) is the commonest slip here.

Page 10 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q2 Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0 (ii) s = –3 (iii) s = 4

(i) 7(0)² − 4(0) + 6 = 0 − 0 + 6 = 6

(ii) 7(−3)² − 4(−3) + 6

= 7(9) + 12 + 6

= 63 + 12 + 6 = 81

(iii) 7(4)² − 4(4) + 6

= 7(16) − 16 + 6

= 112 − 16 + 6 = 102

Why the two minus signs in (ii) do different things: (−3)² = +9, because squaring a
negative gives a positive — so the first term is +63, not −63. But −4(−3) = +12,
because a minus times a minus is a plus. Handling these separately is the whole
difficulty of the question.

Check it yourself: the outputs 6, 81, 102 are not in a linear pattern — unlike Q1,
equal steps in s do not give equal steps here. That is the visible signature of the s²
term.

Q3 The present age of Salil’s mother is three times Salil’s present age. After 5 years,
their ages will add up to 70 years. Find their present ages.

Let Salil’s present age be x years. Then his mother’s present age is 3x years.

After 5 years: Salil = x + 5, mother = 3x + 5

(x + 5) + (3x + 5) = 70

4x + 10 = 70

4x = 60

x = 15

Page 11 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Salil is 15 years old and his mother is 45 years old.
Check: in 5 years they will be 20 and 50, and 20 + 50 = 70. ✓

Why one letter is enough: the first sentence ties the two ages together, so the
mother’s age is not a second unknown — it is 3x. That leaves one unknown and one
condition, and 4x + 10 is a linear polynomial; setting it equal to 70 gives a linear
equation with exactly one solution. Note also that 5 is added twice, once for each
person; forgetting that gives 4x + 5 = 70 and a wrong answer.

Tip: the age gap 45 − 15 = 30 years never changes. You can use that as a second
check on any age problem.

Q4 The difference between two positive integers is 63. The ratio of the two integers is
2:5. Find the two integers.

The ratio 2 : 5 lets both numbers be written with one letter. Let them be 2k and 5k.

5k − 2k = 63

3k = 63

k = 21

2k = 42, 5k = 105

Check: 105 − 42 = 63 ✓ and 42 : 105 = 2 : 5 (divide both by 21). ✓

Why the “k” trick is legitimate: saying the ratio is 2 : 5 means the two numbers are
2 and 5 parts of the same size. Calling that common part k captures the ratio exactly,
and turns two unknowns into one. Every pair in the ratio 2 : 5 — 4 and 10, 6 and 15,
42 and 105 — is of the form (2k, 5k); the difference condition then picks out the
single pair we want.

Check it yourself: the difference of 2k and 5k is 3k, so any such difference must be a
multiple of 3. 63 = 3 × 21 is, which is why whole numbers come out.

Page 12 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q5 Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a
total ₹88, how many coins does she have of each type?

Let the number of five-rupee coins be f. Then the number of two-rupee coins is 3f.

Value of the two-rupee coins = 2 × 3f = 6f

Value of the five-rupee coins = 5f

6f + 5f = 88

11f = 88

f=8

Ruby has 8 five-rupee coins and 24 two-rupee coins (32 coins in all).
Check: 24 × ₹2 = ₹48 and 8 × ₹5 = ₹40; ₹48 + ₹40 = ₹88. ✓

Why you must not add the coins directly: the ₹88 is a total of value, not of coins,
so each count has to be multiplied by what that coin is worth before adding. The
neat consequence is that every “bundle” of 3 two-rupee coins and 1 five-rupee coin is
worth ₹11, and 88 ÷ 11 = 8 tells you there are 8 such bundles.

Tip: Choose the letter for the smaller count (here the five-rupee coins). Then the
other count is 3f, a whole number automatically.

Q6 A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is
four times as long as the shorter piece. How long are the two pieces?

Let the shorter piece be x feet. Then the longer piece is 4x feet.

x + 4x = 300

5x = 300

x = 60

The pieces are 60 feet and 240 feet.

Page 13 of 86

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Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

Check: 60 + 240 = 300 ✓ and 240 = 4 × 60. ✓
co m
e m.
m l as
.co a g
Why 5 appears: the whole fence is made of the shorter piece plus four more of the

s em
same size — 5 equal parts in all. So the shorter piece is one-fifth of 300 and the
a
gl piece is four-fifths. Reading the ratio 1 : 4 as “5 equal parts” is a faster route
alonger
to the same linear equation.

co m
m . ag
l a se
Tip: The question says the pieces are of different sizes, which is consistent: 60 ≠ 240.
agequal answers where it promised different ones, re-
If a problem ever gives you two
read the set-up.

co m
se m.
c o m l a
g and its perimeter is 24
Q7 m If .the length of a rectangle is three more than twice its width
a
l a se cm, what are the dimensions of the rectangle?
ag
m a s
c o agl

.
m(2w + 3) cm.
s e
Let the width be w cm. Then the length is
a
agl
Perimeter = 2(length + width)

co m
2[(2w + 3) + w] = 24
m .
m as e
.co
2(3w + 3) = 24
a g l
s m + 6 = 24
e6w
gl a
a 6w = 18
se m
w=3
com g l a
m . a
ase
Width = 3 cm, length = 2(3) + 3 = 9 cm.
agl
Check: perimeter = 2(9 + 3) = 24 cm ✓ and 9 is three more than twice 3. ✓

. c om
Why the bracket matters: “three more than twice the width”em
m a s is 2w + 3, and the
o
.cperimeter aglWriting 2 × 2w + 3 + w
formula doubles the whole of (length + width).
m instead of 2(2w + 3 + w) leaves the 3 undoubled and gives w = 21/5, which is not a
l a se
ag
c
sensible answer here. The perimeter 6w + 6 is itself a linear polynomial in w, so
m .
setting it to 24 has exactly one solution.
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Check it yourself: the area is 27 cm². Now try the same shape rule with perimeter
36 cm — you should get w = 5, length 13.

Think and Reflect — Page 22
Section 2.3 Exploring linear patterns

Q1 Predict the number of squares in the next three stages of the pattern and write the
sequence of numbers up to Stage 7 of the pattern.

Stage 1 Stage 2 Stage 3 Stage 4

Fig. 2.4, page 21 — the growing pattern of square tiles, Stages 1 to 4.

Stages 1 to 4 of Fig. 2.4 use 1, 3, 5 and 7 tiles. Each new stage adds two tiles — one to each
column — so the next three stages use 9, 11 and 13 tiles.

STAGE 1 2 3 4 5 6 7

Number of square tiles 1 3 5 7 9 11 13

Sequence up to Stage 7: 1, 3, 5, 7, 9, 11, 13.

Page 15 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Stage 5 Stage 6 Stage 7

Stages 5, 6 and 7 drawn out: each is a two-wide column with one extra tile on top, so the growth is
exactly 2 tiles a stage.

Why exactly two each time: look at the shape, not only at the numbers. Stage n is a
block two tiles wide and (n − 1) tiles tall, with a single tile perched on top of the right
column. Moving to the next stage raises the block by one row — that is 2 tiles — and
the lone tile on top just moves up with it. The count is therefore 2(n − 1) + 1 = 2n − 1,
and the constant difference 2 is the coefficient of n.

Tip: Predicting from the picture and from the rule should agree. 2(7) − 1 = 13, which
matches the drawing of Stage 7.

Think and Reflect — Page 22

Page 16 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.3 Exploring linear patterns

Q1 Using the expression 2n – 1, can you find out how many tiles will be there in the
15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles
and 47 tiles?

Stage 1 Stage 2 Stage 3 Stage 4

Fig. 2.4, page 21 — the growing pattern of square tiles, Stages 1 to 4.

Forward — stage given, tiles wanted. Substitute in 2n − 1.

n = 15: 2(15) − 1 = 30 − 1 = 29 tiles

n = 26: 2(26) − 1 = 52 − 1 = 51 tiles

Backward — tiles given, stage wanted. Now 2n − 1 becomes a linear equation.

2n − 1 = 21 → 2n = 22 → n = 11

2n − 1 = 47 → 2n = 48 → n = 24

So Stage 11 has 21 tiles and Stage 24 has 47 tiles.

Why the rule can be run both ways: 2n − 1 is a machine that doubles and subtracts
1. To reverse it you undo those steps in the opposite order — add 1, then halve.
Because doubling and halving are exact inverses, each tile-count comes from one
and only one stage; there is never a choice. This is the same one-solution property
that made the chess-club question have a single answer.

Page 17 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Check it yourself: 2n − 1 is always odd, so no stage can ever hold 20 or 50 tiles. Ask
for 20 and the equation gives n = 10.5, which is not a stage number.

Think and Reflect — Page 23
Section 2.3 Exploring linear patterns (Example 7)

Q1 What amount will be left on the 15th day? How many days will it take for the entire
amount to be spent?

Bela starts with ₹100 and spends ₹5 a day, so after n days she has ₹(100 − 5n).

n = 15: 100 − 5(15) = 100 − 75 = ₹25

The money runs out when the amount left is zero.

100 − 5n = 0

5n = 100
n = 20 days

Why this is decay, and where it stops: the coefficient of n is −5, so every extra day
subtracts a fixed ₹5 — a linear pattern running downwards. The pattern is only
meaningful while the amount left is not negative, i.e. while 100 − 5n ≥ 0, that is n ≤
20. On day 20 the purse is exactly empty; on day 21 the formula would say −₹5,
which the situation does not allow. Real-world models carry such limits even when
the algebra does not.

Tip: the value at which a linear polynomial becomes 0 is worth naming — here it
answers “when does it run out?” in one step.

Think and Reflect — Page 23

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Class 9 Maths Chapter 2 Introduction to Linear Polynomials
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Section 2.3 Exploring linear patterns (Example 8)
co m
e m.
m l as
Q1
.co
For how many km will the fare be ₹130?
m a g
l a se
a g

m
For a journey of n km (with n ≥ 2) the fare is ₹25 + 15(n − 2) = ₹(15n − 5).

. co ag
e m
g l as
a
15n − 5 = 130

15n = 135

co m
m.
n = 9 km

o m l a se
g and ₹15 for each of the
.cother way: for 9 km the meter charges ₹25 for the first 2akm
m
se 7 km: 25 + 15 × 7 = 25 + 105 = ₹130. ✓
Check the
l a
ag
remaining

m a s
agl
Why the formula reads 15n − 5 and not 15n + 25: the ₹15 rate does not start until

m .co
se
after 2 km, so it applies to (n − 2) km, giving 25 + 15n − 30 = 15n − 5. The −5 is what is

g l a
left of the base fare once the two free kilometres have been paid for at the higher
a
rate. This also explains why the rule is stated only for n ≥ 2 — for a 1 km ride the fare

m
is ₹25, not 15(1) − 5 = ₹10.

. co
em
m l as
.co a g
Check it yourself: use the rule for n = 2: 15(2) − 5 = ₹25, which agrees with the table.

se m
a
The formula stitches on correctly at the point where it starts to apply.

ag l
se m
Exercise Set 2.3 — Pages 23–24.co
m g l a
em a
Section 2.3 Exploring linear patternsas
agl
A student has ₹500 in her savings bank account. She gets ₹150 every month as
co m
.
Q1

se the amount she will
pocket money. How much money will she have at the end of every month from the
m
o m l a
.c have in the nth month. ag
second month onwards? Find a linear expression to represent

se m
g l a
a c
m .
m a s e
co agl
She begins with ₹500 and adds ₹150 at the end of each month.

m .
as e
END OF MONTH N
a g l 1 2 3 4 5 …

Amount (₹) 650 800 950 1100 1250 …

co m
m .
m ase
.co


a g l Page 19 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

From the second month onwards the balances are ₹800, ₹950, ₹1100, ₹1250, … — each ₹150
more than the one before.

Amount in the nth month = ₹(500 + 150n)

Check: n = 2 gives 500 + 300 = ₹800. ✓

Why the expression splits this way: ₹500 is there before any pocket money arrives,
so it is the constant term — the value at n = 0. The ₹150 arrives once per month, so it
is multiplied by n. Since the difference between consecutive balances is the fixed
₹150, this is a linear pattern of degree 1, and 150 is its slope.

Try This: when will she first have more than ₹3000? Solve 500 + 150n > 3000, giving n
> 16.67, so in the 17th month.

Q2 A rally starts with 120 members. Each hour, 9 members drop out of the group. How
many members will remain after 1, 2, 3, … hours? Find a linear expression to
represent the number of members at the end of the nth hour.

AFTER N HOURS 1 2 3 4 5 …

Members left 111 102 93 84 75 …

Members at the end of the nth hour = 120 − 9n

Check: n = 3 gives 120 − 27 = 93. ✓

Why the coefficient is negative: members leave, so each hour subtracts a fixed 9
— this is linear decay, and its slope is −9. As in Bela’s pocket money, the model has a
natural end: 120 − 9n ≥ 0 requires n ≤ 13⅓. At n = 13 there are 3 members left, and
within the next hour the rally is over. Beyond that the expression would report a
negative number of people, which is meaningless.

Page 20 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Check it yourself: the count never reaches exactly 0, since 120 is not a multiple of 9.
Solving 120 − 9n = 0 gives n = 13⅓ — useful as an estimate of when the rally empties,
but not an hour number.

Q3 Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm,
(ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.

With the length fixed at 13 cm, area = 13 × breadth.

(i) 13 × 12 = 156 cm²

(ii) 13 × 10 = 130 cm²

(iii) 13 × 8 = 104 cm²

If the breadth is b cm, Area A = 13b cm²

BREADTH B (CM) 8 10 12

Area A = 13b (cm²) 104 130 156

Step for ∆b = 2 — +26 +26

Why area is linear here although area is usually not: area lb involves two
variables and is not linear in general — that was the point of the garden in Example
2. But here the length is held fixed at 13, so only one variable is free, and 13b has
degree 1 in b. Every extra centimetre of breadth adds a strip 13 cm long, i.e. 13 cm²
— a constant, which is exactly what makes the pattern linear.

Tip: the constant term is 0: zero breadth, zero area. Its graph is the line A = 13b
through the origin.

Page 21 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q4 Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the
volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern
representing the volume of the rectangular box.

Length × breadth = 7 × 11 = 77 cm², and that base is fixed. So volume = 77 × height.

(i) 77 × 5 = 385 cm³

(ii) 77 × 9 = 693 cm³

(iii) 77 × 13 = 1001 cm³

If the height is h cm, Volume V = 77h cm³

HEIGHT H (CM) 5 9 13

Volume V = 77h (cm³) 385 693 1001

Step for ∆h = 4 — +308 +308

Why a volume can behave linearly: volume is a product of three lengths, so in
general it is a degree-3 quantity. Fixing two of them turns 7 × 11 into the single
number 77, and the box grows by stacking identical slabs of 77 cm³ — one slab per
centimetre of height. Equal steps in h therefore give equal steps in V, and the step is
4 × 77 = 308 cm³ for every 4 cm.

Check it yourself: 1001 − 693 = 308 and 693 − 385 = 308 — a good arithmetic check
on all three volumes at once.

Page 22 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q5 Sarita is reading a book of 500 pages. She reads 20 pages every day. How many
pages will be left after 15 days? Express this as a linear pattern.

Pages read in 15 days = 20 × 15 = 300

Pages left = 500 − 300 = 200 pages

Pages left after n days = 500 − 20n

DAY N 0 5 10 15 20 25

Pages left 500 400 300 200 100 0

Why it is a linear pattern: the number left drops by the same 20 pages every day,
so consecutive terms differ by the constant −20. That constant difference is the
definition of a linear pattern, and it is the coefficient of n in 500 − 20n. Because the
decrease never speeds up or slows down, we can jump straight to day 15 instead of
subtracting 20 fifteen times.

Tip: Solve 500 − 20n = 0 to find that she finishes the book on the 25th day — and 500
is exactly 25 × 20, so the last day is a full day’s reading.

Think and Reflect — Page 24
Section 2.4 Linear growth and linear decay (Example 9)

Q1 What is the cost for travelling 15 km? For how many kilometres will the cost of the
journey be ₹700?

The cost function is C(d) = 100 + 60d.

Page 23 of 86

ase
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
e m.
C(15) = 100 + 60(15)
m l as
= 100 + 900
m .co a g
l a
= ₹1000se
a g
Now run the machine backwards.
co m
e m . ag
g l as
a
100 + 60d = 700

60d = 600

co m
m.
d = 10 km

m as e
.co a g l
a s emthe ₹100 must be taken out first: the ₹100 is a fixed charge that does not
gl
Why
a depend on the distance — it is the value of C at d = 0. Only the remaining ₹600 was

s
bought at the rate of ₹60 per km. Dividing 700 by 60 without removing the fixed part
m a
.co agl
would give 11.67 km, which is wrong. In the language we are building: 100 is the y-

se m
a
intercept and 60 is the slope.

a g l

m
Check it yourself: C(10) = 100 + 600 = ₹700 ✓, and going from 10 km to 15 km adds
. co
m
5 × ₹60 = ₹300, taking ₹700 to ₹1000.

m as e
.co a g l
se m
g l a
a Think and Reflect — Page 25
se m
com a
Section 2.4 Linear growth and linear decay (Example 10)

. a g l
e m
s the water at the end of 5 months?
What will be the heightaof
Q1
a g l
co m
m .
The height after t months is h(t) = 3 − 0.5t.
m as e
.co a g l
se m h(5) = 3 − 0.5 × 5
g l a
a c
= 3 − 2.5
m .
m a s e
. co agl
= 0.5 m

e m
g l as
a

co m
m .
m ase
.co


a g l Page 24 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why this is decay and how far it can be trusted: the coefficient of t is −0.5, so the
water level drops by a fixed half-metre every month regardless of how full the tank
is. That constant drop is what makes h linear, and its negative sign is what makes it
decay rather than growth. The model also tells us when the tank empties: 3 − 0.5t = 0
gives t = 6 months. Past that the formula would report a negative depth, so the
model stops being physical there.

Check it yourself: extend the book’s table — 3, 2.5, 2, 1.5, 1, 0.5, 0 for t = 0 to 6. The
differences are all −0.5, as a linear pattern demands.

Exercise Set 2.4 — Pages 25–26
Section 2.4 Linear growth and linear decay

Q1 Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. (i) Find the
height after 7 months. (ii) Make a table of values for t varying from 0 to 10 months
and show how the height, h, increases every month. (iii) Find an expression that
relates h and t, and explain why it represents linear growth.

(i) Seven months of growth at 0.5 ft a month adds 3.5 ft.

h = 1.75 + 0.5 × 7

= 1.75 + 3.5

= 5.25 feet

(ii)

MONTH T 0 1 2 3 4 5 6 7 8 9 10

Height h (ft) 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 6.75

(iii)

h = 1.75 + 0.5t

Why this is linear growth: subtract two consecutive heights and see what survives.

Page 25 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

h(t + 1) − h(t) = [1.75 + 0.5(t + 1)] − [1.75 + 0.5t]

= 0.5

The t cancels completely, so the monthly increase is the same fixed 0.5 ft at every
stage — the plant does not grow faster as it gets taller. A fixed positive amount
added over equal intervals is precisely the definition of linear growth. The expression
has degree 1, the constant 1.75 is the height at t = 0, and 0.5 is the slope.

Check it yourself: in the table every step is +0.50 ft, and the 7-month entry 5.25
agrees with part (i).

Q2 A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find
the value of the phone after 3 years. (ii) Make a table of values for t varying from 0
to 8 years and show how the value of the phone, v, depreciates with time. (iii) Find
an expression that relates v and t, and explain why it represents linear decay.

(i)

v = 10000 − 800 × 3

= 10000 − 2400

= ₹7600

(ii)

YEAR T 0 1 2 3 4 5 6 7 8

Value v (₹) 10000 9200 8400 7600 6800 6000 5200 4400 3600

(iii)

v = 10000 − 800t

Page 26 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why this is linear decay: v(t + 1) − v(t) = −800 for every t — a fixed amount is
subtracted over equal intervals, and the coefficient of t is negative. Note what this
model assumes: the phone loses the same ₹800 in its eighth year as in its first. Real
depreciation is often a fixed percentage instead, which would not be linear. The
straight-line assumption is what makes the arithmetic here so simple.

Tip: the model runs out at 10000 − 800t = 0, i.e. t = 12.5 years. Beyond that the value
would go negative, which no second-hand phone does.

Q3 The initial population of a village is 750. Every year, 50 people move from a nearby
city to the village. (i) Find the population of the village after 6 years. (ii) Make a
table of values for t varying from 0 to 10 years and show how the population, P,
increases every year. (iii) Find an expression that relates P and t, and explain why it
represents linear growth.

(i)

P = 750 + 50 × 6

= 750 + 300

= 1050 people

(ii)

YEAR T 0 1 2 3 4 5 6 7 8 9 10

Population P 750 800 850 900 950 1000 1050 1100 1150 1200 1250

(iii)

P = 750 + 50t

Page 27 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why this is linear growth: the yearly increase is P(t + 1) − P(t) = +50, the same
number every year, so equal intervals of time bring equal increases in population. It
is worth seeing why: the 50 arrivals come from the city, not from the village itself, so
the number arriving does not depend on how big the village has become. If instead
the village grew by births at a fixed rate per resident, the yearly increase would itself
grow, and the pattern would not be linear.

Try This: when does the village first cross 2000? Solve 750 + 50t > 2000 to get t > 25,
so in the 26th year.

Q4 A telecom company charges ₹600 for a certain recharge scheme. This prepaid
balance is reduced by ₹15 each day after the recharge. (i) Write an equation that
models the remaining balance b(x) after using the scheme for x days. Explain why it
represents linear decay. (ii) After how many days will the balance run out? (iii) Make
a table of values for x varying from 1 to 10 days and show how the balance b(x),
reduces with time.

(i)

b(x) = 600 − 15x

It is linear decay because b(x + 1) − b(x) = −15 for every x: a fixed amount is removed over equal
intervals, and the coefficient of x is negative. The graph would be a straight line falling from (0,
600).
(ii) The balance runs out when b(x) = 0.

600 − 15x = 0

15x = 600

x = 40 days

(iii)

DAY X 1 2 3 4 5 6 7 8 9 10

Balance b(x) (₹) 585 570 555 540 525 510 495 480 465 450

Page 28 of 86

ase
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Why the zero of the polynomial answers (ii): “the balance runs out” means the

m l a se
output of the machine is 0, so we ask which input sends b to 0. For a linear
o g — here
.c ax + b with a ≠ 0 there is exactly one such input, x = −b/a
a
m
polynomial

l a se = 40. That is why questions of the form “when does it finish?” always have
aag single clean answer in a linear model.
−600/(−15)

o m
Check it yourself: the table drops by m
e
. c ag
s
exactly ₹15 each column, and continuing that
a 30 days reaches ₹0 on day 40.
agl
fall from ₹450 on day 10 for another

co m
se m.
o m
Think and Reflect — Page 27
g l a
m .c Relationships (Example 11) a
se
Section 2.5 Linear

l a
ag Q1 Can you guess what the numbers 20 and 150 in the equation y = 20x + 150
m a s
.co agl
represent?

se m
g l a
a

20 = the charge per GB of data, in rupees
com
m .
m
150 = the fixed monthly fee, in rupees
as e
.co a g l
a s em
agl Why they must mean that: read the equation at two special inputs.
• Put x = 0. Then y = 150 — the bill for a month in which no data is used at all. Only
se m
the fixed fee can survive, so 150 is that fee.
com g l a
m . a
ase
• Increase x by 1. Then y increases by 20(x + 1) + 150 − (20x + 150) = 20, whatever x

a gl
was. One more GB always costs ₹20, so 20 is the rate per GB.
These are exactly the roles of slope and y-intercept: 20 is the slope (change per unit)
and 150 the y-intercept (starting value).
co m
m .
o m l a se
.cCheck it yourself: at 10 GB, 20(10) + 150 = ₹350 and atag20 GB, 20(20) + 150 = ₹550,
m which are the two bills the student actually observed. ✓
l a se
ag
.c
s e m
m a
Exercise Set 2.5 — Page 27m.c
o agl
l a se
ag

co m
m .
m ase
.co


a g l Page 29 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.5 Linear Relationships

Q1 A learning platform charges a fixed monthly fee and an additional cost per digital
learning module accessed. A student observes that when she accessed 10 modules,
her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly
bill y depends on the number of modules accessed, x, according to the relation y =
ax + b, find the values of a and b.

Substitute the two observations into y = ax + b.

400 = 10a + b …(1)

500 = 14a + b …(2)

Subtracting (1) from (2) removes b:

100 = 4a

a = 25

From (1): b = 400 − 10(25) = 400 − 250 = 150

a = 25, b = 150, so y = 25x + 150

Check: x = 10 gives 250 + 150 = ₹400 ✓ and x = 14 gives 350 + 150 = ₹500. ✓

Why two observations are exactly enough: y = ax + b hides two unknowns, so we
need two independent facts. Each bill gives one equation, and subtracting them
eliminates b in one stroke — the difference in the bills, ₹100, is caused only by the
extra 4 modules, so the rate must be 100 ÷ 4 = ₹25 per module. The fixed fee is then
whatever is left over: ₹150. Geometrically, two points determine one straight line.

Tip: Subtracting the equations is quicker than substituting, and it is where the
meaning is: a = (change in y) ÷ (change in x).

Page 30 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q2 A gym charges a fixed monthly fee and an additional cost per hour for using the
badminton court. A student using the gym observed that when she used the
badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her
bill was ₹1100. If the monthly bill y depends on the hours of the use of the
badminton court, x, according to the relation y = ax + b, find the values of a and b.

800 = 10a + b …(1)

1100 = 15a + b …(2)

(2) − (1):

300 = 5a

a = 60

From (1): b = 800 − 10(60) = 800 − 600 = 200

a = 60, b = 200, so y = 60x + 200

Check: 60(15) + 200 = 900 + 200 = ₹1100. ✓

Reading the answer back into the situation: the court costs ₹60 per hour and the
gym charges a fixed ₹200 a month whether the court is used or not. That reading is
what makes the algebra worth doing: the extra ₹300 between the two bills bought
exactly 5 extra hours, so an hour is ₹60; and ₹800 for 10 hours is ₹600 of court time
plus ₹200 that was never about the court.

Try This: what would 20 hours cost? 60(20) + 200 = ₹1400. And for what bill does she
get 25 hours? ₹1700.

Page 31 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q3 Consider the relationship between temperature measured in degrees Celsius (°C)
and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that
ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100
degrees Celsius and 212 degrees Fahrenheit. (Hint: When °C = 0, °F = 32 and when °C
= 100, °F = 212. Use this information to find a and b, and thus, the linear relationship
between °C and °F.)

Here °F is the input and °C the output. The two fixed points give:

0 = 32a + b …(1) (ice melts)

100 = 212a + b …(2) (water boils)

(2) − (1):

100 = 180a

a = 100/180 = 5/9

From (1): b = −32a = −32 × 5/9 = −160/9

a = 5/9, b = −160/9

°C = (5/9)°F − 160/9 = (5/9)(°F − 32)

Check: °F = 32 gives (5/9)(0) = 0 °C ✓; °F = 212 gives (5/9)(180) = 100 °C. ✓

Why the slope is 5/9: between melting and boiling, Celsius moves 100 degrees
while Fahrenheit moves 212 − 32 = 180 degrees. The same physical interval is being
cut into 100 parts on one scale and 180 on the other, so one Fahrenheit degree is
100/180 = 5/9 of a Celsius degree — and that ratio is the slope. The constant −160/9
is there only because the two scales put their zeros in different places; factoring it
out as (5/9)(°F − 32) makes that shift visible.

Try This: a comfortable Delhi winter day at 68 °F is (5/9)(68 − 32) = (5/9)(36) = 20 °C.
And the two scales agree at −40: check that −40 °F is −40 °C.

Think and Reflect — Page 28

Page 32 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.6 Visualising linear relationships

Q1 Identify other points on the line by completing the following table.

8
y-axis
7 B (3, 7)

6

5

4

3

2

1 A (0, 1)
x-axis
−2 −1 0 1 2 3 4 5
−1

Fig. 2.5, page 28 — the straight line y = 2x + 1, with A (0, 1) and B (3, 7) marked on it.

X 1 2 5 7 9 12 20

y 3 15

The line is y = 2x + 1, so double each x and add 1.

X 1 2 5 7 9 12 20

y = 2x + 1 3 5 11 15 19 25 41

So (1, 3), (2, 5), (5, 11), (7, 15), (9, 19), (12, 25) and (20, 41) all lie on the line.

Page 33 of 86

as e
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Why substituting is a genuine test: a point lies on a line exactly when its

m l a se
coordinates satisfy the equation of that line — the equation is nothing but the
o g because 2(7) + 1
.c rule for the set of points on the line. So (7, 15) belongs
a
m
membership

l a s✓,e while (7, 14) does not, and no amount of careful drawing would change that.
g
= 15
aChecking by substitution is exact; reading off a graph is only approximate.

o m
Tip: the two given entries let you find m
e
. c ag
s
the rule even if you had forgotten it: from x = 1
a over 6 steps, so 2 per step — the slope — and
to x = 7 the value climbs 15 − 3 = l12
ag
then 3 = 2(1) + 1 gives the intercept.

co m
em.
com g l as
m .
In-text Questions — Page 29 a
a e Visualising linear relationships (Examples 13, 14)
s2.6
ag l
Section

m 2.7. Join the points (– 3, 6) and (3, – 6) using a s
coFig. agl
Let us plot the points (– 3, 6), (– 2, 4), (0, 0), (1, – 2), (2, – 4), (3, – 6) in the coordinate
plane on a graph paper as shown .in
Q1

s m five points lie on a straight line. Can you guess the
eall
a
gl at the relationship between the x and y coordinates
a ruler. Doing so, observe that
equation of this line byalooking
of each point?

co m
m .
m as e
.co a g l
sem
y-axis

a
(−3, 6) 6

agl (−2, 4) 4

se m
com g l a
.
2

m a
ase
(0, 0) x-axis
−12 −10
agl −6
−8 −4 −2 0 2 4 6 8 10 12
−2 (1, −2)

co m
−4 (2, −4)
m .
m as e
.co a g l
(3, −6)
m
−6

l a se
ag
.c
Fig. 2.7, page 29 — the six points plotted on graph paper.
s e m
m a
e m . co agl
g l as
ANSWER a
m
In every listed point the y-coordinate is −2 times the x-coordinate.

. co
e m
m l as
.co a g
Page 34 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

X −3 −2 0 1 2 3

y 6 4 0 −2 −4 −6

y÷x −2 −2 — −2 −2 −2

Equation of the line: y = −2x

Why this is the right form: the ratio y/x is the same for every point, so y is a
constant multiple of x — that constant is the slope, here −2. Because there is no
constant term, the point (0, 0) satisfies the equation, and indeed the line passes
through the origin. The negative slope shows itself in the picture: as x moves right, y
comes down, so the line falls from upper left to lower right.

Check it yourself: substitute the point that was not used to draw the line, say (−2, 4):
−2(−2) = 4. ✓ A single failed check would mean the guessed equation is wrong.

Q2 Draw the graphs of y = ½x, y = x, y = 2x by selecting suitable points on these lines.
(Hint: In order to graph y = ½x, we could take the points (0, 0) and (4, 2). Can you
verify that these lie on the line?)

Verifying the hint’s two points on y = ½x:

(0, 0): ½ × 0 = 0 ✓

(4, 2): ½ × 4 = 2 ✓

Both satisfy the equation, so both lie on the line — and two points are all a straight line needs.
Convenient points for the three lines:

LINE POINT 1 POINT 2 SLOPE

y = ½x (0, 0) (4, 2) ½

y=x (0, 0) (4, 4) 1

y = 2x (0, 0) (4, 8) 2

Page 35 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why choose x = 0 and x = 4: x = 0 always gives the y-intercept for free, and a
multiple of the denominator (here 4, a multiple of 2) keeps the second coordinate a
whole number so it can be plotted exactly on the grid. Picking x = 3 for y = ½x would
force you to plot 1.5 by eye. Also, taking the two points far apart makes the ruled line
more accurate — a small slip in plotting two nearby points tilts the whole line.

Tip: All three lines pass through the origin because each has the form y = ax with no
constant term. Their difference is only how steeply they climb.

In-text Questions — Page 31

Page 36 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.6 Visualising linear relationships (Fig. 2.9)

Q1 Fig. 2.9 shows all the three graphs on the same axes. Does this help you to conclude
anything about the linear equation y = ax, a > 0 as a varies? What happens when a >
1 and when a < 1? (Hint: You may also plot the equations y = 3x and y = ⅓x on the
same axes.)

y-axis
3
= ½x
y

x
2

=
y
1

x-axis
−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

−1
2x

−2
y=

−3

Fig. 2.9, page 31 — y = 2x, y = x and y = ½x drawn on the same axes.

Three conclusions, all visible at once when the lines share axes.

1. Every line y = ax passes through the origin (0, 0), whatever a is.
2. All of them rise from left to right, since a > 0.
3. The bigger a is, the steeper the line. Taking y = x as the reference: if a > 1 the line is steeper
than y = x; if a < 1 it is less steep.

Page 37 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

6
y y = 3x
5
y=x
4

3

2

1 y = x/3
x
-5 -4 -3 -2 -1 1 2 3 4 5
-1

-2

-3

-4

-5

-6

y = 3x, y = x and y = x/3 on one pair of axes. All three meet at the origin; only the steepness changes
with a.

Page 38 of 86

as e
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Why each conclusion must hold:

m l a se
• Put x = 0 in y = ax and you get y = 0 for every a. So (0, 0) is on all of them — there is
o
.c term to lift the line off the origin. a g
m
no constant

l a se x by 1 and y increases by a. So a is literally the rise per unit of run, i.e. the
g A larger rise over the same run is a steeper line.
• Increase
aslope.
• At x = 4, y = 3x is already at 12 while y = ⅓x has only reached 1⅓. The line y = x, with
com
.
slope 1, makes equal angles with the two axes, so it is the natural dividing case
e m ag
between the two behaviours.
g l as
a
Try This: plot y = 3x and y = ⅓x on the same axes as the hint suggests. They are
co m
mirror images of each other in the line y = x — swapping x and y turns one equation
em.
m l as
.co g
into the other.
m a
l a se
a g
In-text Questions — Page 33
m a s
m.co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.6 Visualising linear relationships (Fig. 2.11)

Q1 Fig. 2.11 shows all the three graphs on the same axes. Does this help you to
conclude anything about the linear equation y = – ax, a > 0, as a varies? What will
happen when a > 1 and when a < 1?

4 y-axis
y
=
−x
3
y=
−⅓x 2

1
x-axis
−7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7
−1

−2 y=

−3
−3x

−4

Fig. 2.11, page 33 — y = −3x, y = −x and y = −⅓x drawn on the same axes.

The picture mirrors the previous one, with one crucial change of sign.

1. Every line y = −ax still passes through the origin.
2. All of them now fall from left to right, because the slope −a is negative — these are pictures
of linear decay.
3. Steepness is decided by the size of a, not its sign: if a > 1 the line falls more steeply than y =
−x; if a < 1 it falls more gently.

Page 40 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

6
y = −3x y
5
y = −x
4

3

2

y = −x/3 1
x
-5 -4 -3 -2 -1 1 2 3 4 5
-1

-2

-3

-4

-5

-6

y = −3x, y = −x and y = −x/3. Compare with the previous figure: each line is its positive-slope partner
reflected in the x-axis.

Why the whole family is just a reflection: replacing a by −a replaces every output y
by −y and leaves x untouched. Geometrically that is a reflection in the x-axis, which
cannot change how steep a line is — only which way it leans. So the pair y = 3x and y
= −3x make equal angles with the x-axis on opposite sides. Increasing x by 1 now
decreases y by a, so a bigger a means a faster fall.

Page 41 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Tip: the useful summary is: the sign of the slope tells you growth or decay; the size of
the slope tells you how fast.

Think and Reflect — Page 33
Section 2.6 Visualising linear relationships

Q1 Differentiate between the graphs of the equations y = 3x + 1, and y = –3x + 1.

Y = 3X + 1 Y = −3X + 1

Slope +3 −3

y-intercept 1, at (0, 1) 1, at (0, 1)

Direction rises to the right (growth) falls to the right (decay)

Cuts the x-axis at (−⅓, 0) (⅓, 0)

Steepness the same — both change y by 3 for a 1-unit change in x

Page 42 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

6
y = −3x + 1 y y = 3x + 1
5

4

3

2
(0, 1)
1
x
-4 -3 -2 -1 1 2 3 4
-1

-2

-3

-4

-5

-6

Both lines cross the y-axis at (0, 1), but with opposite slopes; they are mirror images of each other in
the y-axis.

Page 43 of 86

as e
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Why they meet exactly once, and where: equate them.

m as e
.co a g l
se m
l a
3x + 1 = −3x + 1
g
a6x = 0

com
ag
x = 0, so y = 1
m .
as e
a g l
Their only common point is (0, 1) — the shared y-intercept. The x-intercepts come

m
from 3x + 1 = 0 and −3x + 1 = 0, giving x = −⅓ and x = ⅓: equal distances from the
co
m.
origin on opposite sides, which is the reflection in the y-axis showing up in the

as e
com in the y-axis. l
numbers. Replacing x by −x turns one equation into the other, and that substitution
. a g
em
is reflection

l a s
ag
Check it yourself: at x = 2 the two lines give 7 and −5. They are 12 units apart, and
m a s
.co agl
that gap grows by 6 for every extra unit of x — the difference of the two slopes.

se m
g l a
a
Think and Reflect — Page 35
co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 44 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Section 2.6 Visualising linear relationships (Fig. 2.13)

Q1 Does this help you to conclude anything about the linear equation y = ax + b when a
is fixed but b varies? (Hint: In these equations a = 2, and b takes the values –1, 1 and
5, respectively.)

5 y-axis

4

3
5

1
+

+

1
2
2x

2x

−
2x
y=

y=
1

y=
x-axis
−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7
−1

−2

Fig. 2.13, page 34 — y = 2x + 5, y = 2x + 1 and y = 2x − 1 drawn on the same axes.

Yes. With a = 2 fixed and b = −1, 1, 5 the three lines are parallel: same tilt, different heights.

LINE SLOPE A Y-INTERCEPT B CUTS THE Y-AXIS AT

y = 2x − 1 2 −1 (0, −1)

y = 2x + 1 2 1 (0, 1)

y = 2x + 5 2 5 (0, 5)

Page 45 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

8
y
7

6
y = 2x − 1
5

4
y = 2x + 5
3
y = 2x + 1
2

1
x
-5 -4 -3 -2 -1 1 2 3 4
-1

-2

-3

-4

-5

-6

Changing b slides the line straight up or down; it never tilts, so the three lines never meet.

Page 46 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Why they can never meet: suppose 2x + b₁ = 2x + b₂ at some x. The 2x cancels and
we are left with b₁ = b₂. So two lines with the same slope and different intercepts have
no common point at all — they are parallel. Better still, at any given x the vertical gap
between y = 2x + 5 and y = 2x + 1 is (2x + 5) − (2x + 1) = 4, the same everywhere. A
constant gap is exactly what “parallel” means.

Tip: Adding b to a linear expression shifts its graph b units vertically — up if b is
positive, down if negative. The slope, being the rate of change, is untouched by
adding a constant.

Exercise Set 2.6 — Page 36
Section 2.6 Visualising linear relationships

Q1 (i) Draw the graphs of the following sets of lines. In each case, reflect on the role of
‘a’ and ‘b’. (i) y = 4x, y = 2x, y = x

Each has b = 0, so plot the origin and one more point.

LINE POINTS USED A (SLOPE) B (Y-INTERCEPT)

y = 4x (0, 0), (1, 4) 4 0

y = 2x (0, 0), (2, 4) 2 0

y=x (0, 0), (4, 4) 1 0

Page 47 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

6
y = 4x y = 2x
y
5

4

3 y=x

2

1
x
-4 -3 -2 -1 1 2 3 4
-1

-2

-3

-4

-5

-6

y = 4x, y = 2x and y = x: same y-intercept 0, increasing slopes, increasing steepness.

Page 48 of 86

ase
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
m.
Role of a and b here: b = 0 for all three, so all three are pinned to the origin — that

m l a se
is what a zero y-intercept means. The only thing left to vary is a, and it controls the
o
c 1 unit lifts y = 4x by 4, y = 2x by 2 and y = x by 1. Sinceagall slopes are
tilt: a run .of
m
se all three represent linear growth, and the three lines fan out from the origin
g l a
positive,
awithout ever being parallel.

o m
Check it yourself: at x = 3 the heightsm
e
. c ag
s
are 12, 6 and 3. The line with the biggest

a g
slope is always the highest one to lathe right of the origin — and the lowest one to the
left.

co m
se m.
Q1 (ii) .c o m l a
g reflect on the role of
m a
Draw the graphs of the following sets of lines. In each case,
e ‘a’ and ‘b’. (ii) y = – 6x, y = – 3x, y = – x
ag las
s
m a
em
.co agl
a s
LINE POINTS USED A (SLOPE) B

ag l
y = −6x (0, 0), (1, −6) −6 0

com
.
y = −3x (0, 0), (1, −3) −3 0

e m
m l as
.co g
0
a
y = −x (0, 0), (3, −3) −1

se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 49 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

6
y = −3x y
5

4

y = −x 3

2

1
x
-4 -3 -2 -1 1 2 3 4
-1

-2

-3

-4

-5

-6
y = −6x

All three fall from left to right through the origin; the more negative the slope, the steeper the fall.

Page 50 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Role of a and b here: again b = 0, so all three pass through the origin. Every slope is
negative, so all three are pictures of linear decay: raising x by 1 lowers y by 6, 3 and 1
respectively. Steepness is governed by how far a is from 0, not by its sign — y = −6x is
the steepest of the three and is exactly as steep as y = 6x would be, only leaning the
other way.

Tip: comparing this set with (i): y = −x is the mirror image of y = x in the x-axis, and
likewise for the other pairs.

Q1 (iii) Draw the graphs of the following sets of lines. In each case, reflect on the role of
‘a’ and ‘b’. (iii) y = 5x, y = –5x

LINE POINTS USED A B

y = 5x (0, 0), (1, 5) 5 0

y = −5x (0, 0), (1, −5) −5 0

Page 51 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

6
y
y = −5x 5 y = 5x

4

3

2

1
x
-4 -3 -2 -1 1 2 3 4
-1

-2

-3

-4

-5

-6

Equal and opposite slopes: the two lines cross at the origin and make equal angles with the x-axis.

Page 52 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Role of a and b here: the two slopes have the same size, 5, and opposite signs,
while b = 0 for both. So the lines are equally steep and cross at their common point,
the origin. One shows growth, the other decay. Replacing x by −x in y = 5x gives y =
−5x, which says the pair is symmetric about the y-axis; replacing y by −y gives the
same pair, so they are symmetric about the x-axis too.

Check it yourself: at x = 2 the lines are at y = 10 and y = −10 — equally far from the
x-axis, on opposite sides.

Q1 (iv) Draw the graphs of the following sets of lines. In each case, reflect on the role of
‘a’ and ‘b’. (iv) y = 3x – 1, y = 3x, y = 3x + 1

LINE POINTS USED A B

y = 3x − 1 (0, −1), (1, 2) 3 −1

y = 3x (0, 0), (1, 3) 3 0

y = 3x + 1 (0, 1), (1, 4) 3 1

Page 53 of 86

ase
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
em.
m l as
m .co a g
l a se 6
y
ag
5
m −1
ym=.co3x ag
l a se 4
a g
y = 3x
3 co m
e m.
m l as
.co y = 3x + 21 a g
a s em
agl
1 s
m a
em
.co x agl
a s
a gl
-4 -3 -2 -1 1 2 3 4
-1 co m
m .
m as e
.co a g l
a s em -2
agl
-3 se m
co m g l a
m . a
ase
agl
-4

-5 co m
m .
m as e
.co a g l
a s em -6
agl
.c
s e m
m a
. co agl
Same slope 3, intercepts −1, 0 and 1: three parallel lines, each 1 unit above the last.

e m
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Role of a and b here: this is the opposite experiment to (i). Now a is held fixed and b
is varied, so the tilt cannot change — the lines are parallel. At any chosen x, the
vertical gap between y = 3x + 1 and y = 3x is (3x + 1) − 3x = 1, the same everywhere,
so the three lines never converge. Changing b slides the whole line vertically;
changing a would rotate it.

Tip: two lines are parallel exactly when their slopes are equal and their intercepts are
not. Equal slopes and equal intercepts would make them the same line.

Q1 (v) Draw the graphs of the following sets of lines. In each case, reflect on the role of
‘a’ and ‘b’. (v) y = –2x – 3, y = –2x, y = 2x + 3

LINE POINTS USED A B

y = −2x − 3 (0, −3), (−2, 1) −2 −3

y = −2x (0, 0), (2, −4) −2 0

y = 2x + 3 (0, 3), (1, 5) 2 3

Page 55 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

7
y
y = −2x − 3 6

5

4

3

2

1
x
-5 -4 -3 -2 -1 1 2 3 4
-1

-2

-3

-4
y = 2x + 3 -5

-6 y = −2x

The first two are parallel (slope −2); the third has slope +2, so it crosses both.

Page 56 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Role of a and b here: the first two share the slope −2 and differ only in b, so they are
parallel, with a constant vertical gap of 3. The third line has slope +2, a different
number, so it is not parallel to them — it cuts each of them exactly once. Solving 2x +
3 = −2x gives 4x = −3, x = −¾, so it meets y = −2x at (−¾, 1½); solving 2x + 3 = −2x − 3
gives x = −½, the point (−½, 2).

Did you know? Sets (i) to (iv) each vary one thing at a time. Set (v) deliberately mixes
them, so that you have to decide parallelism by comparing slopes rather than by the
look of the printed equations. Had the third line been y = −2x + 3, all three would
have been parallel — compare the shape of the answer in that case.

End-of-Chapter Exercises — Pages 36–39
Chapter 2 Introduction to Linear Polynomials

Q1 Write a polynomial of degree 3 in the variable x, in which the coefficient of the x²
term is –7.

For example: x³ − 7x² + 2x + 5

Other correct answers: 4x³ − 7x², −x³ − 7x² + 1, 2x³ − 7x² − 3x.

The two conditions that must be met:
• Degree 3 means the coefficient of x³ must be non-zero. Writing −7x² + 2x + 5 would
satisfy the second condition but has degree 2, not 3.
• The x² term must be exactly −7x², sign included.
Everything else is free: the x term and the constant may be any numbers at all,
including 0. This is why the answer is not unique — the conditions fix two of the four
coefficients and leave two to choose.

Check it yourself: in x³ − 7x² + 2x + 5 the coefficients in order are 1, −7, 2, 5. The first
is not zero (degree 3 ✓) and the second is −7 ✓.

Page 57 of 86

Class 9 Maths Chapter 2 Introduction to Linear Polynomials AglaSem · NCERT Solutions

Q2 Find the values of the following polynomials at the indicated values of the
variables. (i) 5x² – 3x + 7 if x = 1 (ii) 4t³ – t² + 6 if t = a

(i) Substitute x = 1.

5(1)² − 3(1) + 7

=5−3+7

=9

(ii) Substitute t = a. The input is a letter, so the output is an expression.

4a³ − a² + 6

= 4a³ − a² + 6

Why (ii) is not a trick question: evaluating a polynomial means replacing the
variable by whatever is given — a number, another letter, even a whole expression.
Nothing says the input must be numerical. Here every t becomes an a and no
arithmetic can be done, so the answer stays in symbols. Note also that in (i) the
value at x = 1 equals the sum of the coefficients, 5 + (−3) + 7 = 9, because every
power of 1 is 1 — a quick check worth remembering.

Try This: evaluate the same polynomial in (ii) at t = 2a. You should get 4(2a)³ − (2a)² +
6 = 32a³ − 4a² + 6 — brackets are essential.

Q3 If we multiply a number by 5/2 and add 2/3 to the product, we get –7/12. Find the
number.

Let the number be x. The description translates directly into a linear equation.

Page 58 of 86

ase
Class 9 Maths Chapter 2 Introduction to Linear Polynomials
a g l AglaSem · NCERT Solutions

co m
e m.
(5/2)x + 2/3 = −7/12
m l as
.co
(5/2)x = −7/12 − 2/3
m a g
l a se
g
= −7/12 − 8/12
a= −15/12 = −5/4

co m
. ag
x = (−5/4) × (2/5)
em
x = −10/20 = −1/2
g l as
a
m
Check: (5/2)(−1/2) = −5/4; and −5/4 + 2/3 = −15/12 + 8/12 = −7/12. ✓
co
se m.
o m l a
Why the LCM 12 is the right common denominator: to subtract 2/3 from −7/12
g LCM of 12 and 3.
m .c be written over the same denominator, and 12 is the a
se dividing by 5/2 is the same as multiplying by its reciprocal 2/5 — that is what
both must

g l a
Then
a “undoing a multiplication” means for fractions. The polynomial (5/2)x + 2/3 is linear,

m a s
agl
so this equation has exactly one solution.
.co
a s em
l fractions at the start: multiply the whole equation by
Tip: an alternative is to cleargall
a
12 to get 30x + 8 = −7, so 30x = −15 and x = −½. Same answer, no fraction arithmetic
in the middle.
co m
m .
m as e
.co a g l
a s em A positive number is 5 times another number. If 21 is added to both the numbers,
agl Q4
then one of the new numbers becomes twice the other new number. What are the
numbers?
se m
com g l a
m . a
ase
agl

Let the smaller number be x. The other is 5x. After adding 21 they become x + 21 and 5x + 21.

m
Since 5x + 21 is the larger of the two, it must be the one that is twice the other.

. co
em
m l as
.co a g
5x + 21 = 2(x + 21)

se m 5x + 21 = 2x + 42
g l a
a c
3x = 21
m .
m a s e
. co agl
x=7

e m
g l as
The numbers are 7 and 35.
a
Check: adding 21 gives 28 and 56, and 56 = 2 × 28. ✓

co m
m .
m ase
.co


a g l Page 59 of 86