Introduction to Linear Polynomials: Class 9 Ganita Manjari · Chapter 2

Ganita Manjari Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

96 min
1 simulation
43 practice questions
31 MCQs

What you will learn in Introduction to Linear Polynomials

Name the terms, variables, coefficients and constant term of an algebraic expression, and find the degree of a polynomial in one variable.
Recognise linear polynomials, evaluate them as input-output rules and solve linear equations that come from everyday situations.
Spot a linear pattern from its constant difference and write a rule for its nth term.
Model linear growth and linear decay with rules such as V = 200 + 40t, and say where such a model stops making sense.
Find a linear relationship y = ax + b from two measurements and explain what a and b mean in context.
Draw the graph of y = ax + b from two points and read its slope, its y-intercept and which lines are parallel.

Original companion text aligned to NCERT Mathematics Class 9, Chapter 2, pp. 16–40. Not affiliated with or endorsed by NCERT.

2.1 Introduction

Watch the scorer at a school cricket match. She does not invent a new calculation for every batter. She keeps one rule in her head: a four is worth 4 runs, a six is worth 6, and runs taken between the wickets are added on. Algebra lets us write a rule like this once, with letters standing for the numbers that change from one innings to the next.

The scorer's rule is an algebraic expression. In this chapter we sort such expressions into families and then spend most of our time on the simplest useful family, the linear polynomials (रैखिक बहुपद). We will read them as rules, find them hiding in growing patterns, use them to describe steady change, and finally draw them as straight lines.

Example 1NCERT Example 1 · p. 16

A batter's score

In an inter-house match, Tanvi hits $f$ fours and $s$ sixes. She also scores 11 runs by running between the wickets. Write an expression for her total score. What is her score if she hits 3 fours and 2 sixes?

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

Let us name the parts of $4 f + 6 s + 11$ . The pieces that are added are its terms (पद). The letters $f$ and $s$ are variables (चर), because their values vary from one innings to another; you may have called them letter-numbers before. The number that multiplies a variable is its coefficient (गुणांक), and a term with no variable is a constant term.

Table 2.1The parts of an algebraic expression

Part	In $4 f + 6 s + 11$	Why it has this name
term	$4 f$ , $6 s$ and $11$	the three pieces joined by plus signs
variable	$f$ and $s$	they take different values in different innings
coefficient	4 for $f$ , 6 for $s$	each multiplies a variable
constant term	$11$	it has no variable, so it never changes

Example 2NCERT Example 2 · p. 16

Seats for a school trip

A school books $b$ buses for a trip. Each bus has $s$ seats, and 2 seats in every bus are kept for teachers. Write an expression for the number of seats left for students. How many are there when $b = 3$ and $s = 40$ ?

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

This expression also uses two variables, but its first term, $b s$ , is a product of two variables. No term of $4 f + 6 s + 11$ multiplies two variables together, and that changes how the two expressions behave. If you double both $b$ and $s$ , the term $b s$ becomes four times as large, while $4 f$ and $6 s$ only double when $f$ and $s$ double.

Example 3NCERT Example 3 · p. 17

A plot twice as long as it is wide

A farmer marks out a rectangular vegetable plot whose length is twice its width. If the width is $x$ metres, write expressions for the perimeter and the area of the plot. Find both when $x = 5$ .

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

The plot needs only one letter, $x$ , where the score and the seats each needed two. Its area also brings in a power: $2 x^{2}$ means $2 \times x \times x$ . The rest of this chapter sticks to one letter at a time, as in $6 x$ and $2 x^{2}$ .

Quick check

1 of 3

b s - 2 b

, what is the coefficient of

b

in the second term?

s

2

1

- 2

Show answer

Answer (D) $- 2$

The sign travels with the term: the second term is

- 2 b

, so its coefficient is

- 2

. The letter

s

belongs to the first term,

b s

Polynomials in one variable

Take the expression $2 x^{3} - x^{2} + 7 x - 5$ , a polynomial (बहुपद) in $x$ . Every term is a number multiplied by a whole-number power of $x$ : $2 x^{3}$ , $- x^{2}$ , $7 x$ and $- 5$ . The number in front of each power is its coefficient; notice that $- x^{2}$ has the coefficient $- 1$ , even though no 1 is written. The term without $x$ is the constant term, and the highest power of $x$ , here 3, is the degree (घात). Fig. 2.1 takes the expression apart one idea at a time.

5/5The highest power is 3, so the degree is 3.

Fig. 2.1Terms, coefficients and powers of

2 x^{3} - x^{2} + 7 x - 5

Step 1 shows the polynomial 2x cubed, minus x squared, plus 7x, minus 5, written as one expression. Step 2 draws a box around each of the four terms; the constant term minus 5 has an orange box. Step 3 adds a row of coefficients under the terms: 2, minus 1 (circled, because the 1 is not written), 7, and minus 5 marked as the constant. Step 4 adds a row with the power of x in each term: 3, 2, 1 and 0. Step 5 circles the power 3 and shows a banner saying that the highest power is 3, so the degree is 3.

Definition

Polynomial in one variable

An expression whose terms are each a number multiplied by a whole-number power (0, 1, 2, 3, …) of one variable. Mathematicians also call it a univariate polynomial, from 'uni' (one) and 'variate' (variable).

Example $2 x^{3} - x^{2} + 7 x - 5$ , $y^{4} - 9$ and $3 m + 2$ are polynomials in one variable.

Definition

Degree of a polynomial

The largest power of the variable that appears in the polynomial (with a non-zero coefficient).

Example $2 x^{3} - x^{2} + 7 x - 5$ has degree 3; $3 m + 2$ has degree 1.

Why do we say the constant term has power 0? Any non-zero number raised to the power 0 is 1, so $- 5 = - 5 \times x^{0}$ . For the same reason a polynomial that is just a number, such as $6$ , has degree 0.

A power that does not appear has coefficient 0. In $3 x^{3} + 2 x - 1$ there is no $x^{2}$ term, so the coefficient of $x^{2}$ is 0; we could write the polynomial as $3 x^{3} + 0 x^{2} + 2 x - 1$ without changing its value.

Polynomials of low degree have names of their own, listed in Table 2.2. Not every expression in one variable is a polynomial, though. In $\frac{1}{x} = x^{- 1}$ the power is $- 1$ , and in $\sqrt{x}$ it is one half. Neither power is a whole number, so neither expression is a polynomial.

Table 2.2Naming polynomials by their degree

Degree	Name	Examples
0	constant polynomial	$6$ , $- \frac{3}{4}$
1	linear polynomial	$5 m - 2$ , $9 - x$
2	quadratic polynomial	$t^{2} - 4 t$ , $3 - 2 x^{2}$
3	cubic polynomial	$y^{3} + 1$ , $4 x - x^{3}$

Quick check

1 of 4

What is the degree of

6 - 5 y^{2} + y^{4} - y

1

2

4

6

Show answer

Answer (C) $4$

The powers of

y

are 0, 2, 4 and 1, so the degree is 4. It is not 6: that is the constant term, and it is not the power of the first term written.

Practice · 5 questions

Practice 2.1: questions with answers

1
Short answerEasyNCERT Exercise Set 2.1, Q1
Write the degree of each polynomial.
1. $7 + x - 4 x^{2}$
  Show answer
  Answer 2
  The highest power of $x$ is 2.
2. $5 p - 2 p^{4} + p^{3}$
  Show answer
  Answer 4
  The powers are 1, 4 and 3; the highest is 4.
3. $12$
  Show answer
  Answer 0
  A non-zero constant has degree 0, since $12 = 12 p^{0}$ .
4. $- 3 + 9 w$
  Show answer
  Answer 1
  $9 w$ has power 1 and $- 3$ has power 0, so the degree is 1.
2
Short answerMediumNCERT Exercise Set 2.1, Q2
Write one polynomial of each kind. Many answers are possible.
1. Degree 1 in the variable $z$ , with constant term 6.
  Show solution
  AnswerFor example, $4 z + 6$ .
  Any $a z + 6$ with $a \neq 0$ works, such as $4 z + 6$ or $6 - z$ .
2. Degree 2, with no term in $x$ (only an $x^{2}$ term and a constant).
  Show solution
  AnswerFor example, $3 x^{2} - 5$ .
  Choose a non-zero coefficient for $x^{2}$ and a constant, and leave out $x$ : $3 x^{2} - 5$ or $x^{2} + 1$ .
3. Degree 3, with exactly two terms.
  Show solution
  AnswerFor example, $x^{3} + 8$ or $2 t^{3} - t$ .
  One term must be a cube of the variable; the other can be any lower power, such as $x^{3} + 8$ or $2 t^{3} - t$ .
3
Short answerEasyNCERT Exercise Set 2.1, Q3
Look at the polynomial $2 x^{5} - x^{4} + 9 x^{2} - x + 3$ .
1. What is the coefficient of $x^{4}$ ?
  Show answer
  Answer -1
  $- x^{4} = - 1 \times x^{4}$ , so the coefficient is $- 1$ .
2. What is the coefficient of $x^{2}$ ?
  Show answer
  Answer 9
  The term is $9 x^{2}$ .
3. What is the coefficient of $x^{3}$ ?
  Show answer
  Answer 0
  There is no $x^{3}$ term, so its coefficient is 0.
Short answerEasyNCERT Exercise Set 2.1, Q4
Give the coefficient of $m$ in $m^{4} + 3 m^{2} - 7$ .
Show answer
Answer 0
The polynomial has no $m$ term: it is $m^{4} + 0 m^{3} + 3 m^{2} + 0 m - 7$ . So the coefficient of $m$ is 0.
5
Short answerMediumNCERT Exercise Set 2.1, Q5
Find the constant term of each polynomial.
1. $12 - 5 k + k^{3}$
  Show answer
  Answer 12
  The term without $k$ is 12.
2. $(k - 2) (k + 3)$
  Show answer
  Answer -6
  $(k - 2) (k + 3) = k^{2} + 3 k - 2 k - 6 = k^{2} + k - 6$ , so the constant term is $- 6$ .

2.2 Linear Polynomials

From here on we focus on polynomials of degree 1. Every such polynomial can be written as $a x + b$ , where $a$ and $b$ are numbers and $a \neq 0$ . If $a$ were 0, the $x$ would vanish and only a constant would be left. So $3 x + 2$ , $5 - 4 t$ and $0.5 m$ are all linear polynomials.

Definition

Linear polynomial

A polynomial of degree 1. It can always be written as $a x + b$ , where $a \neq 0$ .

Example $7 y - 1$ , $4 - x$ and $0.5 n$ are linear; $x^{2} + 1$ and $8$ are not.

Linear polynomials appear whenever something changes by the same amount at every step. Here are two everyday cases.

Example 4NCERT Example 4 · p. 19

Kilometres on a litre

Arjun's scooter runs about 45 km on one litre of petrol. Write the distance it can cover on $L$ litres. Find the distance for $L = 1, 2, 3$ and $4$ . How much farther can it go for each extra litre? And for an extra 1.5 litres?

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

Example 5NCERT Example 5 · p. 19

Binding a project file

A shop near school binds project files. It charges ₹25 for the cover and binding, plus ₹2 for each printed page. Write the bill for $p$ pages and make a table for 1 to 5 pages. Riya's bill was ₹85. How many pages did she print?

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

Line up the bills from Fig. 2.2: 27, 29, 31, 33, 35. Subtract each bill from the next and you get 2 every time, because every extra page costs the same ₹2. Linear polynomials always behave like this: feed them inputs that go up in equal steps, and the outputs change in equal steps too. A list of numbers whose jumps are all equal is called a linear pattern, and Section 2.3 studies such lists.

The areas of the plot in Example 3 behave differently. As the width goes 1, 2, 3, 4, 5, the area $2 x^{2}$ goes 2, 8, 18, 32, 50, and the jumps keep growing. Equal jumps are a test we can use: if they are not equal, the rule cannot be linear.

4/4The jumps 6, 10, 14, 18 keep growing, so $2 x^{2}$ is not linear.

Fig. 2.2Equal jumps come from a linear rule; growing jumps do not.

Step 1 shows a table of pages 1 to 5 and the binding bills 27, 29, 31, 33 and 35 rupees. Step 2 adds solid green arcs linking neighbouring bills, each labelled plus 2. Step 3 adds a second table for the plot, with widths 1 to 5 and areas 2, 8, 18, 32 and 50 square metres for the rule A equals 2 x squared. Step 4 adds dashed orange arcs linking neighbouring areas labelled plus 6, plus 10, plus 14 and plus 18, which are not equal.

Linear equations

Riya's bill led to $25 + 2 p = 85$ : a linear polynomial on one side and a number on the other. An equation of this kind is a linear equation (रैखिक समीकरण). Solving it means finding the value of the variable that makes the two sides equal. We may add or subtract the same number on both sides, or multiply or divide both sides by the same non-zero number, and the two sides stay equal.

Example 6NCERT Example 6 · p. 20

Chairs for the annual day

For the annual day, the chairs in a school hall are set out in two equal blocks, with 14 extra chairs in front for guests. There are 150 chairs in all. How many chairs are in each block?

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

Fig. 2.3 shows the same two moves on a balance, with a box for each block of chairs and a bag for the 14 guest chairs. Taking the same amount off both pans, or keeping half of each pan, never tips it. Here $2 x + 14$ is the linear polynomial and 150 is the number it is set equal to.

3/3Keep half of each side: $x = 68$ .

Fig. 2.3Doing the same thing to both sides keeps the balance level.

Step 1 shows a level balance: the left pan holds two boxes marked x, one for each block of chairs, and a bag marked 14, and the right pan holds a bag marked 150; the equation 2x plus 14 equals 150 is written below. Step 2 lifts the 14 bag off the left pan and changes the right bag to 136, with the label take 14 off on both sides; the equation becomes 2x equals 136. Step 3 fades one x box and splits the right side into two bags of 68, one faded, with the label keep half; the equation reads x equals 68.

Polynomials as input–output rules

A polynomial can also be read as a rule that turns one number into another. Fig. 2.4 builds the rule $p (x) = 3 x - 4$ from two stages in a row: the first stage triples whatever it receives, and the second subtracts 4. Follow two numbers through it. The input 6 becomes 18 and then 14, since $3 \times 6 - 4 = 14$ . The input $- 2$ becomes $- 6$ and then $- 10$ .

5/5 $3 \times (- 2) = - 6$ and $- 6 - 4 = - 10$ , so $p (- 2) = - 10$ .

Fig. 2.4The rule

p (x) = 3 x - 4

as a machine with two stages.NCERT Fig. 2.3 · p. 20

Two tall machine boxes stand in a row, the first labelled times 3 and the second labelled minus 4, with column headings x, 3x and 3x minus 4. In step 1 all the number slots are empty. Step 2 puts the input 6 into the first slot. Step 3 shows 18 coming out of the times 3 box. Step 4 shows 14 coming out of the minus 4 box. Step 5 runs a second input, minus 2, through the lower row, giving minus 6 and then minus 10.

Each input comes out as exactly one output, and a rule with this property is called a function (फलन). We record an input together with its output by writing $p (6) = 14$ , read 'p of 6 equals 14', or $p (- 2) = - 10$ . Naming the rule $p$ lets us ask for its value at any number: $p (0) = - 4$ and $p (10) = 26$ .

The area machine gives $A (10) = 2 \times 100 = 200$ , and its outputs 2, 8, 18, 32 jump by 6, 10 and 14. So $p (x) = 3 x - 4$ is a linear function, with equal jumps, while $A (x) = 2 x^{2}$ is a quadratic function, whose jumps grow.

Quick check

1 of 4

Which of these is a linear polynomial?

5 - 3 y

y^{2} - 3

\frac{3}{y} + 1

4

Show answer

Answer (A) $5 - 3 y$

5 - 3 y

has degree 1.

y^{2} - 3

is quadratic,

\frac{3}{y}

has a negative power so it is not a polynomial, and

4

is a constant.

Practice · 7 questions

Practice 2.2: questions with answers

1
Short answerEasyNCERT Exercise Set 2.2, Q1
Evaluate the linear polynomial $4 - 3 x$ at
1. $x = 0$
  Show answer
  Answer 4
  $4 - 3 (0) = 4$ .
2. $x = - 2$
  Show answer
  Answer 10
  $4 - 3 (- 2) = 4 + 6 = 10$ .
3. $x = 5$
  Show answer
  Answer -11
  $4 - 3 (5) = 4 - 15 = - 11$ .
2
Short answerEasyNCERT Exercise Set 2.2, Q2
Work out $2 t^{2} + 5 t - 3$ , a quadratic polynomial, for
1. $t = 0$
  Show answer
  Answer -3
  $0 + 0 - 3 = - 3$ .
2. $t = - 2$
  Show answer
  Answer -5
  $2 (4) + 5 (- 2) - 3 = 8 - 10 - 3 = - 5$ .
3. $t = 3$
  Show answer
  Answer 30
  $2 (9) + 5 (3) - 3 = 18 + 15 - 3 = 30$ .
3
Short answerMediumNCERT Exercise Set 2.2, Q3
A water tank holds four times as much water as a drum. If 30 litres are drawn from the tank and 6 litres are poured into the drum, they hold the same amount. How much did each hold at first?
1. How much was in the drum at first?
  L
  Show answer
  Answer 12 L
  $4 x - 30 = x + 6$ gives $3 x = 36$ , so $x = 12$ .
2. How much was in the tank at first?
  L
  Show answer
  Answer 48 L
  $4 \times 12 = 48$ litres. Check: $48 - 30 = 18$ and $12 + 6 = 18$ .
Short answerMediumNCERT Exercise Set 2.2, Q4
Two rolls of cloth have lengths in the ratio $3 : 7$ . Together they are 150 m long. How long is each roll?
Show solution
Answer45 m and 105 m.
Let the lengths be $3 k$ and $7 k$ . Their total is $3 k + 7 k = 10 k = 150$ , so $k = 15$ . The rolls are $3 \times 15 = 45$ m and $7 \times 15 = 105$ m. Check: $45 + 105 = 150$ and $45 : 105 = 3 : 7$ .
5
Short answerMediumNCERT Exercise Set 2.2, Q5
At a school play, adult tickets cost ₹50 and child tickets cost ₹20. In all, 54 tickets were sold and the takings were ₹1,620. How many tickets of each kind were sold?
1. Adult tickets
  tickets
  Show answer
  Answer 18 tickets
  $50 a + 20 (54 - a) = 1620$ gives $30 a + 1080 = 1620$ , so $30 a = 540$ and $a = 18$ .
2. Child tickets
  tickets
  Show answer
  Answer 36 tickets
  $54 - 18 = 36$ . Check: $50 \times 18 + 20 \times 36 = 900 + 720 = 1620$ .
Short answerMediumNCERT Exercise Set 2.2, Q6
A prize fund of ₹1,500 is shared between the first and second teams in a quiz. The first team gets ₹300 more than twice the second team's share. Find each share.
Show solution
AnswerSecond team ₹400, first team ₹1,100.
The first team gets $2 x + 300$ , so $x + (2 x + 300) = 1500$ . Then $3 x = 1200$ and $x = 400$ . The first team gets $2 \times 400 + 300 = 1100$ rupees. Check: $400 + 1100 = 1500$ .
7
Short answerMediumNCERT Exercise Set 2.2, Q7
In an isosceles triangle, the base is 5 cm shorter than each of the two equal sides. The perimeter is 70 cm. Find the lengths of the sides.
1. Each equal side
  cm
  Show answer
  Answer 25 cm
  $s + s + (s - 5) = 70$ gives $3 s - 5 = 70$ , so $3 s = 75$ and $s = 25$ .
2. The base
  cm
  Show answer
  Answer 20 cm
  $25 - 5 = 20$ cm. Check: $25 + 25 + 20 = 70$ .

2.3 Exploring linear patterns

Rangoli designs often grow outward from a centre. Fig. 2.5 shows a simple design built from square tiles. Stage 1 is a centre tile with one tile on each of its four sides. To make each new stage, the artist adds one more tile to the end of every arm.

5/5Four arms of 4 tiles plus the centre: $4 \times 4 + 1 = 17$ .

Fig. 2.5Each stage adds 4 tiles, one at the end of each arm.NCERT Fig. 2.4 · p. 21

A plus shape of square tiles grows around a dark centre tile, with a panel on the right giving the stage number, the tile count and the running list of counts. Step 1 shows stage 1: the centre and one tile on each side, 5 tiles. Steps 2, 3 and 4 add one tile to the end of each arm, shown in orange as new tiles, giving 9, 13 and 17 tiles, and the list grows 5, 9, 13, 17 with plus 4 marked between neighbours. Step 5 colours the four arms differently and explains 4 arms of 4 tiles plus 1 centre tile, 4 times 4 plus 1 equals 17.

Table 2.3Tiles needed at each stage of the rangoli

Stage $n$	1	2	3	4	5	6	7
Tiles	5	9	13	17	21	25	29

There are two good ways to find a rule for stage $n$ . Growth view: stage 1 has 5 tiles and each later stage adds 4, so stage $n$ has $5 + 4 (n - 1)$ tiles. Structure view: stage $n$ has four arms of $n$ tiles each, plus the centre, so it has $4 n + 1$ tiles. The two rules must agree, and algebra confirms it: $5 + 4 (n - 1) = 5 + 4 n - 4 = 4 n + 1$ .

The counts 5, 9, 13, 17, … climb by the same 4 every time, and the rule behind them, $4 n + 1$ , is a polynomial of degree 1. These two facts always come as a pair: the property below proves that every rule of degree 1 makes equal jumps.

PropertyEqual steps in, equal steps out

For any linear polynomial $p (n) = a n + b$ , the values at consecutive whole numbers always differ by $a$ : $p (n + 1) - p (n) = a$ for every $n$ .

Rangoli check

1 of 3

How many tiles does stage 20 of the rangoli need?

tiles

Show answer

Answer 81 tiles

4 \times 20 + 1 = 81

Try this

Design your own growing pattern

You’ll need

about 20 coins, buttons or bottle caps
a sheet of paper

Use 5 coins, buttons or bottle caps to make stage 1 of any shape you like.
Grow it by adding exactly 3 objects to make each new stage.
Build stages 2, 3 and 4 and record the counts in the table.
Write a rule for stage $n$ in two ways: from the growth, and from the shape of your design.

Stage	1	2	3	4
Objects	5

Your entries are saved automatically.

Example 7NCERT Example 7 · p. 22

Rides on a metro card

Rehana's metro card has ₹250 on it. Each ride on her usual route costs ₹30. How much is left after $n$ rides? After how many rides will ₹70 be left?

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

Example 8NCERT Example 8 · p. 23

Sending a parcel

A courier charges ₹60 for any parcel up to 3 kg. Each kilogram above 3 kg costs ₹25 more, and weights are rounded up to whole kilograms. Make a table for 1 to 7 kg. Find a rule for the charge on a parcel of $n$ kg, where $n \geq 3$ . What does a 12 kg parcel cost?

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

The charges 60, 60, 60, 85, 110, 135, 160 do not form a linear pattern from the very start, because the first jumps are 0. From 3 kg onwards, though, every jump is 25, and the rule $25 n - 15$ fits. We can also run the rule backwards: a parcel charged ₹210 weighs 9 kg, since $25 n - 15 = 210$ gives $25 n = 225$ .

Definition

Linear pattern

A list of numbers in which each term is the previous term plus the same fixed number. That fixed number can be negative.

Example5, 9, 13, 17, … (add 4) and 250, 220, 190, … (add $- 30$ ).

Compare

Linear or not?

Linear patterns

7, 11, 15, 19, … (jumps of 4)
40, 34, 28, 22, … (jumps of $- 6$ )
2.5, 3, 3.5, 4, … (jumps of 0.5)

Not linear

1, 4, 9, 16, … (jumps 3, 5, 7)
3, 6, 12, 24, … (each term doubles)
60, 60, 60, 85, … (jumps 0, 0, 25)

Look back at the rules we found: $4 n + 1$ , $250 - 30 n$ and $25 n - 15$ . Each is a polynomial of degree 1 in $n$ , so the property above guarantees their equal jumps. The link also works the other way. A list that starts at $f$ and changes by a fixed $d$ each time has $n$ th term $f + d (n - 1)$ , and this is linear in $n$ whenever $d \neq 0$ . For the rangoli, $5 + 4 (n - 1) = 4 n + 1$ .

Quick check

1 of 4

Which of these lists are linear patterns?

7, 11, 15, 19

2, 4, 8, 16

40, 34, 28, 22

1, 3, 6, 10

Select all that apply

Show answer

Answer

(A) $7, 11, 15, 19$
(C) $40, 34, 28, 22$

7, 11, 15, 19

adds 4 each time and

40, 34, 28, 22

adds

- 6

. Doubling (2, 4, 8, 16) gives jumps 2, 4, 8, and 1, 3, 6, 10 has jumps 2, 3, 4; neither is constant.

Practice · 5 questions

Practice 2.3: questions with answers

1
Short answerEasyNCERT Exercise Set 2.3, Q1
A school library starts the year with 320 books and adds 45 new books every month.
1. How many books are there after 1, 2, 3 and 4 months?
  Show solution
  Answer365, 410, 455 and 500.
  Add 45 each time: $320 + 45 = 365$ , then 410, 455 and 500.
2. Write an expression for the number of books after $n$ months.
  Show solution
  Answer $320 + 45 n$
  After $n$ months, $n$ lots of 45 have been added to 320, so there are $320 + 45 n$ books.
3. How many books will there be after a full year?
  books
  Show answer
  Answer 860 books
  $320 + 45 \times 12 = 320 + 540 = 860$ .
2
Short answerEasyNCERT Exercise Set 2.3, Q2
A school bus leaves its first stop with 56 students. At every later stop 4 students get off and nobody gets on.
1. How many students are on the bus after 1, 2 and 3 more stops?
  Show solution
  Answer52, 48 and 44.
  Subtract 4 each time: 52, 48, 44.
2. Write an expression for the number of students after $n$ stops.
  Show solution
  Answer $56 - 4 n$
  $n$ stops take away $4 n$ students, leaving $56 - 4 n$ .
3. After how many stops is the bus empty?
  stops
  Show answer
  Answer 14 stops
  $56 - 4 n = 0$ gives $4 n = 56$ , so $n = 14$ .
3
Short answerEasyNCERT Exercise Set 2.3, Q3
A triangle has a base of 8 cm. Its height $h$ cm can change.
1. Find the area when the height is 2 cm, 3 cm and 4 cm. What linear pattern connects $h$ and the area?
  Show solution
  Answer8, 12 and 16 square cm; area $= 4 h$ .
  Area $= \frac{1}{2} \times 8 \times h = 4 h$ square centimetres. For $h = 2, 3, 4$ this gives 8, 12 and 16. Each extra centimetre of height adds 4 square centimetres, so the areas form a linear pattern with rule $4 h$ .
2. What height gives an area of 36 square centimetres?
  cm
  Show answer
  Answer 9 cm
  $4 h = 36$ , so $h = 9$ cm.
4
Short answerMediumNCERT Exercise Set 2.3, Q4
A glass fish tank has a rectangular base 40 cm long and 25 cm wide. Water is poured in slowly, so its depth $d$ cm rises.
1. Write the volume of water, in cubic centimetres, for a depth of $d$ cm.
  Show solution
  Answer $V = 1000 d$
  The base has area $40 \times 25 = 1000$ square centimetres, so $V = 1000 d$ cubic centimetres: 10000 for 10 cm, 15000 for 15 cm, and so on, a linear pattern.
2. How many litres does each extra centimetre of depth add?
  L
  Show answer
  Answer 1 L
  Each centimetre adds 1000 cubic centimetres, which is 1 litre.
3. How deep is the water when the tank holds 18 litres?
  cm
  Show answer
  Answer 18 cm
  18 litres is 18000 cubic centimetres, and $1000 d = 18000$ gives $d = 18$ cm.
5
Short answerEasyNCERT Exercise Set 2.3, Q5
A teacher has a pile of 180 answer sheets to mark and marks 12 sheets every hour.
1. How many sheets are still unmarked after 7 hours?
  sheets
  Show answer
  Answer 96 sheets
  $180 - 12 \times 7 = 180 - 84 = 96$ .
2. Write the linear pattern for the sheets left after $n$ hours, and find when she finishes.
  Show solution
  Answer $180 - 12 n$ ; after 15 hours.
  Sheets left $= 180 - 12 n$ : 168, 156, 144, … She finishes when $180 - 12 n = 0$ , that is, after $n = 15$ hours.

2.4 Linear growth and linear decay

Many quantities change steadily with time: water filling a tank, the charge in a phone battery, the length of a road being built. When the change is the same in every equal stretch of time, a linear polynomial in the time is a good model.

Example 9NCERT Example 9 · p. 24

Filling the overhead tank

A pump fills an overhead tank that already holds 200 litres. It adds 40 litres every minute, so the volume after $t$ minutes is $V (t) = 200 + 40 t$ litres. Make a table for $t = 0$ to 5. How much water is there after 12 minutes? The tank is full at 1000 litres. When does it fill up?

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

Example 10NCERT Example 10 · p. 24

Streaming on a phone

Kiran's phone battery shows 92% when she starts watching videos, and it drops by 8 percentage points every hour. The charge after $t$ hours is $B (t) = 92 - 8 t$ per cent. Make a table for $t = 0$ to 4. What is the charge after 6 hours? The phone warns her at 20%. When does the warning come?

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

Fig. 2.6 shows the two situations side by side, one bar for each minute or hour. The tank's bars climb by the same amount each time: this is linear growth (रैखिक वृद्धि). The battery's bars shrink by the same amount each time: this is linear decay (रैखिक ह्रास). Join the tops of the bars and you get straight lines, one rising and one falling.

6/6The tops of the bars lie on straight lines: one rising, one falling.

Fig. 2.6Equal changes in equal times: linear growth on the left, linear decay on the right.

Two bar charts sit side by side: on the left the water in a tank in litres against time in minutes, on the right a phone battery in per cent against time in hours. Step 1 shows the first bars, 200 litres and 92 per cent. Steps 2 to 5 add one bar each time: the tank bars read 240, 280, 320 and 360, each 40 higher, and the battery bars read 84, 76, 68 and 60, each 8 lower. Step 6 draws a dashed straight line along the tops of the bars in each chart, rising for the tank and falling for the battery.

Definition

Linear growth

Growth by equal additions: over every stretch of time of the same length, the quantity gains the same fixed amount.

ExampleThe tank gains 40 litres every minute.

Definition

Linear decay

Decline by equal subtractions: over every stretch of time of the same length, the quantity loses the same fixed amount.

ExampleThe battery loses 8 percentage points every hour.

In a rule of the form starting value + rate × time, the sign of the rate tells you the kind of change. A rate of $+ 40$ litres per minute means growth, and a rate of $- 8$ per hour means decay. Every model has limits, too. The battery rule gives $B (12) = 92 - 96 = - 4$ , which is impossible. The rule only makes sense while the battery lasts, from $t = 0$ up to $t = 11.5$ hours, when $B$ reaches 0.

Quick check

1 of 3

Which situation shows linear decay?

A bus pass that costs ₹5 more each yearBacteria that double every hourA savings jar that gets ₹10 each weekA 20 L can of oil from which 1.5 L is used each day

Show answer

Answer (D) A 20 L can of oil from which 1.5 L is used each day

The oil drops by the same 1.5 L every day, which is linear decay. The bus pass and the jar show linear growth, and doubling is not linear at all.

Practice · 4 questions

Practice 2.4: questions with answers

1
Short answerEasyNCERT Exercise Set 2.4, Q1
An electric kettle heats 1 litre of water that starts at 24 °C. While it heats, the temperature rises by 6 °C every minute.
1. Fill in the temperature every 2 minutes, from $t = 0$ to $t = 12$ .
  Show solution
  Answer24, 36, 48, 60, 72, 84, 96
  $t$ (min) 0 2 4 6 8 10 12
  $T$ (°C) 24 36 48 60 72 84 96
2. Write a rule for the temperature $T$ °C after $t$ minutes. Which feature of your table shows that this is linear growth?
  Show solution
  Answer $T = 24 + 6 t$
  $T = 24 + 6 t$ . In the table, every 2 minutes add the same 12 °C, which is 6 °C in each minute. Equal rises in equal times make this linear growth.
3. After how many minutes does the water reach 90 °C?
  min
  Show answer
  Answer 11 min
  $24 + 6 t = 90$ gives $6 t = 66$ , so $t = 11$ . The rule stops working at 100 °C, when the water boils, a little before $t = 13$ .
2
Short answerEasyNCERT Exercise Set 2.4, Q2
A family's cooking-gas cylinder holds 14.2 kg of gas. The family uses about 0.4 kg a day.
1. Write a rule for the gas left, $g$ kg, after $t$ days, and use it to fill in a table for $t = 0, 5, 10, \dots, 35$ .
  Show solution
  Answer $g = 14.2 - 0.4 t$ : 14.2, 12.2, 10.2, 8.2, 6.2, 4.2, 2.2, 0.2
  $g = 14.2 - 0.4 t$ . Every 5 days use $5 \times 0.4 = 2$ kg, so each entry is 2 less than the one before.
  $t$ (days) 0 5 10 15 20 25 30 35
  $g$ (kg) 14.2 12.2 10.2 8.2 6.2 4.2 2.2 0.2
2. The family books a refill as soon as less than 2 kg is left. After how many days does that happen?
  days
  Show answer
  Answer 31 days
  After 30 days $g = 14.2 - 12 = 2.2$ kg, which is not yet below 2. After 31 days $g = 14.2 - 12.4 = 1.8$ kg, so they book after 31 days.
3. Why is this linear decay? If they never refilled, when would the gas run out?
  Show solution
  AnswerThe same 0.4 kg is used every day; it would run out after 35.5 days.
  The gas falls by the same 0.4 kg in every day, so it is linear decay. It runs out when $14.2 - 0.4 t = 0$ , that is, $t = 35.5$ days.
3
Short answerEasyNCERT Exercise Set 2.4, Q3
A new metro line has 18 km of track already laid. The builders add 2.5 km of track every month.
1. Which rule gives the length of track, $L$ km, after $t$ months?
  $L = 2.5 + 18 t$ $L = 20.5 t$ $L = 18 + 2.5 t$ $L = 18 - 2.5 t$
  Show answer
  Answer (C) $L = 18 + 2.5 t$
  At the start ( $t = 0$ ) there are 18 km, and each month adds 2.5 km, so $L = 18 + 2.5 t$ . The rule $2.5 + 18 t$ swaps the start and the rate, $20.5 t$ gives 0 km at the start, and $18 - 2.5 t$ makes the track shrink.
2. How much track will there be after 8 months?
  km
  Show answer
  Answer 38 km
  $18 + 2.5 \times 8 = 18 + 20 = 38$ km.
3. Make a table of $L$ for every second month up to $t = 12$ . The finished line will be 55.5 km long. When will it be complete, and why is this linear growth?
  Show solution
  Answer18, 23, 28, 33, 38, 43, 48; complete after 15 months.
  $t$ (months) 0 2 4 6 8 10 12
  $L$ (km) 18 23 28 33 38 43 48
  The same 2.5 km is added in every month, so the length grows linearly. $18 + 2.5 t = 55.5$ gives $2.5 t = 37.5$ , so $t = 15$ months.
4
Short answerEasyNCERT Exercise Set 2.4, Q4
A school office has 2400 sheets of printer paper and uses 150 sheets a day.
1. Let $P (d)$ be the number of sheets left after $d$ days. Write a rule for $P (d)$ . How does the rule show that this is decay rather than growth?
  Show solution
  Answer $P (d) = 2400 - 150 d$
  $P (d) = 2400 - 150 d$ . The number multiplying $d$ is negative: each day takes away the same 150 sheets, so this is linear decay.
2. The office reorders paper when 600 sheets are left. After how many days does it reorder?
  days
  Show answer
  Answer 12 days
  $2400 - 150 d = 600$ gives $150 d = 1800$ , so $d = 12$ .
3. Make a table of $P (d)$ for $d = 0, 4, 8, 12, 16$ . When does the paper run out?
  Show solution
  Answer2400, 1800, 1200, 600, 0; after 16 days.
  $d$ (days) 0 4 8 12 16
  $P (d)$ 2400 1800 1200 600 0
  Every 4 days use $4 \times 150 = 600$ sheets. The stock reaches 0 at $d = 16$ , since $2400 \div 150 = 16$ .

$t$ (min)	0	2	4	6	8	10	12
$T$ (°C)	24	36	48	60	72	84	96

$t$ (days)	0	5	10	15	20	25	30	35
$g$ (kg)	14.2	12.2	10.2	8.2	6.2	4.2	2.2	0.2

$t$ (months)	0	2	4	6	8	10	12
$L$ (km)	18	23	28	33	38	43	48

$d$ (days)	0	4	8	12	16
$P (d)$	2400	1800	1200	600	0

2.5 Linear Relationships

So far each rule had one variable. Often, though, we want to connect two quantities that change together: the stage number and the tile count, the minutes and the litres, the distance and the fare. When a quantity $y$ depends on a quantity $x$ through a rule $y = a x + b$ , we say that $x$ and $y$ are in a linear relationship (रैखिक संबंध).

Definition

Linear relationship

A link between two changing quantities in which one is a fixed multiple of the other plus a fixed number: $y = a x + b$ .

ExampleFor the rangoli, $y = 4 x + 1$ , where $x$ is the stage and $y$ the number of tiles.

Two rules from earlier sections already have this form. The tank in Example 9 follows $V = 40 t + 200$ : time $t$ plays the part of $x$ , the volume $V$ plays the part of $y$ , $a = 40$ and $b = 200$ . The rangoli in Section 2.3 gives $y = 4 x + 1$ when $x$ is the stage and $y$ the tile count. Often, though, we are handed a few readings rather than the rule. If we know the relationship is linear, two readings are enough to recover $a$ and $b$ .

Example 11NCERT Example 11 · p. 26

Working out a taxi app's charges

A taxi app charges a fixed booking fee plus a fixed amount for each kilometre. A 4 km ride cost ₹92 and a 10 km ride cost ₹170. If the fare $y$ rupees for $x$ km is $y = a x + b$ , find $a$ and $b$ .

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

Both numbers have a clear meaning. The extra 6 km cost ₹78 more, so each kilometre costs $78 \div 6 = 13$ rupees: $a$ is the rate per kilometre. At 0 km the rule gives ₹40: $b$ is the booking fee, which you pay however short the ride. Fig. 2.7 shows the same reasoning on a graph.

4/4Extend the line to 0 km: the fare is ₹40, so $b = 40$ .

Fig. 2.7The rise over the run gives the rate; running back to 0 km gives the booking fee.

Axes show distance in kilometres from 0 to 12 across and fare in rupees from 0 to 200 up. Step 1 plots two points, (4, 92) and (10, 170). Step 2 draws a dashed right-angled path from the first point to the second, 6 km across and 78 rupees up. Step 3 adds a card reading 78 divided by 6 equals 13 per km. Step 4 joins the points with a line and extends it, dashed, back to the fare axis, where a ring marks 40 rupees at 0 km, the booking fee.

The graph suggests a shortcut you can use whenever a relationship is linear: $a = \frac{change in y}{change in x} = \frac{170 - 92}{10 - 4} = 13$ . Once you know $a$ , work back from either measurement to find $b$ . The machine below hides a linear rule; use the same idea to crack it.

InteractiveCrack the machine's rule

A machine with a hidden rule of the form y equals a x plus b. You type any number as the input and press Feed it; the output appears and each input-output pair is added to a table sorted by input, with the change from the row above. You enter guesses for a and b and press Check; a wrong guess is shown an input where it disagrees with the machine. Buttons give hints, show the rule and start a new machine, and a menu picks one of three levels: positive whole numbers, negatives allowed, or a slope that ends in a half, such as 1.5 or minus 0.5.

Fig. 2.8Feed the machine numbers, watch its outputs, then guess its rule

y = a x + b

Quick check

1 of 3

A plumber charges

y = 150 x + 200

rupees for

x

hours of work. What does the 200 stand for?

A fixed visiting chargeThe charge for each hourThe number of hours workedThe bill for 2 hours

Show answer

Answer (A) A fixed visiting charge

x = 0

hours the charge is still 200, so it is a fixed charge for the visit. The charge for each hour is 150, the coefficient of

x

Practice · 3 questions

Practice 2.5: questions with answers

1
Short answerMediumNCERT Exercise Set 2.5, Q1
An electrician charges a fixed visiting fee plus an hourly rate. A 2-hour job cost ₹650 and a 5-hour job cost ₹1,250. The charge for $x$ hours is $y = a x + b$ rupees. Find $a$ and $b$ .
1. $a$ , the hourly rate
  rupees per hour
  Show answer
  Answer 200 rupees per hour
  $a = \frac{1250 - 650}{5 - 2} = \frac{600}{3} = 200$ .
2. $b$ , the visiting fee
  rupees
  Show answer
  Answer 250 rupees
  $650 = 2 \times 200 + b$ , so $b = 250$ . Check: $5 \times 200 + 250 = 1250$ .
2
Short answerMediumNCERT Exercise Set 2.5, Q2
A spring hangs from a hook. With a 200 g mass it is 17 cm long, and with a 500 g mass it is 23 cm long. Its length $L$ cm and the mass $m$ g follow $L = a m + b$ .
1. Find $a$ .
  cm per g
  Show answer
  Answer 0.02 cm per g
  $a = \frac{23 - 17}{500 - 200} = \frac{6}{300} = 0.02$ cm for each gram.
2. Find $b$ . What does it tell you about the spring?
  Show solution
  Answer $b = 13$ : the spring is 13 cm long with no mass on it.
  $17 = 0.02 \times 200 + b = 4 + b$ , so $b = 13$ . Putting $m = 0$ gives $L = 13$ , so $b$ is the length of the spring with nothing hanging from it.
3
Short answerHardNCERT Exercise Set 2.5, Q3
A candle burns down steadily. After burning for 2 hours it is 21 cm tall, and after 5 hours it is 12 cm tall. Its height $h$ cm after $t$ hours follows $h = a t + b$ .
1. Find $a$ .
  cm per hour
  Show answer
  Answer -3 cm per hour
  $a = \frac{12 - 21}{5 - 2} = \frac{- 9}{3} = - 3$ . It is negative because the height goes down.
2. Find $b$ . What do $a$ and $b$ tell you about the candle?
  Show solution
  Answer $b = 27$ : the candle was 27 cm tall when it was lit, and it loses 3 cm every hour.
  $21 = - 3 \times 2 + b$ , so $b = 27$ . Check: $- 3 \times 5 + 27 = 12$ . At $t = 0$ the height is $b = 27$ cm, the height of the new candle. The slope $a = - 3$ says it loses 3 cm in every hour: linear decay.
3. After how many hours of burning is the candle used up?
  hours
  Show answer
  Answer 9 hours
  $27 - 3 t = 0$ gives $t = 9$ hours.

2.6 Visualising linear relationships

In Example 11 a graph of the taxi fares made the rate and the booking fee easy to see. We now draw such graphs for any relationship $y = a x + b$ . The key fact is simple: plot every pair $(x, y)$ that fits a linear relationship, and all the points lie on one straight line. That is why these polynomials are called linear.

Two points fix a line

Through two points you can draw exactly one straight line. So to draw the graph of $y = a x + b$ , work out two points, plot them, rule a line through them and carry it on past both points. A third point makes a good check. Fig. 2.9 does this for $y = 2 x - 3$ : $x = 0$ gives $y = - 3$ , and $x = 2$ gives $y = 1$ .

4/4 $C (4, 5)$ fits the equation and lies on the line; $D (5, 4)$ does not.

Fig. 2.9Two points fix the line; a third point checks it.NCERT Fig. 2.5 · p. 28

A coordinate grid on the left and a small table of x and y values on the right. Step 1 shows the table with two rows: x equals 0 gives y equals minus 3, labelled A, and x equals 2 gives y equals 1, labelled B. Step 2 plots A at (0, minus 3) and B at (2, 1). Step 3 joins them with a straight line extended across the grid and labelled y equals 2x minus 3. Step 4 adds C at (4, 5), which lies on the line and is ticked, and D at (5, 4), which lies below the line and is crossed; the table shows that x equals 5 would need y equals 7, not 4.

Here is the test for any point: put its $x$ -coordinate into the rule. If the rule returns the point's $y$ -coordinate, the point is on the line; otherwise it is not. $C (4, 5)$ passes, since $2 \times 4 - 3 = 5$ . $D (5, 4)$ fails, since $2 \times 5 - 3 = 7$ , not 4, so $D$ sits below the line.

Think and reflectNCERT Think and Reflect · p. 28

More points on the same line

Complete the table for $y = 2 x - 3$ . Then plot the points on graph paper. Do they all land on the line through $A$ and $B$ ?

$x$	−1	1	3	4	6	10
$y$	−5			5

Your entries are saved automatically.

Quick check

1 of 3

On the line

y = 2 x - 3

, what is

y

when

x = 10

Show answer

Answer 17

2 \times 10 - 3 = 17

, so

(10, 17)

is on the line.

We can also go the other way: start from plotted points and find the rule. Fig. 2.10 shows two sets of points to try.

4/4Each $y$ is $- 1.5$ times its $x$ : $y = - 1.5 x$ .

Fig. 2.10Points that lie on one line follow one linear rule.NCERT Figs 2.6–2.7 · p. 29

Two small grids, each from minus 4 to 4 on both axes. Step 1 shows Set A: five round points at (minus 4, minus 2), (minus 2, minus 1), (0, 0), (2, 1) and (4, 2). Step 2 draws a straight line through them labelled y equals 0.5x. Step 3 shows Set B on the second grid: five square points at (minus 2, 3), (minus 1, 1.5), (0, 0), (1, minus 1.5) and (2, minus 3). Step 4 draws a dashed falling line through them labelled y equals minus 1.5x.

Example 12NCERT Example 12 · p. 28

Reading a rule from points

The points $(- 4, - 2)$ , $(- 2, - 1)$ , $(0, 0)$ , $(2, 1)$ and $(4, 2)$ form Set A in Fig. 2.10. Check that they lie on one straight line and find its equation.

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

Example 13NCERT Example 13 · p. 29

A line that falls

Set B in Fig. 2.10 contains $(- 2, 3)$ , $(- 1, 1.5)$ , $(0, 0)$ , $(1, - 1.5)$ and $(2, - 3)$ . Find the equation of the line through them.

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

Lines through the origin: slope

Example 14NCERT Example 14 · p. 29

Steep and gentle lines

Draw $y = x$ , $y = 2.5 x$ and $y = 0.4 x$ on the same axes. How do the three lines differ?

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

4/4With $a < 0$ the lines fall from left to right.

Fig. 2.11Every line

y = a x

passes through the origin;

a

decides its steepness and direction.NCERT Figs 2.8–2.11 · p. 30

A grid from minus 6 to 6 on both axes with a legend on the right. Step 1 shows the dotted line y equals x. Step 2 adds the steeper solid line y equals 2.5x with the point (2, 5) marked. Step 3 adds the gentler dashed line y equals 0.4x with the point (5, 2) marked, the mirror image of (2, 5) in the line y equals x. Step 4 adds three falling lines, y equals minus x, y equals minus 2.5x and y equals minus 0.4x, listed under the heading falling.

The three lines differ only in how far they climb for each step of 1 to the right: 2.5 for $y = 2.5 x$ , 1 for $y = x$ and 0.4 for $y = 0.4 x$ . That climb is the number $a$ itself, and it is called the slope (ढाल) of the line. A slope bigger than 1 gives a line steeper than $y = x$ , and a slope between 0 and 1 gives a gentler one. The lines $y = 2.5 x$ and $y = 0.4 x$ are also mirror images in the line $y = x$ . The point $(2, 5)$ on one matches $(5, 2)$ on the other, and their slopes $\frac{5}{2}$ and $\frac{2}{5}$ are reciprocals.

Definition

Slope

The number $a$ in $y = a x + b$ . It tells how much $y$ changes when $x$ increases by 1: $y$ goes up when $a > 0$ and down when $a < 0$ .

Example $y = 2.5 x$ has slope 2.5; $y = - 0.4 x$ has slope $- 0.4$ .

Example 15NCERT Example 15 · p. 31

Falling lines

Draw $y = - x$ , $y = - 2.5 x$ and $y = - 0.4 x$ on the same axes and compare them with the lines in Example 14.

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

So the sign of the slope gives the direction and its size gives the steepness. This ties back to Section 2.4. The tank, which grows linearly, has a rising graph with a positive slope. The battery, which decays linearly, has a falling graph with a negative slope.

The slope also matches the constant difference of a linear pattern. Fig. 2.12 plots the rangoli counts from Section 2.3, with the stage number across and the tile count up. From one stage to the next the points step 1 to the right and 4 up. So 4 is both the jump in 5, 9, 13, 17 and the slope of $y = 4 x + 1$ . The two axes use different scales here, so do not judge the steepness by eye.

3/3All the points lie on $y = 4 x + 1$ , with slope 4.

Fig. 2.12Each step of 1 across goes 4 up: the slope equals the common jump.

Axes show the stage number from 0 to 4 across and the number of tiles from 0 to 20 up. Step 1 plots four points at heights 5, 9, 13 and 17 with dotted lines down to the stage axis. Step 2 draws a staircase between them: a blue step of plus 1 across and a green step of plus 4 up between each pair. Step 3 draws the straight line y equals 4x plus 1 through all four points and extends it back to the tile axis, where a ring marks (0, 1).

SimulationEqual steps along a line

A coordinate plane from minus 2 to 8 across and minus 3 to 10 up with four draggable points, A (0, minus 1), B (2, 2), C (4, 5) and D (6, 8), which all lie on the line y equals 1.5x minus 1. Segments join A to B, B to C and C to D, and each is measured; all three start at root 13. Pressing and holding a point, dragging it or focusing it with the keyboard shows the across and up legs, delta x and delta y, of the segments that meet at it. Points snap to half units, and the lengths update as you drag.

Fig. 2.13Points

A

B

C

and

D

start on the line

y = 1.5 x - 1

, joined by segments.

Try this

Press and hold $D$ (or select it with the Tab key) to see the across and up legs of $C D$ . Read $Δ x$ and $Δ y$ : what is $Δ y \div Δ x$ ? Do the same with $A$ for $A B$ .
Drag $D$ up to $(6, 10)$ . Where does the path now bend, and what happens to the length of $C D$ ? Where must $D$ go to straighten it again?
Put $D$ back at $(6, 8)$ and drag $C$ down to $(3, 3.5)$ . The steps are now different sizes. Press and hold $C$ : is $Δ y \div Δ x$ still 1.5 for $B C$ and for $C D$ ?

Changing $b$ : parallel lines

Example 16NCERT Example 16 · p. 33

Three lines with the same slope

Draw $y = - x - 2$ , $y = - x + 1$ and $y = - x + 4$ on the same axes. What do you notice?

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

4/4The gaps stay 3 everywhere, so the lines are parallel.

Fig. 2.14Same slope, different

b

: the line slides up without turning.NCERT Figs 2.12–2.13 · p. 33

A grid from minus 5 to 6 across and minus 6 to 7 up, with a legend on the right. Step 1 shows the solid line y equals minus x minus 2. Step 2 slides a copy of it up by 3 units to become the dashed line y equals minus x plus 1. Step 3 slides another copy up 3 more units to become the dotted line y equals minus x plus 4. Step 4 marks the points where the lines cross the y-axis, (0, minus 2), (0, 1) and (0, 4), and draws arrows at x equals minus 2 showing equal vertical gaps of 3 between neighbouring lines.

Read down any column of the table in Example 16: the three values of $y$ always differ by 3, the same as the differences between the values of $b$ . Adding a number to the rule lifts every point of the line by that amount, so the whole line moves up without tilting. Lines with the same slope therefore keep a fixed gap and never cross.

Where a line meets the $y$ -axis

Every point on the $y$ -axis has $x = 0$ . Putting $x = 0$ into $y = a x + b$ gives $y = a \times 0 + b = b$ . So the line $y = a x + b$ always meets the $y$ -axis at $(0, b)$ , and the number $b$ is called its y-intercept (y-अंतःखंड). Fig. 2.15 shows three lines with different values of $b$ .

4/4On the $y$ -axis $x = 0$ , so $y = b$ .

Fig. 2.15The line

y = a x + b

crosses the

y

-axis at

(0, b)

.NCERT Fig. 2.14 · p. 35

A grid from minus 5 to 5 across and minus 5 to 6 up, with a legend on the right. Step 1 draws the solid line y equals 2x plus 4 and rings its crossing with the y-axis at (0, 4). Step 2 adds the dashed line y equals minus 0.5x plus 1, crossing at (0, 1). Step 3 adds the dotted line y equals x minus 3, crossing at (0, minus 3). Step 4 highlights the whole y-axis and notes that on the y-axis x equals 0, so y equals b.

Definition

y-intercept

The number $b$ in $y = a x + b$ . It tells you where the line meets the $y$ -axis: at $(0, b)$ .

Example $y = x - 3$ has $y$ -intercept $- 3$ : it crosses the $y$ -axis 3 units below the origin.

The sign of $b$ tells you which side of the origin the crossing is on. The line $y = 2 x + 4$ crosses 4 units above $O$ , while $y = x - 3$ crosses 3 units below it, at $(0, - 3)$ . So the $y$ -intercept is a signed number, not just a distance. When $b = 0$ the crossing point is the origin itself, as for the lines in Lines through the origin: slope.

PropertyReading a line from its equation

Take lines of the form $y = a x + b$ . (i) Each one crosses the $y$ -axis at $(0, b)$ , and each step of 1 to the right changes $y$ by $a$ . (ii) Lines with the same $a$ but different values of $b$ are parallel: each is a copy of the others, slid up or down. (iii) Lines with the same $b$ all pass through $(0, b)$ , and changing $a$ turns the line about that point.

(i) At $x = 0$ , $y = a \times 0 + b = b$ , so the line crosses the $y$ -axis at $(0, b)$ . Moving from $x$ to $x + 1$ changes $y$ by $a (x + 1) + b - (a x + b) = a$ , which is exactly what the slope measures.

(ii) Take $y = a x + b_{1}$ and $y = a x + b_{2}$ with $b_{1} \neq b_{2}$ . At every $x$ the second line is $b_{2} - b_{1}$ above the first: the same gap everywhere. If the lines met at some $x$ , we would have $a x + b_{1} = a x + b_{2}$ , which gives $b_{1} = b_{2}$ . That is false, so the lines never meet: they are parallel.

(iii) Every line $y = a x + b$ with the same $b$ passes through $(0, b)$ , whatever the value of $a$ . Only its steepness and direction change, so the lines turn about that point.

The Line lab puts all of this in your hands. Move the sliders for $a$ and $b$ , keep copies of lines to compare, and then play Match the line.

InteractiveLine lab: slope and intercept

An interactive graph of y equals a x plus b on a grid from minus 6 to 6. Sliders and plus and minus buttons change the slope a from minus 4 to 4 and the y-intercept b from minus 6 to 6. The line is drawn with its crossing point (0, b) ringed and a small step of 1 across and a up. A table shows y for x from minus 2 to 3 and the constant change a between neighbours, and a status box says whether the line rises or falls, whether it is steeper than y equals x, and whether it is parallel to a kept line or meets it on the y-axis. You can show the line y equals x, keep up to four copies of lines to compare, and switch to a Match the line game that draws a dashed target line for you to match, with hints.

Fig. 2.16Change

a

and

b

y = a x + b

and watch the line, its

y

-intercept, its 1-across step and its table of values.

Try this

Explore with the Line lab

Set $b = 0$ and tick Show the line y = x. Slide $a$ from 0.4 up to 4. Which point never moves, and for which values of $a$ is your line steeper than $y = x$ ?
Press Keep this line, then change only $b$ . What does the status box say about the two lines? Now clear the kept lines, set $a = 1.5$ , $b = 2$ , keep it, and try $a = - 1.5$ with the same $b$ . What does the status box say now?
Switch to Match the line and match three dashed lines. Which do you find first, $a$ or $b$ , and how?

Recap

Introduction to Linear Polynomials: key points and summary

A polynomial in one variable is a sum of terms, each a number times a whole-number power of the variable. Its degree is the highest power present (Polynomials in one variable).
Polynomials of degree 0, 1, 2 and 3 are called constant, linear, quadratic and cubic. A missing power has coefficient 0 (Table 2.2).
A linear polynomial can be written $a x + b$ with $a \neq 0$ . Setting it equal to a number gives a linear equation (Section 2.2).
A polynomial can be read as an input–output rule, or function, such as $p (x) = 3 x - 4$ with $p (6) = 14$ (Polynomials as input–output rules).
Equal steps in the input of a linear polynomial give equal steps in the output: $p (n + 1) - p (n) = a$ (Property).
A linear pattern adds the same number each time, and its $n$ th term is linear in $n$ , like $4 n + 1$ for the rangoli (Section 2.3).
Linear growth adds the same amount in each equal time interval and linear decay takes the same amount away. Every model has limits (Section 2.4).
Two variables are in a linear relationship when $y = a x + b$ . Two measurements are enough to find $a$ and $b$ (Section 2.5).
The graph of $y = a x + b$ is a straight line. Two points fix it, and a point is on it exactly when its coordinates fit the equation (Two points fix a line).
The slope $a$ is the change in $y$ for each step of 1 in $x$ . Positive slopes rise, negative slopes fall, and a larger size means a steeper line (Lines through the origin: slope).
The line $y = a x + b$ crosses the $y$ -axis at $(0, b)$ , and $b$ is its y-intercept: positive above the origin, negative below it, and 0 for a line through the origin (Where a line meets the $y$-axis).
Same slope, different $y$ -intercept: the two lines are parallel and never meet (Changing $b$: parallel lines).

Flash cards · 8

Key terms

1 / 8

The steel truss in Fig. 2.17 is used in question 12 of the exercises below.

4/4Stage 4: 9 rods.

Fig. 2.17Each new triangle shares one rod with the triangle before it and adds two new rods.NCERT End-of-chapter Exercises, Q12 · p. 38

Four stages of a footbridge truss made from equal steel rods bolted at the joints, drawn side by side. Step 1 shows stage 1, a single triangle of 3 rods. Step 2 adds stage 2, two triangles sharing a rod, with the 2 new rods at its right-hand end drawn in blue and marked plus 2, 5 rods in all. Step 3 adds stage 3, three triangles pointing up, down and up, 7 rods. Step 4 adds stage 4, four triangles, 9 rods. Each label gives the stage number and the number of rods.

Practice · 18 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

Short answerEasyNCERT End-of-chapter Exercises, Q1
Make up a degree-4 polynomial in $y$ that has constant term 3, has $- 2$ as the coefficient of $y$ , and has no $y^{3}$ term.
Show solution
AnswerFor example, $y^{4} + 5 y^{2} - 2 y + 3$ .
The $y^{4}$ term fixes the degree at 4. Leave out $y^{3}$ , include $- 2 y$ and the constant 3. Any $y^{2}$ term may be added, so $y^{4} + 5 y^{2} - 2 y + 3$ , $2 y^{4} - 2 y + 3$ and many others are correct.
2
Short answerEasyNCERT End-of-chapter Exercises, Q2
Find the value of each polynomial.
1. $3 m^{2} + 2 m - 8$ when $m = - 2$
  Show answer
  Answer 0
  $3 (- 2)^{2} + 2 (- 2) - 8 = 12 - 4 - 8 = 0$ .
2. $2 r^{3} - 5 r + 1$ when $r = k$
  Show solution
  Answer $2 k^{3} - 5 k + 1$
  Replace every $r$ by $k$ : $2 k^{3} - 5 k + 1$ . Substituting a letter gives an expression, not a number.
Short answerMediumNCERT End-of-chapter Exercises, Q3
Three-quarters of a number, minus one-half, equals five-eighths. Find the number.
Show answer
Answer 1.5
$\frac{3}{4} x = \frac{5}{8} + \frac{1}{2} = \frac{9}{8}$ , so $x = \frac{9}{8} \times \frac{4}{3} = \frac{3}{2} = 1.5$ . Check: $\frac{3}{4} \times \frac{3}{2} - \frac{1}{2} = \frac{9}{8} - \frac{4}{8} = \frac{5}{8}$ .
4
Short answerMediumNCERT End-of-chapter Exercises, Q4
Aditi owns three times as many stickers as Irfan. After she gives Irfan 10 of her stickers, she has exactly twice as many as he has. How many stickers did each of them have at first?
1. Irfan at first
  stickers
  Show answer
  Answer 30 stickers
  $3 x - 10 = 2 (x + 10)$ gives $3 x - 10 = 2 x + 20$ , so $x = 30$ .
2. Aditi at first
  stickers
  Show answer
  Answer 90 stickers
  $3 \times 30 = 90$ . Check: after the gift Aditi has 80 and Irfan has 40, and $80 = 2 \times 40$ .
5
Short answerEasyNCERT End-of-chapter Exercises, Q5
A tree-planting drive along a highway has already planted 1,200 saplings, and it adds 175 more each month.
1. How many saplings will have been planted after 8 months?
  saplings
  Show answer
  Answer 2600 saplings
  $1200 + 175 \times 8 = 1200 + 1400 = 2600$ .
2. After how many months will the total first be more than 5,000?
  months
  Show answer
  Answer 22 months
  After 21 months the total is $1200 + 175 \times 21 = 4875$ , which is not enough. After 22 months it is $1200 + 175 \times 22 = 5050$ .
3. Write the linear pattern for the number planted after $n$ months.
  Show solution
  Answer $1200 + 175 n$
  Each month adds 175, so after $n$ months the total is $1200 + 175 n$ : 1375, 1550, 1725, … a linear pattern with jump 175.
Short answerHardNCERT End-of-chapter Exercises, Q6
In a two-digit number, the tens digit is twice the units digit. Swapping the two digits gives a number that is 36 less than the original. Find the number.
Show answer
Answer 84
The number is $21 u$ and the swapped number is $10 u + 2 u = 12 u$ . So $21 u - 12 u = 36$ , which gives $9 u = 36$ and $u = 4$ . The number is 84; swapping gives 48, and $84 - 48 = 36$ .
7
DiagramMediumNCERT End-of-chapter Exercises, Q7
For each line, find the slope and the $y$ -intercept, and the point where it crosses the $y$ -axis. Then draw the lines.
1. $2 y = 3 x - 8$
  Show solution
  AnswerSlope 1.5, $y$ -intercept $- 4$ , crossing point $(0, - 4)$ .
  Divide by 2: $y = 1.5 x - 4$ . A second point is $(2, - 1)$ .
2. $y = 5 - x$
  Show solution
  AnswerSlope $- 1$ , $y$ -intercept 5, crossing point $(0, 5)$ .
  Rewrite it as $y = - x + 5$ . A second point is $(5, 0)$ .
3. $4 y = 6 x + 2$
  Show solution
  AnswerSlope 1.5, $y$ -intercept 0.5, crossing point $(0, 0.5)$ .
  Divide by 4: $y = 1.5 x + 0.5$ . A second point is $(1, 2)$ .
4. $3 y + 6 x = 9$
  Show solution
  AnswerSlope $- 2$ , $y$ -intercept 3, crossing point $(0, 3)$ .
  Subtract $6 x$ from both sides: $3 y = - 6 x + 9$ . Divide by 3: $y = - 2 x + 3$ . A second point is $(1, 1)$ .
5. Which two of these lines are parallel?
  (b) and (d)(a) and (c)(a) and (d)(b) and (c)
  Show answer
  Answer (B) (a) and (c)
  Lines (a) and (c) both have slope 1.5 and different $y$ -intercepts, so they are parallel. Lines (b) and (d) both fall, but their slopes, $- 1$ and $- 2$ , are different.
8
Short answerMediumNCERT End-of-chapter Exercises, Q8
On a clear day, the air temperature $T$ (°C) at a height of $h$ km above a town is roughly $T = 30 - 6.5 h$ .
1. Find the temperature at a height of 2 km.
  °C
  Show answer
  Answer 17 °C
  $30 - 6.5 \times 2 = 30 - 13 = 17$ °C.
2. At what height is the air at $- 9$ °C?
  km
  Show answer
  Answer 6 km
  $30 - 6.5 h = - 9$ gives $6.5 h = 39$ , so $h = 6$ km.
3. As the height increases, the temperature shows
  linear growthlinear decayno steady change
  Show answer
  Answer (B) linear decay
  It falls by the same 6.5 °C for every kilometre climbed, which is linear decay.
9
DiagramMediumNCERT End-of-chapter Exercises, Q9
A cyclist rides at a steady 5 m/s. The distance $d$ metres covered in $t$ seconds is $d = 5 t$ . Draw the graph of $d$ against $t$ for $0 \leq t \leq 10$ .
1. What distance is covered in 3 seconds? Check it on your graph.
  m
  Show answer
  Answer 15 m
  $d = 5 \times 3 = 15$ m. On the graph, go up from $t = 3$ to the line and across to 15.
2. How long does it take to cover 40 m?
  s
  Show answer
  Answer 8 s
  $5 t = 40$ gives $t = 8$ seconds.
3. Why does the graph pass through the origin?
  Show solution
  AnswerBecause $d = 0$ when $t = 0$ .
  The rule has no constant term ( $b = 0$ ): at time 0 no distance has been covered, so the line starts at $(0, 0)$ . Its slope, 5, is the speed.
10
Short answerHardNCERT End-of-chapter Exercises, Q10
A linear polynomial $p (x)$ has $p (2) = 1$ and $p (5) = 10$ .
1. Find $p (x)$ .
  $3 x - 5$ $3 x + 1$ $9 x - 17$ $\frac{1}{3} x + \frac{1}{3}$
  Show answer
  Answer (A) $3 x - 5$
  Slope $= \frac{10 - 1}{5 - 2} = 3$ . Then $1 = 3 \times 2 + b$ gives $b = - 5$ , so $p (x) = 3 x - 5$ . Check: $3 \times 5 - 5 = 10$ .
2. Find $p (0)$ . What does it tell you about the graph?
  Show answer
  Answer -5
  $p (0) = 3 \times 0 - 5 = - 5$ , so the graph crosses the $y$ -axis at $(0, - 5)$ .
3. Solve $p (x) = 0$ . Which point of the graph does this give?
  Show answer
  Answer 1.6667
  $3 x - 5 = 0$ gives $x = \frac{5}{3} \approx 1.67$ : the graph crosses the $x$ -axis at $(\frac{5}{3}, 0)$ .
4. Plot $(2, 1)$ and $(5, 10)$ , draw the line through them, and use it to check your answers to (b) and (c).
  Show solution
  AnswerA rising line through $(0, - 5)$ , $(\frac{5}{3}, 0)$ , $(2, 1)$ and $(5, 10)$ .
  Plot $(2, 1)$ and $(5, 10)$ and join them. Extended, the line meets the $y$ -axis at $(0, - 5)$ and the $x$ -axis between 1 and 2, at $x = \frac{5}{3}$ .
11
Long answerHardNCERT End-of-chapter Exercises, Q11
Two linear polynomials $f (x) = a x + b$ and $g (x) = c x + d$ are linked in two ways: the graph of $g$ passes through $(0, 1)$ and $(2, 7)$ , and $f (x) - g (x) = x - 5$ for every value of $x$ .
1. Find $g (x)$ and $f (x)$ .
  Show solution
  Answer $g (x) = 3 x + 1$ and $f (x) = 4 x - 4$ .
  The graph of $g$ passes through $(0, 1)$ , so $d = 1$ . From $(0, 1)$ to $(2, 7)$ it rises 6 for 2 across, so $c = 6 \div 2 = 3$ . Hence $g (x) = 3 x + 1$ .
  Then $f (x) = g (x) + (x - 5) = 3 x + 1 + x - 5 = 4 x - 4$ . Check: $(4 x - 4) - (3 x + 1) = x - 5$ .
2. At what value of $x$ do the graphs of $f$ and $g$ meet?
  Show answer
  Answer 5
  The graphs meet where $f (x) = g (x)$ , that is, where $f (x) - g (x) = x - 5 = 0$ . So $x = 5$ , and both give $y = 16$ : the graphs meet at $(5, 16)$ .
12
Short answerMediumNCERT End-of-chapter Exercises, Q12
Fig. 2.17 shows how a footbridge truss is built from equal steel rods. Each new triangle shares one rod with the triangle before it and needs two new rods.
1. Why does each stage need exactly 2 more rods than the stage before it?
  Show solution
  AnswerThe new triangle has 3 sides, but one of them is a rod of the previous triangle.
  A triangle needs 3 rods. The new triangle shares one side with the last triangle of the previous stage, so only its other 2 sides are new rods.
2. Complete the table for stages 1 to 5, and find a rule for the number of rods in stage $n$ .
  Show solution
  Answer3, 5, 7, 9, 11, …; $2 n + 1$
  Stage 1 2 3 4 5 … $n$
  Rods 3 5 7 9 11 … $2 n + 1$
  Start with 3 and add 2 for each later stage: $3 + 2 (n - 1) = 2 n + 1$ . Or count by direction: $n$ triangles have $n + 1$ slanting rods and $n$ level rods, and $(n + 1) + n = 2 n + 1$ .
3. How many rods does a truss of 40 triangles need?
  rods
  Show answer
  Answer 81 rods
  $2 \times 40 + 1 = 81$ .
4. A workshop has 100 rods. What is the largest stage it can build?
  Show answer
  Answer 49
  Stage 49 needs $2 \times 49 + 1 = 99$ rods, leaving 1 over, and stage 50 would need 101. Every stage uses an odd number of rods, so no stage uses exactly 100.
13
Long answerHardNCERT End-of-chapter Exercises, Q13
Line $L_{1}$ goes through $(1, - 1)$ and $(4, 5)$ . Line $L_{2}$ is parallel to $L_{1}$ and crosses the $x$ -axis at $(- 4, 0)$ .
1. Find the equation of $L_{1}$ in the form $y = a x + b$ .
  Show solution
  Answer $y = 2 x - 3$
  Slope $= \frac{5 - (- 1)}{4 - 1} = \frac{6}{3} = 2$ . Then $- 1 = 2 \times 1 + b$ gives $b = - 3$ .
2. Find the equation of $L_{2}$ .
  Show solution
  Answer $y = 2 x + 8$
  Parallel, so its slope is also 2. It passes through $(- 4, 0)$ : $0 = 2 \times (- 4) + b$ gives $b = 8$ .
3. At what value of $x$ does $L_{1}$ cross the $x$ -axis?
  Show answer
  Answer 1.5
  $2 x - 3 = 0$ gives $x = 1.5$ , so $L_{1}$ crosses at $(1.5, 0)$ .
4. Measured straight up, how far above $L_{1}$ is $L_{2}$ ?
  units
  Show answer
  Answer 11 units
  At every $x$ the gap is $(2 x + 8) - (2 x - 3) = 11$ . For example, at $x = 0$ the lines are at 8 and $- 3$ .
14
Short answerHardNCERT End-of-chapter Exercises, Q14
The lines $y = x - 1$ , $y = 3 x - 5$ and $y = - 0.5 x + 2$ all pass through one point.
1. Which point is it?
  $(0, 2)$ $(4, 3)$ $(2, 1)$ $(0, - 1)$
  Show answer
  Answer (C) $(2, 1)$
  At $x = 2$ : $2 - 1 = 1$ , $3 \times 2 - 5 = 1$ and $- 0.5 \times 2 + 2 = 1$ . The points $(4, 3)$ and $(0, - 1)$ fit only $y = x - 1$ , and $(0, 2)$ fits only $y = - 0.5 x + 2$ .
2. Explain why every line $y = a x + 1 - 2 a$ , with $a \neq 0$ , passes through this point. Which feature changes from one of these lines to another?
  Show solution
  AnswerThey all pass through $(2, 1)$ ; only the slope $a$ changes.
  $y = a x + 1 - 2 a = a (x - 2) + 1$ . At $x = 2$ this is $a \times 0 + 1 = 1$ , whatever $a$ is, so every such line passes through $(2, 1)$ . The three lines above are the cases $a = 1$ , $3$ and $- 0.5$ . Only the slope changes, so the lines turn about $(2, 1)$ , just as lines with the same $b$ turn about $(0, b)$ .
Quick answerEasy
A linear pattern starts 12, 9, 6, … What is its 10th term?
$- 15$ $- 18$ $- 3$ $- 12$
Show answer
Answer (A) $- 15$
The jump is $- 3$ , so the $n$ th term is $15 - 3 n$ , and the 10th term is $15 - 30 = - 15$ .
Quick answerEasy
Every line of the form $y = a x$ passes through the origin.
TrueFalse
Show answer
Answer True
Putting $x = 0$ gives $y = 0$ , whatever $a$ is.
Quick answerMedium
The line $y = a x + 4$ passes through the point $(2, 10)$ . Find $a$ .
Show answer
Answer 3
$10 = 2 a + 4$ gives $2 a = 6$ , so $a = 3$ .
18
Quick answerMedium
Put these lines in order from the least steep to the steepest.
1. $y = x - 7$
2. $y = - 3 x$
3. $y = - 0.5 x + 1$
4. $y = 0.2 x$
Show answer
Answer
$y = 0.2 x$
$y = - 0.5 x + 1$
$y = x - 7$
$y = - 3 x$
Steepness depends on the size of the slope, not its sign or $b$ : 0.2, 0.5, 1 and 3.

Stage	1	2	3	4	5	…	$n$
Rods	3	5	7	9	11	…	$2 n + 1$

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 Set 2.1, Q1	Practice 2.1 · Q1
NCERT Exercise Set 2.1, Q2	Practice 2.1 · Q2
NCERT Exercise Set 2.1, Q3	Practice 2.1 · Q3
NCERT Exercise Set 2.1, Q4	Practice 2.1 · Q4
NCERT Exercise Set 2.1, Q5	Practice 2.1 · Q5
NCERT Exercise Set 2.2, Q1	Practice 2.2 · Q1
NCERT Exercise Set 2.2, Q2	Practice 2.2 · Q2
NCERT Exercise Set 2.2, Q3	Practice 2.2 · Q3
NCERT Exercise Set 2.2, Q4	Practice 2.2 · Q4
NCERT Exercise Set 2.2, Q5	Practice 2.2 · Q5
NCERT Exercise Set 2.2, Q6	Practice 2.2 · Q6
NCERT Exercise Set 2.2, Q7	Practice 2.2 · Q7
NCERT Exercise Set 2.3, Q1	Practice 2.3 · Q1
NCERT Exercise Set 2.3, Q2	Practice 2.3 · Q2
NCERT Exercise Set 2.3, Q3	Practice 2.3 · Q3
NCERT Exercise Set 2.3, Q4	Practice 2.3 · Q4
NCERT Exercise Set 2.3, Q5	Practice 2.3 · Q5
NCERT Exercise Set 2.4, Q1	Practice 2.4 · Q1
NCERT Exercise Set 2.4, Q2	Practice 2.4 · Q2
NCERT Exercise Set 2.4, Q3	Practice 2.4 · Q3
NCERT Exercise Set 2.4, Q4	Practice 2.4 · Q4
NCERT Exercise Set 2.5, Q1	Practice 2.5 · Q1
NCERT Exercise Set 2.5, Q2	Practice 2.5 · Q2
NCERT Exercise Set 2.5, Q3	Practice 2.5 · Q3
NCERT Exercise Set 2.6, Q1	Practice 2.6 · Q1
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q14	End-of-chapter exercises · Q14

Quick check: MCQs on Introduction to Linear Polynomials

2.1 Introduction3 questions
Polynomials in one variable4 questions
Polynomials as input–output rules4 questions
Rangoli check3 questions
2.3 Exploring linear patterns4 questions
2.4 Linear growth and linear decay3 questions
2.5 Linear Relationships3 questions
Two points fix a line3 questions
Where a line meets the $y$-axis4 questions

Glossary · 15

Key terms (English–Hindi)

coefficientगुणांकgunank: The number that multiplies a variable or a power of it. In $- x^{2}$ the coefficient is $- 1$ .First seen in the chapter →
constant termअचर पदachar pad: A term with no variable in it, so its value never changes, like $- 5$ in $2 x^{3} - x^{2} + 7 x - 5$ .First seen in the chapter →

degreeघातghaat: In a polynomial, the largest power of the variable that appears. $2 x^{3} - x^{2} + 7 x - 5$ has degree 3.First seen in the chapter →

functionफलनphalan: A rule that gives exactly one output for each input, such as $p (x) = 3 x - 4$ .First seen in the chapter →

linear decayरैखिक ह्रासrekhik hraas: Decline by equal subtractions: in every time step of the same length, the quantity loses the same fixed amount.First seen in the chapter →
linear equationरैखिक समीकरणrekhik samikaran: An equation made by setting a linear polynomial equal to a number, such as $25 + 2 p = 85$ .First seen in the chapter →
linear growthरैखिक वृद्धिrekhik vriddhi: Growth by equal additions: in every time step of the same length, the quantity gains the same fixed amount.First seen in the chapter →
linear patternरैखिक प्रतिरूपrekhik pratiroop: A list of numbers in which the same number is added each time to get the next term.First seen in the chapter →
linear polynomialरैखिक बहुपदrekhik bahupad: A polynomial of degree 1, which can be written as $a x + b$ with $a \neq 0$ .First seen in the chapter →
linear relationshipरैखिक संबंधrekhik sambandh: A link between two changing quantities of the form $y = a x + b$ ; its graph is a straight line.First seen in the chapter →

polynomialबहुपदbahupad: A sum of terms, each a number times a whole-number power of a variable, such as $2 x^{3} - x^{2} + 7 x - 5$ .First seen in the chapter →

slopeढालdhaal: The number $a$ in $y = a x + b$ : how much $y$ changes when $x$ goes up by 1.First seen in the chapter →

termपदpad: One of the parts of an expression that are added together, such as $4 f$ , $6 s$ and $11$ in $4 f + 6 s + 11$ .First seen in the chapter →

variableचरchar: A letter that stands for a number that can change, such as $x$ or $t$ .First seen in the chapter →

y-intercepty-अंतःखंडy-antahkhand: The number $b$ in $y = a x + b$ ; the line meets the $y$ -axis at $(0, b)$ .First seen in the chapter →

Introduction to Linear Polynomials: Class 9 Ganita Manjari · Chapter 2

What you will learn in Introduction to Linear Polynomials

2.1 Introduction

Polynomials in one variable

Practice 2.1: questions with answers

2.2 Linear Polynomials

Linear equations

Polynomials as input–output rules

Practice 2.2: questions with answers

2.3 Exploring linear patterns

Practice 2.3: questions with answers

2.4 Linear growth and linear decay

Practice 2.4: questions with answers

2.5 Linear Relationships

Practice 2.5: questions with answers

2.6 Visualising linear relationships

Two points fix a line

Lines through the origin: slope

Changing $b$ : parallel lines

Where a line meets the $y$ -axis

Practice 2.6: questions with answers

Introduction to Linear Polynomials: key points and summary

Practice questions with answers (NCERT “End-of-chapter exercises”)

Find your NCERT exercise question

Quick check: MCQs on Introduction to Linear Polynomials

Key terms (English–Hindi)

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2.1 Introduction

Polynomials in one variable

2.2 Linear Polynomials

Linear equations

Polynomials as input–output rules

2.3 Exploring linear patterns

2.4 Linear growth and linear decay

2.5 Linear Relationships

2.6 Visualising linear relationships

Two points fix a line

Lines through the origin: slope

Changing b: parallel lines

Where a line meets the y-axis

Find your NCERT exercise question

Quick check: MCQs on Introduction to Linear Polynomials

You reached the end

Changing $b$ : parallel lines

Where a line meets the $y$ -axis