MathsCBSE Class 9 · Chapter 2

Introduction to Linear Polynomials Class 9: Concepts, Graphs and Solved Questions

Updated 15 October 2026 · Fact-checked

Introduction to Linear Polynomials in Class 9 covers algebraic expressions, polynomials in one variable and their degree, and linear polynomials such as 2x + 3. You use them to model linear patterns, linear growth and decay, find a relationship y = ax + b from two facts, and draw it as a straight line with slope a and y-intercept b.

What Introduction to Linear Polynomials is about

This chapter of *Ganita Manjari* Part I is CBSE's Introduction to Polynomials (Unit II, Algebra). It starts from everyday algebraic expressions, such as the cost of fencing a garden, and narrows down to expressions in one variable. You meet the words term, coefficient, constant and degree, and then focus on polynomials of degree 1: linear polynomials.

The big idea is that linear polynomials describe situations that change by the same amount at each step: a chess club fee that goes up by ₹50 a match, a pocket-money balance that goes down by ₹5 a day, an auto fare that rises ₹15 per km. These are linear patterns. When the amount goes up it is linear growth; when it goes down it is linear decay.

The chapter then joins algebra to the coordinate plane of Chapter 1:

a linear relationship is written $y = a x + b$ , and two facts are enough to find a and b
its graph is a straight line; a is the slope (how steep, and which way) and b is the y-intercept (where it cuts the y-axis)
lines with the same slope are parallel

Expect 1-mark questions on degree and coefficients, 2- and 3-mark word problems that lead to a linear equation, and longer questions that ask you to model a situation, find $y = a x + b$ and draw or read its graph.

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 marks in the school annual exam, shared with Sequences and Progressions, Exploring Algebraic Identities and Linear Equations in Two Variables.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit II Algebra

Key concepts

Algebraic expressions: terms, variables, coefficients and constants

NCERT section 2.1 Introduction

An algebraic expression joins numbers and variables (letter-numbers such as x, y) with operations. In $40 x + 15 y + 20$ :

the terms are $40 x$ , $15 y$ and $20$
the variables are x and y
the coefficients are 40 (of x) and 15 (of y)
the constant is 20

An expression can have one variable, like $x (10 - x) = 10 x - x^{2}$ , or several, like $200 l + 160 w + 50 l w$ . In this chapter we work with expressions in one variable.

Worked example

A stationery shop sells notebooks at ₹40 each and pens at ₹15 each, and adds a fixed ₹20 delivery charge. Write the bill for x notebooks and y pens. Name its terms, variables, coefficients and constant, and find the bill for 3 notebooks and 4 pens.

Try it first, then show the solutionSolution

Cost of notebooks $= 40 x$ , cost of pens $= 15 y$ , delivery $= 20$ .
Bill $= 40 x + 15 y + 20$ .
Terms: $40 x$ , $15 y$ , $20$ . Variables: x, y. Coefficients: 40 and 15. Constant: 20.
For $x = 3$ , $y = 4$ : $40 \times 3 + 15 \times 4 + 20 = 120 + 60 + 20 = 200$ .

Answer: Bill $= ₹ (40 x + 15 y + 20)$ ; for 3 notebooks and 4 pens it is ₹200.

Polynomials in one variable and their degree

NCERT section 2.1 Introduction

An expression in one variable whose terms are numbers times whole-number powers of that variable is a polynomial in one variable (also called a univariate polynomial). The degree is the highest power of the variable.

Degree	Name	Example
0	constant	$8$ (think of it as $8 x^{0}$ )
1	linear	$3 z + 7$
2	quadratic	$x^{2} + 5 x + 1$
3	cubic	$5 y^{3} + y^{2} + 2 y - 1$

The coefficient of a power is the number multiplying it, sign included. If a power is missing, its coefficient is 0. Write the polynomial in descending powers before you read off the degree.

Worked example

For $p (x) = 3 - 4 x + x^{3} - 6 x^{2}$ : (i) write it in descending powers, (ii) give its degree and name, (iii) give the coefficient of $x^{2}$ , the coefficient of x and the constant term.

Try it first, then show the solutionSolution

(i) Descending powers: $x^{3} - 6 x^{2} - 4 x + 3$ .
(ii) The highest power is 3, so the degree is 3: a cubic polynomial.
(iii) Coefficient of $x^{2}$ is $- 6$ ; coefficient of x is $- 4$ ; constant term is $3$ .

Answer: $x^{3} - 6 x^{2} - 4 x + 3$ ; degree 3 (cubic); coefficients $- 6$ and $- 4$ ; constant 3.

Linear polynomials as input-output machines

NCERT section 2.2 Linear Polynomials

A linear polynomial has degree 1, so it looks like $a x + b$ with $a \neq = 0$ . Examples: $4 x$ (perimeter of a square of side x), $200 + 50 m$ (club fee for m matches).

You can think of $p (x) = a x + b$ as a machine: put in a value of x, get out one value. This is called a function of x. To find the output, substitute and simplify; this is called finding the value of the polynomial, written $p (k)$ .

Key feature: when x goes up by 1, a linear polynomial always changes by the same amount, a.

Worked example

For $p (x) = 3 x - 7$ , find $p (0)$ , $p (- 2)$ and $p (\frac{5}{3})$ . By how much does $p (x)$ change when x increases by 1?

Try it first, then show the solutionSolution

$p (0) = 3 (0) - 7 = - 7$ .
$p (- 2) = 3 (- 2) - 7 = - 6 - 7 = - 13$ .
$p (\frac{5}{3}) = 3 \times \frac{5}{3} - 7 = 5 - 7 = - 2$ .
$p (x + 1) - p (x) = 3 (x + 1) - 7 - (3 x - 7) = 3$ .

Answer: $p (0) = - 7$ , $p (- 2) = - 13$ , $p (\frac{5}{3}) = - 2$ ; p increases by 3 for every increase of 1 in x.

From a word problem to a linear equation

NCERT section 2.2 Linear Polynomials

Setting a linear polynomial equal to a number gives a linear equation, which has one solution. The method for word problems:
1. Choose the unknown and write 'Let ... be x'.
2. Write every other quantity in terms of x.
3. Form the equation from the fact given in the question.
4. Solve, then check the answer in the words of the question.

Age problems, number problems, coin problems and perimeter problems all follow this pattern.

Worked example

A mother is now 4 times as old as her son. After 6 years, the sum of their ages will be 72 years. Find their present ages.

Try it first, then show the solutionSolution

Let the son's present age be x years. The mother's present age is $4 x$ years.
After 6 years: son $x + 6$ , mother $4 x + 6$ .
$(x + 6) + (4 x + 6) = 72$ , so $5 x + 12 = 72$ .
$5 x = 60$ , so $x = 12$ .
Check: ages 12 and 48; after 6 years 18 and 54, and $18 + 54 = 72$ .

Answer: The son is 12 years old and the mother is 48 years old.

Linear patterns

NCERT section 2.3 Exploring linear patterns

In a linear pattern, each term is the previous one plus the same fixed number. Because of that constant difference, the n-th term is a linear polynomial in n.

To find the rule:
1. Find the constant difference d. The rule starts with $d \times n$ .
2. Compare $d \times n$ with the actual terms and add or subtract the same number to match.

For tiles 1, 3, 5, 7, ... the difference is 2; $2 n$ gives 2, 4, 6, ..., which is 1 too many each time, so the rule is $2 n - 1$ . Use the rule both ways: to find a term from its position, and to test whether a number appears in the pattern (n must come out as a counting number).

Worked example

Squares are made in a row with matchsticks, each new square sharing a side with the last. Stage 1 uses 4 sticks, Stage 2 uses 7, Stage 3 uses 10. (i) Find a rule for Stage n. (ii) How many sticks are in Stage 20? (iii) Which stage uses 151 sticks?

Try it first, then show the solutionSolution

The difference is $7 - 4 = 3$ each time, so the rule starts with $3 n$ .
$3 n$ gives 3, 6, 9, which is 1 less than 4, 7, 10. So the rule is $3 n + 1$ .
Stage 20: $3 (20) + 1 = 61$ sticks.
$3 n + 1 = 151$ , so $3 n = 150$ and $n = 50$ .

Answer: (i) $3 n + 1$ (ii) 61 sticks (iii) Stage 50

Linear growth and linear decay

NCERT section 2.4 Linear growth and linear decay

Linear expressions model quantities that change by a fixed amount over equal intervals of time or distance.

Linear growth: increases by a fixed amount each interval, e.g. journey cost $C (d) = 100 + 60 d$ .
Linear decay: decreases by a fixed amount each interval, e.g. water height $h (t) = 3 - 0.5 t$ .

General form: value $=$ starting value $\pm$ (change per interval) $\times$ (number of intervals). In decay problems, the time when the quantity runs out is found by setting the expression equal to 0. Making a table of values for a few intervals is a quick check that the change really is constant.

Worked example

A storage tank holds 5000 L of water and leaks 250 L every day. (i) Write the volume V after t days. (ii) Is this growth or decay? (iii) Find the volume after 8 days and when the tank will be empty.

Try it first, then show the solutionSolution

(i) $V = 5000 - 250 t$ .
(ii) The volume falls by the same 250 L each day, so it is linear decay.
(iii) After 8 days: $V = 5000 - 250 \times 8 = 3000$ L.
Empty when $5000 - 250 t = 0$ , so $t = 20$ days.

Answer: $V = 5000 - 250 t$ (linear decay); 3000 L after 8 days; empty after 20 days.

Finding a linear relationship y = ax + b from two facts

NCERT section 2.5 Linear Relationships

A linear relationship between two variables is $y = a x + b$ . If you know two pairs of values, you can find a and b:
1. Substitute each pair to get two equations in a and b.
2. Subtract one from the other to remove b and find a.
3. Put a back into either equation to find b.

In a bill, a is the rate (cost per unit) and b is the fixed charge (the bill when x = 0). Shortcut: $a = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ and $b = y_{1} - a x_{1}$ .

Worked example

A taxi company charges a fixed amount plus a rate per km. A ride of 8 km costs ₹260 and a ride of 12 km costs ₹360. Find the relation $y = a x + b$ , and say what a and b mean.

Try it first, then show the solutionSolution

$260 = 8 a + b$ and $360 = 12 a + b$ .
Subtract: $100 = 4 a$ , so $a = 25$ .
$b = 260 - 8 \times 25 = 260 - 200 = 60$ .
So $y = 25 x + 60$ . Check: $25 \times 12 + 60 = 360$ .

Answer: $y = 25 x + 60$ : ₹25 per km plus a fixed charge of ₹60.

Graph of y = ax + b: slope and y-intercept

NCERT section 2.6 Visualising linear relationships

Every linear relationship $y = a x + b$ has a straight-line graph. Two points are enough to draw it; a third point is a useful check. A point lies on the line only if its coordinates satisfy the equation.

Slope a: how steep the line is. $a > 0$ : the line rises to the right (growth). $a < 0$ : it falls to the right (decay). For $y = a x$ , a bigger $∣ a ∣$ means steeper; $a = 1$ gives the line $y = x$ , equally inclined to both axes.
y-intercept b: the line cuts the y-axis at $(0, b)$ . Lines $y = a x$ pass through the origin.
Changing b with a fixed slides the line up or down: lines with equal slopes and different intercepts are parallel.

To find where a line cuts the x-axis, put $y = 0$ .

Worked example

For the line $y = - 2 x + 3$ : (i) find two points on it and describe its slope and y-intercept; (ii) find where it cuts the x-axis; (iii) check whether $(4, - 5)$ lies on it.

Try it first, then show the solutionSolution

(i) $x = 0$ gives $y = 3$ , so $(0, 3)$ ; $x = 2$ gives $y = - 1$ , so $(2, - 1)$ .
Slope $a = - 2$ (negative, so the line falls to the right); y-intercept $b = 3$ .
(ii) Put $y = 0$ : $- 2 x + 3 = 0$ , so $x = 1.5$ . It cuts the x-axis at $(1.5, 0)$ .
(iii) $- 2 (4) + 3 = - 5$ , which equals the y-coordinate, so $(4, - 5)$ lies on the line.

Answer: Points $(0, 3)$ and $(2, - 1)$ ; slope $- 2$ , y-intercept 3; x-axis at $(1.5, 0)$ ; yes, $(4, - 5)$ is on the line.

Maths in real life: Introduction to Linear Polynomials

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Linear polynomial as an input-output rule

Suppose a cyber café charges a fixed ₹20 to log in plus ₹30 for every hour you use a computer.

Write the charge for h hours and find what you pay for 3.5 hours.

Try it first, then show the solutionSolution

Charge $= 20 + 30 h$ .
For $h = 3.5$ : $20 + 30 \times 3.5 = 20 + 105 = 125$ .

Answer: ₹125 for 3.5 hours.

Memory hook: Fixed part plus rate times units: every bill is a linear polynomial.

Linear growth

Suppose you plant a 40 cm sapling in the school garden and it grows a steady 6 cm every month.

Write its height after t months and find when it reaches 1 m.

Try it first, then show the solutionSolution

Height $= 40 + 6 t$ cm.
1 m = 100 cm, so $40 + 6 t = 100$ .
$6 t = 60$ , so $t = 10$ months.

Answer: $h = 40 + 6 t$ ; it reaches 1 m after 10 months.

Memory hook: Growth adds the same amount every month.

Linear decay

Suppose your phone starts at 100% charge and loses 8% every hour while you watch videos.

Find the charge left after 4 hours and when the battery would be flat.

Try it first, then show the solutionSolution

Charge $= 100 - 8 t$ .
After 4 h: $100 - 32 = 68%$ .
Flat when $100 - 8 t = 0$ : $t = 12.5$ h.

Answer: 68% after 4 hours; flat after 12.5 hours.

Memory hook: Decay takes away the same amount every hour; set it to zero for 'runs out'.

Finding y = ax + b from two facts

Suppose a print shop charges ₹350 for 100 copies of your project and ₹650 for 250 copies, with a fixed set-up charge plus a rate per copy.

Find the rate per copy and the fixed charge.

Try it first, then show the solutionSolution

$350 = 100 a + b$ and $650 = 250 a + b$ .
Subtract: $300 = 150 a$ , so $a = 2$ .
$b = 350 - 200 = 150$ .

Answer: ₹2 per copy plus a fixed ₹150: $y = 2 x + 150$ .

Memory hook: Two bills are enough to unmask the rate and the fixed charge.

Comparing slopes

Suppose taxi A charges ₹50 plus ₹15 per km and taxi B charges ₹80 plus ₹12 per km.

For what distance do they cost the same, and which is cheaper for a 20 km trip?

Try it first, then show the solutionSolution

$15 x + 50 = 12 x + 80$ gives $3 x = 30$ , $x = 10$ km.
At 20 km: A $= 350$ , B $= 320$ .

Answer: Same cost at 10 km (₹200); for 20 km, taxi B is cheaper by ₹30.

Memory hook: Smaller slope wins on long trips; smaller intercept wins on short ones.

Topics in this chapter

Introduction to Linear Polynomials formula sheet

Polynomial in one variable: $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$; Powers are whole numbers; the degree is the highest power n with $a_{n} \neq = 0$ .
Types by degree: $0 : constant, 1 : linear, 2 : quadratic, 3 : cubic$; A constant like 8 has degree 0 because $8 = 8 x^{0}$ .
Linear polynomial: $p (x) = a x + b, a \neq = 0$; Degree exactly 1.
Value of a polynomial: $p (k) = value of p (x) when x = k$; Substitute, keeping negative numbers in brackets.
Constant change: $p (x + 1) - p (x) = a$; A linear polynomial changes by a for every step of 1 in x.
Linear pattern rule: $T_{n} = d n + c$; d is the constant difference; choose c so that the first term fits.
Linear growth: $V = V_{0} + r t$; Starting value $V_{0}$ , fixed increase r per interval.
Linear decay: $V = V_{0} - r t$; Runs out when $V = 0$ , that is $t = \frac{V _{0}}{r}$ .
Linear relationship: $y = a x + b$; a: slope (rate); b: y-intercept (fixed part).
a and b from two points: $a = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}, b = y_{1} - a x_{1}$; Same as solving the two equations $y_{1} = a x_{1} + b$ , $y_{2} = a x_{2} + b$ .
Where the line meets the axes: $y-axis : (0, b); x-axis : (- \frac{b}{a}, 0)$; Put $x = 0$ or $y = 0$ .
Parallel lines: $y = a x + b_{1} ∥ y = a x + b_{2} (b_{1} \neq = b_{2})$; Equal slopes, different intercepts.
Lines through the origin: $y = a x$; b = 0; steeper than $y = x$ when $∣ a ∣ > 1$ .

How the school exam asks Introduction to Linear Polynomials

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Degree, coefficient and constant	MCQs or 1-mark questions: the degree of a polynomial, the coefficient of a given power, the constant term, or which expression is linear.	Rewrite in descending powers. Read the highest power for the degree. Keep the sign with each coefficient. A missing power has coefficient 0.
Value of a polynomial	Find $p (- 2)$ or $p (\frac{1}{2})$ , or find x when $p (x)$ has a given value.	Substitute with brackets, do powers first, then multiplication, then addition. For 'find x', set the polynomial equal to the value and solve.
Word problem to linear equation	Ages, two numbers, coins, a cut fence, or a rectangle with a given perimeter.	Write 'Let ... be x', express every quantity in x, form one equation, solve and check in words. State the final answer with units.
Linear pattern rule	A tile, matchstick or table pattern; find the n-th term, a given term, or whether a number can appear.	Find the constant difference d, write $d n + c$ , fix c from Stage 1. For 'can it appear', solve $d n + c =$ number and check that n is a counting number.
Growth and decay models	Savings, depreciation, a leaking tank, a recharge balance; write the expression, make a table, find a value or when it runs out.	Value $=$ start $\pm$ rate $\times$ time. Say 'growth' if the rate is added and 'decay' if it is subtracted. Set the expression to 0 for 'runs out'.
Finding y = ax + b and reading its graph	Two bills or two readings are given; find a and b, interpret them, and draw or compare lines.	Form two equations, subtract to get a, substitute to get b. Plot two points plus a check point. Read slope and intercept; equal slopes mean parallel lines.

Tricks to remember Introduction to Linear Polynomials

Short memory aids. Say each one aloud once, then use it on a question the same day.

Degree = the top floor
Think of the powers as floors of a building. The degree is the top floor that has someone living in it (non-zero coefficient).
Use it when: Avoids counting terms instead of reading the highest power.
a is the attitude, b is the base
In $y = a x + b$ , a is the line's attitude (how steep and which way it leans) and b is its base on the y-axis.
Use it when: Read slope and y-intercept correctly even when the equation is written $y = 4 - 3 x$ .
Same slope, same lane
Lines with the same a are cars in parallel lanes: they never meet.
Use it when: Spot parallel lines in MCQs and graph questions.
Difference first, then fix
For a pattern rule: the constant difference goes in front of n, then fix the first term.
Use it when: Writing $T_{n} = d n + c$ quickly.
Grow adds, decay takes away
Growth: start + rate × time. Decay: start − rate × time.
Use it when: Setting up depreciation and savings models without sign mistakes.

Introduction to Linear Polynomials: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Degree (MCQ) · 1 mark · MCQExam level

The degree of $7 - 2 x + 5 x^{3} - x^{2}$ is
(a) 1
(b) 2
(c) 3
(d) 4

Try it first, then show the solutionSolution

The powers present are 0, 1, 3 and 2.
The highest is 3.

Answer: (c) 3

Example 2: Spot the linear polynomial · 1 mark · MCQExam level

Which of these is a linear polynomial?
(a) $x^{2} + 1$
(b) $3 - 4 y$
(c) $5$
(d) $x (x + 1)$

Try it first, then show the solutionSolution

(a) has degree 2; (c) has degree 0; (d) is $x^{2} + x$ , degree 2.
(b) is $- 4 y + 3$ , degree 1.

Answer: (b) $3 - 4 y$

Example 3: Parallel lines (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): The lines $y = 3 x + 2$ and $y = 3 x - 5$ are parallel.
Reason (R): Lines with the same slope and different y-intercepts are parallel.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Both lines have slope 3 and intercepts 2 and $- 5$ .
R is the rule that makes A true.

Answer: (a)

Example 4: Value at a negative number · 2 marks · Very short answerExam level

Find the value of $4 t^{2} - 3 t + 1$ when $t = - 2$ .

Try it first, then show the solutionSolution

$4 (- 2)^{2} - 3 (- 2) + 1 = 4 \times 4 + 6 + 1$ .
$= 16 + 6 + 1 = 23$ .

Answer: 23

Example 5: Two numbers · 3 marks · Short answerHarder exam level

One number is 3 times another. If 12 is added to both, the larger becomes twice the smaller. Find the numbers.

Try it first, then show the solutionSolution

Let the numbers be x and $3 x$ .
$3 x + 12 = 2 (x + 12)$ .
$3 x + 12 = 2 x + 24$ , so $x = 12$ .
Check: 24 and 48; $48 = 2 \times 24$ .

Answer: 12 and 36

Example 6: Fixed fee and rate from two bills · 3 marks · Short answerHarder exam level

A family uses 15 kL of water and pays ₹420; next month it uses 25 kL and pays ₹620. The bill is $y = a x + b$ for x kL. Find a and b, explain them, and find the bill for 32 kL.

Try it first, then show the solutionSolution

$420 = 15 a + b$ and $620 = 25 a + b$ .
Subtract: $10 a = 200$ , so $a = 20$ ; $b = 420 - 300 = 120$ .
a is ₹20 per kL; b is the fixed charge ₹120.
Bill for 32 kL $= 20 \times 32 + 120 = 760$ .

Answer: $y = 20 x + 120$ ; ₹760 for 32 kL.

Example 7: Rectangle from its perimeter · 3 marks · Short answerHarder exam level

The length of a rectangle is 4 cm more than twice its width, and its perimeter is 50 cm. Find its length, width and area.

Try it first, then show the solutionSolution

Let the width be w cm; the length is $2 w + 4$ cm.
$2 (2 w + 4 + w) = 50$ , so $6 w + 8 = 50$ and $w = 7$ .
Length $= 2 (7) + 4 = 18$ cm. Check: $2 (18 + 7) = 50$ .
Area $= 18 \times 7 = 126$ cm².

Answer: Width 7 cm, length 18 cm, area 126 cm².

Example 8: Triangle matchstick pattern · 5 marks · Long answerHarder exam level

Triangles are made in a row from matchsticks, each new triangle sharing a side with the last. Stage 1 uses 3 sticks, Stage 2 uses 5, Stage 3 uses 7.
(i) Find the rule for Stage n.
(ii) How many sticks does Stage 25 need?
(iii) Can exactly 100 sticks make a stage? Justify.
(iv) Is the pattern linear growth or decay? What does the slope of its graph represent?

Try it first, then show the solutionSolution

Difference $= 2$ each time, so the rule starts $2 n$ ; $2 n$ gives 2, 4, 6, one less than 3, 5, 7. Rule: $2 n + 1$ .
Stage 25: $2 (25) + 1 = 51$ sticks.
$2 n + 1 = 100$ gives $n = 49.5$ , which is not a counting number, so 100 sticks cannot make a stage (it would need an odd number).
The number of sticks increases by 2 at every stage, so it is linear growth.
On the graph of $y = 2 x + 1$ , the slope 2 is the number of sticks added at each stage.

Answer: (i) $2 n + 1$ (ii) 51 (iii) No, n would be 49.5 (iv) Linear growth; the slope 2 is the sticks added per stage.

Example 9: Two parallel lines · 5 marks · Long answerHarder exam level

Draw $y = 2 x - 3$ and $y = 2 x + 1$ on the same axes. (i) Write the slope and y-intercept of each. (ii) Find where each line cuts the x-axis. (iii) Explain why the lines never meet.

Try it first, then show the solutionSolution

$y = 2 x - 3$ : $(0, - 3)$ , $(1, - 1)$ , $(3, 3)$ . $y = 2 x + 1$ : $(0, 1)$ , $(1, 3)$ , $(- 1, - 1)$ .
(i) Slopes are both 2; intercepts are $- 3$ and $1$ .
(ii) $2 x - 3 = 0$ gives $(1.5, 0)$ ; $2 x + 1 = 0$ gives $(- 0.5, 0)$ .
(iii) For the same x, the second line is always $1 - (- 3) = 4$ units higher, because both rise by 2 for each step in x.
So the gap never closes: the lines are parallel.

Answer: Both have slope 2; intercepts $- 3$ and 1; x-axis crossings $(1.5, 0)$ and $(- 0.5, 0)$ ; parallel, always 4 units apart vertically.

Example 10: Auto fare with a minimum charge · 3 marks · Short answerHOTS / competency

An auto charges ₹30 for the first 2 km and ₹14 for every km after that. (i) Write the fare for n km, $n \geq 2$ , as a linear polynomial. (ii) Find the fare for 12 km. (iii) How far can you go for ₹198?

Try it first, then show the solutionSolution

(i) Fare $= 30 + 14 (n - 2) = 14 n + 2$ .
(ii) $14 (12) + 2 = 170$ , so ₹170.
(iii) $14 n + 2 = 198$ , so $14 n = 196$ and $n = 14$ km.
Check (iii): $30 + 14 \times 12 = 198$ .

Answer: (i) $14 n + 2$ (ii) ₹170 (iii) 14 km

Example 11: Choosing a data plan · 4 marks · Case-basedHOTS / competency

Plan A costs ₹150 a month plus ₹10 per GB. Plan B costs ₹250 a month plus ₹6 per GB. (i) Write the monthly cost of each for x GB. (ii) For what usage do they cost the same? (iii) Which plan should a student who uses about 30 GB a month choose, and why?

Try it first, then show the solutionSolution

(i) A: $y = 10 x + 150$ ; B: $y = 6 x + 250$ .
(ii) $10 x + 150 = 6 x + 250$ , so $4 x = 100$ and $x = 25$ GB; both cost ₹400.
(iii) At 30 GB: A $= 450$ , B $= 430$ .
Above 25 GB, B grows more slowly (smaller slope), so B is cheaper.

Answer: (i) $10 x + 150$ and $6 x + 250$ (ii) at 25 GB, ₹400 each (iii) Plan B: ₹430 against ₹450, and the gap widens as usage grows.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Filling the school water tank

A school's overhead tank holds 2200 L when full. At 6 a.m. it has 400 L of water in it. A pump then adds water at a steady 150 L per hour until the tank is full.

(i) · 1 mark
Write an expression for the volume of water V after t hours of pumping.
Answer
Answer: $V = 400 + 150 t$
Start 400 L, adding 150 L each hour: $V = 400 + 150 t$ .
(ii) · 1 mark
How much water is in the tank after 5 hours?
Answer
Answer: 1150 L
$400 + 150 \times 5 = 1150$ L.
(iii) · 2 marks
At what time will the tank be full? Is this growth or decay, and what do the numbers 400 and 150 mean on the graph of V against t?
Answer
Answer: Full at 6 p.m.; linear growth; 400 is the y-intercept, 150 is the slope.
$400 + 150 t = 2200$ gives $t = 12$ hours, so at 6 p.m. (1 mark). The volume rises by a fixed 150 L each hour, so it is linear growth; on the graph, 400 is where the line starts on the V-axis (y-intercept) and 150 is the slope (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the value of $p (x) = 2 x^{2} - 5 x + 3$ at $x = - 1$ and at $x = \frac{3}{2}$ . (2 marks)

Step	Marks
$p (- 1) = 2 (- 1)^{2} - 5 (- 1) + 3 = 2 + 5 + 3 = 10$	1
$p (\frac{3}{2}) = 2 \times \frac{9}{4} - \frac{15}{2} + 3 = 4.5 - 7.5 + 3 = 0$	1

A sports hall charges a fixed monthly fee plus a fee per hour. For 6 hours the bill is ₹540 and for 10 hours it is ₹780. Find the relation $y = a x + b$ and explain what a and b stand for. (3 marks)

Step	Marks
Two equations: $540 = 6 a + b$ and $780 = 10 a + b$	1
Subtracting: $240 = 4 a$ , so $a = 60$ ; then $b = 540 - 360 = 180$	1
$y = 60 x + 180$ : ₹60 per hour, fixed fee ₹180	1

Draw the graphs of $y = 2 x + 1$ and $y = 2 x - 3$ on the same axes. Write the slope and y-intercept of each and state how the two lines are related. Where does $y = 2 x - 3$ cut the x-axis? (5 marks)

Step	Marks
Table for $y = 2 x + 1$ : $(0, 1)$ , $(1, 3)$ , $(2, 5)$	1
Table for $y = 2 x - 3$ : $(0, - 3)$ , $(1, - 1)$ , $(2, 1)$	1
Both lines plotted correctly, with axes labelled and lines extended	1
Slopes both 2; y-intercepts 1 and $- 3$ ; equal slopes, so the lines are parallel	1
$2 x - 3 = 0$ gives $x = 1.5$ : the line cuts the x-axis at $(1.5, 0)$	1

Where students lose marks in Introduction to Linear Polynomials

The slips examiners see most often, and what to write instead.

Mistake: Counting terms instead of finding the highest power
Degree is confused with the number of terms.
Wrong working
$x^{4} + 3$ has 2 terms, so its degree is 2.
Right working
The highest power of x is 4, so the degree is 4.
Do this: Look only at the exponents and pick the largest.
Usually costs: The 1-mark answer.
Mistake: Squaring a negative number wrongly
The bracket is dropped when substituting.
Wrong working
$2 x^{2}$ at $x = - 3$ : $2 \times - 3^{2} = - 18$
Right working
$2 (- 3)^{2} = 2 \times 9 = 18$
Do this: Always substitute negative values inside brackets.
Usually costs: The value mark and the final answer.
Mistake: Ignoring the starting condition in a fare or fee
The fixed part covers some units, but students multiply the rate by all units.
Wrong working
Fare for 10 km with ₹25 for the first 2 km and ₹15 per km after: $25 + 15 \times 10 = 175$
Right working
$25 + 15 \times (10 - 2) = 25 + 120 = 145$
Do this: Ask: from which unit does the rate start? Multiply the rate by only those units.
Usually costs: 1 to 2 marks in a 3-mark model.
Mistake: Reading the slope from the wrong number
The equation is not in the form $y = a x + b$ .
Wrong working
$y = 4 - 3 x$ : slope 4, intercept $- 3$
Right working
Rewrite as $y = - 3 x + 4$ : slope $- 3$ , intercept 4
Do this: Rearrange to $y = a x + b$ first; a is the number multiplying x.
Usually costs: 1 mark each for the slope and intercept.
Mistake: Writing the pattern rule as d × n only
The constant difference is spotted but the first term is not checked.
Wrong working
Pattern 1, 3, 5, 7: rule $2 n$
Right working
$2 n$ gives 2, 4, 6, which is 1 too many, so the rule is $2 n - 1$ ; check $n = 1$ : $2 - 1 = 1$
Do this: Always test your rule on the first two terms.
Usually costs: The rule mark and every answer that uses it.
Mistake: Plotting a line from two points with an arithmetic slip
Two wrong points still make a straight line, so the slip is not noticed.
Wrong working
$y = 2 x - 3$ : points $(0, - 3)$ and $(2, 2)$
Right working
$x = 2$ gives $y = 1$ ; use $(0, - 3)$ , $(2, 1)$ and a check point $(3, 3)$
Do this: Plot three points; if they are not in a line, recheck.
Usually costs: The graph marks (often 2).

NCERT exercise guide: Introduction to Linear Polynomials

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise Set 2.1

What it tests: Degree of a polynomial, writing polynomials of a given degree, and reading coefficients and the constant term.

How to approach it: Arrange the terms in descending powers. The degree is the top power; the coefficient of a power includes its sign; the constant term is the term with no variable.

Like the degree and coefficient questions in Exercise Set 2.1

For $q (y) = 6 - 2 y^{3} + y^{5} - 8 y^{2}$ , find (i) the degree, (ii) the coefficient of $y^{3}$ , (iii) the coefficient of $y^{4}$ , (iv) the constant term.

Show the solutionSolution

Descending powers: $y^{5} - 2 y^{3} - 8 y^{2} + 6$ .
(i) Highest power 5, so the degree is 5.
(ii) The coefficient of $y^{3}$ is $- 2$ .
(iii) There is no $y^{4}$ term, so its coefficient is 0.
(iv) The constant term is 6.

Answer: Degree 5; coefficient of $y^{3}$ is $- 2$ ; coefficient of $y^{4}$ is 0; constant term 6.

Exercise Set 2.2

What it tests: Values of linear and quadratic polynomials, and word problems (ages, two numbers in a ratio, coins, cutting a fence, a rectangle's perimeter) that lead to a linear equation.

How to approach it: For values, substitute in brackets. For word problems, let the unknown be x, write every quantity in x, form one equation and check the answer.

Like the coin and ratio problems in Exercise Set 2.2

Meera has twice as many ₹5 coins as ₹10 coins. The coins are worth ₹220 in all. How many coins of each kind does she have?

Show the solutionSolution

Let the number of ₹10 coins be x. Then there are $2 x$ coins of ₹5.
Value: $10 x + 5 (2 x) = 220$ .
$20 x = 220$ , so $x = 11$ .
Check: 11 coins of ₹10 make ₹110 and 22 coins of ₹5 make ₹110; total ₹220.

Answer: 11 coins of ₹10 and 22 coins of ₹5.

Exercise Set 2.3

What it tests: Writing linear expressions for savings that grow each month, a group that shrinks each hour, and areas or volumes when one length changes in equal steps.

How to approach it: Make a short table for the first few steps, find the constant change, and write starting value $\pm$ change $\times$ n. Check with one row of the table.

Like the savings and reading problems in Exercise Set 2.3

Arjun is reading a 640-page book and reads 25 pages each day. Write the number of pages left after n days, find the pages left after 18 days, and say on which day he finishes.

Show the solutionSolution

Pages left after n days $= 640 - 25 n$ .
After 18 days: $640 - 25 \times 18 = 640 - 450 = 190$ .
He finishes when $640 - 25 n \leq 0$ , that is $n \geq 25.6$ .
After 25 days 15 pages are left, so he finishes on day 26.

Answer: $640 - 25 n$ ; 190 pages left after 18 days; he finishes on the 26th day.

Exercise Set 2.4

What it tests: Linear growth (a plant, a village's population) and linear decay (a phone's value, a recharge balance), with tables and expressions.

How to approach it: Identify the start value and the fixed change per interval. Growth adds, decay subtracts. Make the table from t = 0, and set the expression to 0 for when something runs out.

Like the depreciation problem in Exercise Set 2.4

A laptop is bought for ₹48,000 and its value falls by ₹4,500 every year. (i) Write its value v after t years. (ii) Find its value after 4 years. (iii) After how many years will it be worth ₹21,000?

Show the solutionSolution

(i) $v = 48000 - 4500 t$ , linear decay.
(ii) $v = 48000 - 18000 = 30000$ .
(iii) $48000 - 4500 t = 21000$ , so $4500 t = 27000$ and $t = 6$ .

Answer: (i) $v = 48000 - 4500 t$ (ii) ₹30,000 (iii) after 6 years

Exercise Set 2.5

What it tests: Finding a and b in $y = a x + b$ from two readings: platform fees, gym fees and the Celsius–Fahrenheit relation.

How to approach it: Substitute both readings, subtract to remove b, find a, then b. Interpret a as the rate and b as the fixed part, and check with the second reading.

Like the fixed-fee problems in Exercise Set 2.5

A water supplier charges a fixed monthly amount plus a rate per kilolitre. A family using 15 kL pays ₹420, and using 25 kL pays ₹620. Find $y = a x + b$ and the bill for 32 kL.

Show the solutionSolution

$420 = 15 a + b$ and $620 = 25 a + b$ .
Subtract: $200 = 10 a$ , so $a = 20$ .
$b = 420 - 300 = 120$ , so $y = 20 x + 120$ .
For 32 kL: $20 \times 32 + 120 = 760$ .

Answer: $y = 20 x + 120$ ; the bill for 32 kL is ₹760.

Exercise Set 2.6

What it tests: Drawing families of lines to see the role of a and b: $y = a x$ with different a, positive and negative slopes, and lines with the same slope but different intercepts.

How to approach it: For each line take $x = 0$ and two other easy values. Draw each family on one set of axes, then describe: what changes when a changes, and what changes when b changes.

Like the line families in Exercise Set 2.6

Draw $y = 3 x$ , $y = 3 x + 2$ and $y = - 3 x$ on the same axes. What do you notice?

Show the solutionSolution

$y = 3 x$ : $(0, 0)$ , $(1, 3)$ , $(- 1, - 3)$ .
$y = 3 x + 2$ : $(0, 2)$ , $(1, 5)$ , $(- 1, - 1)$ .
$y = - 3 x$ : $(0, 0)$ , $(1, - 3)$ , $(- 1, 3)$ .
$y = 3 x$ and $y = 3 x + 2$ have the same slope, so they are parallel; the second is shifted up by 2.
$y = 3 x$ and $y = - 3 x$ both pass through the origin; one rises and one falls, equally steeply.

Answer: $y = 3 x$ and $y = 3 x + 2$ are parallel (same slope 3); $y = - 3 x$ is a mirror image of $y = 3 x$ in the y-axis, through the origin.

End-of-chapter Exercises

What it tests: Mixed questions on degree and values, number problems, savings patterns, and (starred, enrichment) slopes and intercepts, temperature scales, matchstick patterns and finding linear polynomials from conditions.

How to approach it: Identify which skill each question needs: value, equation, pattern rule, or $y = a x + b$ . Starred questions are not assessed by NCERT, but working through them builds speed.

Like the number problems at the end of the chapter

One number is 3 times another. If 12 is added to each, the larger new number is twice the smaller new number. Find the numbers.

Show the solutionSolution

Let the smaller number be x; the larger is $3 x$ .
After adding 12: $3 x + 12 = 2 (x + 12)$ .
$3 x + 12 = 2 x + 24$ , so $x = 12$ .
Numbers: 12 and 36. Check: $36 + 12 = 48$ and $2 (12 + 12) = 48$ .

Answer: 12 and 36

Practice questions on Introduction to Linear Polynomials

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The coefficient of $x^{2}$ in $2 x^{4} - 5 x^{2} + x - 8$ is
(a) 2
(b) $- 5$
(c) 5
(d) $- 8$
Hint
Keep the sign with the number.
Answer
(b) $- 5$
Q2 · 1 mark · MCQExam level
The y-intercept of the line $y = 4 - 3 x$ is
(a) $- 3$
(b) 3
(c) 4
(d) $\frac{4}{3}$
Hint
Rewrite as $y = a x + b$ .
Answer
(c) 4. In $y = - 3 x + 4$ , b = 4.
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): 5 is a polynomial of degree 0.
Reason (R): 5 can be written as $5 x^{0}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What is $x^{0}$ ?
Answer
(a). $x^{0} = 1$ , so $5 = 5 x^{0}$ and the highest power is 0.
Q4 · 2 marks · Very short answerExam level
For $p (x) = 7 - 2 x$ , find $p (- 3)$ and the value of x for which $p (x) = 0$ .
Hint
Substitute for the first; solve an equation for the second.
Answer
$p (- 3) = 7 + 6 = 13$ ; $7 - 2 x = 0$ gives $x = 3.5$ .
Q5 · 3 marks · Short answerHarder exam level
The length of a rectangle is 4 more than twice its width, and its perimeter is 50 cm. Find its length and width.
Hint
Let the width be w.
Answer
$2 (2 w + 4 + w) = 50$ gives $w = 7$ ; length 18 cm, width 7 cm.
Q6 · 3 marks · Short answerHarder exam level
A scooter bought for ₹72,000 loses ₹6,000 in value every year. Write its value after t years. When will it be worth ₹30,000? After how many years would the model give a value of zero?
Hint
Linear decay: start minus rate times time.
Answer
$v = 72000 - 6000 t$ ; $v = 30000$ when $t = 7$ years; $v = 0$ when $t = 12$ years.
Q7 · 3 marks · Short answerHarder exam level
The graph of a linear polynomial $p (x)$ passes through $(2, 7)$ and $(5, 16)$ . Find $p (x)$ and the points where its graph cuts both axes.
Hint
Write $p (x) = a x + b$ and use the two points.
Answer
$7 = 2 a + b$ , $16 = 5 a + b$ , so $a = 3$ , $b = 1$ : $p (x) = 3 x + 1$ . It cuts the y-axis at $(0, 1)$ and the x-axis at $(- \frac{1}{3}, 0)$ .
Q8 · 2 marks · Very short answerHarder exam level
Kavya has ₹1,200 and saves ₹350 every month. Write the amount after n months. After how many months will she have ₹5,400?
Hint
Linear growth.
Answer
$1200 + 350 n$ ; $350 n = 4200$ , so $n = 12$ months.
Q9 · 2 marks · Very short answerHarder exam level
Which of the points $(1, - 1)$ , $(3, 5)$ and $(- 2, - 8)$ lie on the line $y = 3 x - 4$ ? Show your check for each.
Hint
Substitute x and compare with y.
Answer
$(1, - 1)$ : $3 - 4 = - 1$ , yes. $(3, 5)$ : $9 - 4 = 5$ , yes. $(- 2, - 8)$ : $- 6 - 4 = - 10 \neq = - 8$ , no.
Q10 · 5 marks · Long answerHarder exam level
Draw $y = - x + 4$ and $y = - x - 2$ on the same axes. State the slope and y-intercept of each, the points where each cuts the x-axis, and how the lines are related.
Hint
Take $x = 0$ , 2 and 4 for both.
Answer
Both have slope $- 1$ (falling to the right). Intercepts 4 and $- 2$ . x-axis crossings $(4, 0)$ and $(- 2, 0)$ . Same slope, so the lines are parallel, 6 units apart vertically.
Q11 · 3 marks · Short answerHOTS / competency
Pentagons are made in a row with matchsticks, each new pentagon sharing one side with the last. Stage 1 uses 5 sticks, Stage 2 uses 9, Stage 3 uses 13. Find the rule for Stage n. Can 150 sticks make a stage? Can 149?
Hint
Each new pentagon adds 4 sticks.
Answer
Rule $4 n + 1$ . $4 n + 1 = 150$ gives $n = 37.25$ , so no. $4 n + 1 = 149$ gives $n = 37$ , so yes: Stage 37.
Q12 · 4 marks · Case-basedHOTS / competency
Two candles are lit at the same time. Candle P is 30 cm tall and burns 1.5 cm per hour; candle Q is 24 cm tall and burns 1 cm per hour. After how many hours will they be the same height, and what is that height? Which candle burns out first?
Hint
Set $30 - 1.5 t = 24 - t$ .
Answer
$30 - 1.5 t = 24 - t$ gives $0.5 t = 6$ , so $t = 12$ h; height 12 cm. P burns out at $t = 20$ h and Q at $t = 24$ h, so P burns out first.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the degree of a polynomial?Show answer
The highest power of the variable that has a non-zero coefficient.
2.What is the degree of the constant polynomial 8?Show answer
0, because $8 = 8 x^{0}$ .
3.Give the general form of a linear polynomial.Show answer
$a x + b$ with $a \neq = 0$ .
4.What makes a pattern linear?Show answer
The difference between consecutive terms is constant.
5.What is the difference between linear growth and linear decay?Show answer
Growth increases by a fixed amount over equal intervals; decay decreases by a fixed amount over equal intervals.
6.In $y = a x + b$ , what do a and b tell you about the graph?Show answer
a is the slope; b is the y-intercept, so the line cuts the y-axis at $(0, b)$ .
7.Which lines pass through the origin?Show answer
Lines of the form $y = a x$ .
8.When are two lines $y = a x + b$ parallel?Show answer
When they have the same slope a and different intercepts b.
9.How do you check whether a point lies on a line?Show answer
Substitute its coordinates; the equation must be true.
10.How many points do you need to draw a straight line?Show answer
Two, but plot a third as a check.

Best order to study it

Revise terms, coefficients and constants with two or three everyday expressions.
Learn degree and the four names (constant, linear, quadratic, cubic), then do Exercise Set 2.1.
Practise finding values of polynomials, especially at negative and fractional values.
Solve the word problems of Exercise Set 2.2, checking each answer in words.
Work through linear patterns, growth and decay with tables (Sets 2.3 and 2.4).
Finish with $y = a x + b$ : finding a and b from two facts, then drawing and comparing graphs (Sets 2.5 and 2.6).

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you name the terms, coefficients and constant of $7 x - 2 y + 5$ ?
Can you give the degree of $3 - x^{2} + 4 x^{5}$ ?
Can you write examples of constant, linear, quadratic and cubic polynomials?
Can you find $p (- 2)$ for $p (x) = x^{2} - 3 x + 1$ ?
Can you turn an age or coin problem into a linear equation and solve it?
Can you write the rule for the pattern 5, 9, 13, 17, ...?
Can you decide whether 100 can appear in that pattern?
Can you write a depreciation model and say when the value reaches a given amount?
Can you find $y = a x + b$ from two bills?
Can you draw $y = - x + 2$ and state its slope and intercepts?

Introduction to Linear Polynomials: frequently asked questions

Is this chapter the same as 'Introduction to Polynomials' in the CBSE syllabus?

Yes. CBSE's 2026-27 Class 9 syllabus lists 'Introduction to Polynomials' under Unit II Algebra, and the NCERT chapter that teaches it is 'Introduction to Linear Polynomials' in Ganita Manjari Part I.

Are the remainder theorem and factor theorem in Class 9 now?

No. They were in the old Class 9 syllabus, but the 2026-27 CBSE content for this chapter lists expressions, polynomials, degree, linear polynomials, patterns, growth and decay, linear relationships, and slope and y-intercept. It does not mention the remainder or factor theorem.

Do I need to know the word 'function'?

Only as an idea: a polynomial takes an input x and gives one output, like a machine. NCERT says you will study functions properly in later grades, so you will not be asked for a formal definition.

Are the starred questions at the end of the chapter in the exam?

NCERT says questions marked with an asterisk (*) are for enrichment and are not to be assessed. They are still good practice, and the ideas behind them (finding p(x) from two points, slopes and intercepts) are in the main chapter.

What is the difference between slope and y-intercept in a word problem?

The slope is the rate: how much y changes for each unit of x, such as ₹60 per hour. The y-intercept is the starting or fixed value: the value of y when x is 0, such as a ₹180 fixed fee.

How is this chapter linked to other chapters?

Linear patterns lead into Sequences and Progressions (arithmetic progressions are linear patterns), and the graph of y = ax + b leads into Two Variables, One Line, where you study linear equations in two variables.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part I, Mathematics Textbook for Grade 9 (First Edition 2026)