NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials

NCERT Solutions help students understand concepts clearly, strengthen problem-solving skills, and prepare effectively for school and competitive exams. ALLEN's NCERT Solutions for Class 9 Maths Chapter 2 – Introduction to Linear Polynomials are based on the latest NCERT syllabus (2026–27) and are fully CBSE-aligned. Our expert-created, chapter-wise solutions provide step-by-step answers to every NCERT question, making learning simple and systematic. With free PDF downloads, students can learn anytime and build a strong foundation in algebra.

1.0Download NCERT Class 9 Maths Chapter 2 Introduction to Linear Polynomials Solutions

Class 9 Maths Chapter 2 – Introduction to Linear Polynomials introduces the fundamentals of polynomials, including variables, coefficients, constant terms, degree, linear patterns, functions, and simple linear relationships. Students can download the free PDF of NCERT Solutions for Class 9 Maths Chapter 2 for step-by-step solutions to every exercise, prepared according to the latest NCERT 2026–27 syllabus.

NCERT Solutions Class 9 Maths Chapter 2

2.0Learning Outcomes - NCERT Class 9 Maths Chapter 2 Solutions

Identify the variable, coefficient, constant term, and degree of a polynomial.
Distinguish between constant, linear, quadratic, and cubic polynomials based on their degree.
Understand the characteristics of a linear polynomial.
Find the value of a polynomial by substituting the given value of the variable.
Recognise and represent linear relationships using algebraic expressions.
Understand functions as an input-output process through simple examples.
Model real-life situations using linear growth and linear decay.
Represent and interpret simple linear relationships using tables and graphs.

3.0Detailed Class 9 Maths Chapter 2 Introduction to Linear Polynomials Solutions

Exercise : 2.1

1.Find the degrees of the following polynomials: (i) $2 x^{2} - 5 x + 3$ (ii) $y^{3} + 2 y - 1$ (iii) -9 (iv) $4 z - 3$

Sol. (i) 2 (ii) 3 (iii) 0 (iv) 1

2. Write polynomials of degrees 1, 2 and 3 .

Sol. Degree 1 polynomials $= 2 x + 3, 3 x, 9 x - 4$ etc.

Degree 2 polynomials $= 2 x^{2} + 3, 3 x^{2} + 2 x$ , $9 x^{2} + 2 x + 3$ etc.

Degree 3 polynomials $= 3 x^{3}, 4 x^{3} + 2 x^{2}$ , $3 x^{3} + 2 x^{2} + x + 5$ etc.

3. What are the coefficients of $x^{2}$ and $x^{3}$ in the polynomial $x^{4} - 3 x^{3} + 6 x^{2} - 2 x + 7$ ?

Sol. Coefficient of $x^{2} = 6$ Coefficient of $x^{3} = - 3$

4. What is the coefficient of $z$ in the polynomial $4 z^{3} + 5 z^{2} - 11$ ?

Sol. Coefficient of $z = 0$

5. What is the constant term of the polynomial $9 x^{3} + 5 x^{2} - 8 x - 10$ ?

Sol. Constant term $= - 10$

Exercise: 2.2

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

Sol. (i) Put $x = 0$

$5 (0) - 3 = - 3$

(ii) Put $x = - 1$ $5 (- 1) - 3 = - 5 - 3 = - 8$ (iii) Put $x = 2$ $5 (2) - 3 = 10 - 3 = 7$

2. Find the value of the quadratic polynomial $7 s^{2} - 4 s + 6$ if: (i) $s = 0$ (ii) $s = - 3$ (iii) $s = 4$

Sol. (i) Put s $= 0$

$7 (0)^{2} - 4 (0) + 6 = 6$

(ii) Puts $= - 3$

$7 (- 3)^{2} - 4 (- 3) + 6 = 63 + 12 + 6 = 81$

(iii) Put $s = 4$

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

Sol. Let present age of Salil $= x$ years and present age of Salil's mother will be $=$ 3x years

After 5 years: $(x + 5) + (3 x + 5) = 70$ $4 x + 10 = 70$ $4 x = 60$ $x = 15$ Hence present age of Salil is 15 years and Salil's mother is 45 years.

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

Sol. Let two positive integers are x and $x + 63$ . ATQ, $\frac{x}{x + 63} = \frac{2}{5}$ $5 x = 2 x + 126$ $3 x = 126$ $x = 42$ Hence the two integers are 42 and 105.

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

Sol. Let number of five Rupee coins be $= x$ and number of two Rupee coins will be $= 3 x$ ATQ,

Value of five rupee coins $= 5 x$ and value of two rupee wins $= 2 (3 x) = 6 x$ Given that $5 x + 6 x = 88$ $11 x = 88$ $x = 8$ Hence number of five rupee coins are 8 and number of two rupee coins are 24.

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

Sol. Let length of shorter piece $= x$ feet So, length of longer piece will be $= 4 x$ feet ATQ, $4 x + x = 300$ $5 x = 300$ $x = 60$ Hence, length of shorter piece is 60 feet and length of longer piece is 240 feet.

7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm , what are the dimensions of the rectangle? Sol. Let width of rectangle be $= x cm$ So length of rectangle will be $= 2 x + 3 cm$ Now, perimeter of rectangle $= 2 (ℓ + b)$ $24 = 2 [(2 x + 3) + x]$ $24 = 2 [3 x + 3]$ $3 x + 3 = 12$ $3 x = 9$ $x = 3$ Hence length of rectangle is 9 cm and width of rectangle is 3 cm .

Exercise: 2.3

1.A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the $n^{th}$ month. Sol. Initial balance $=$ ₹500 Monthly addition = ₹150 Month $1 = 500 + 150 =$ ₹650 Month $2 = 650 + 150 =$ ₹800 Month $3 = 800 + 150 =$ ₹950 Month $4 = 950 + 150$ = ₹1100 So, the expression is : $A = 500 + 150 n$ , where A is amount and n is number of months.

2.A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, $2, 3, \dots$ hours? Find a linear expression to represent the number of members at the end of the $n^{th}$ hour.

Sol. Rally starts with 120 members and 9 members leave each hour. So,

After 1 hour = 120-9(1) = 111 members After 2 hours = 120-9(2)=102 members After 3 hours $= 120 - 9 (3) = 93$ members After 4 hours $= 120 - 9 (4) = 84$ members After 5 hours $= 120 - 9 (5) = 75$ members So, the expression is : $M = 120 - 9 n$ , where M is members and n is number of hours.

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

Sol. Length of Rectangle $= 13 cm$ (Given) Area of Rectangle $= ℓ \times b$ (i) $b = 12 cm$

Area $= 13 \times 12 = 156 cm^{2}$ (ii) $b = 10 cm$

Area $= 13 \times 10 = 130 cm^{2}$ (iii) $b = 8 cm$

Area $= 13 \times 8 = 104 cm^{2}$ So, the linear pattern is $A = 13 b$ , where $A =$ Area and $b =$ breadth.

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

Sol. Given, Length of Rectangular box $= 7 cm$ and breadth of Rectangular box $= 11 cm$ Volume of rectangle $= ℓ \times b \times h$ (i) $h = 5 cm$

Volume $= 7 \times 11 \times 5 = 385 cm^{3}$ (ii) $h = 9 cm$

Volume $= 7 \times 11 \times 9 = 693 cm^{3}$ (iii) $h = 13 cm$

Volume $= 7 \times 11 \times 13 = 1001 cm^{3}$ So, the linear pattern is $V = 77 h$ , where $V =$ Volume and $h =$ height.

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

Sol. Total pages $= 500$ Sarita reads every day = 20 pages So, after 15 days $= 500 - 20 (15)$ $= 500 - 300$ $= 200$ So, the linear pattern is $y = 500 - 20 x$ , where $y =$ pages left and $x =$ number of days.

Exercise : 2.4

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

Sol. Initial height $= 1.75$ feet Growth per month $= 0.5$ feet (i) Height after 7 months $= 1.75 + 0.5 \times 7$ $= 1.75 + 3.5$ $= 5.25$ feet (ii)

Time $(t)$ in months	Height $(h)$ in feet
0 month	1.75 feet
1 month	2.25 feet
2 months	2.75 feet
3 months	3.25 feet
4 months	3.75 feet
5 months	4.25 feet
6 months	4.75 feet
7 months	5.25 feet
8 months	5.75 feet
9 months	6.25 feet
10 months	6.75 feet

(iii) Expressions is $h = 1.75 + 0.5 t$ , where $h =$ height and $t =$ months.

It represents linear growth because the plant grows by the same fixed amount every month. 2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find the value of the phone after 3 years. (ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time. (iii) Find an expression that relates $v$ and t , and explain why it represents linear decay.

Sol. Initial cost of mobile phone $= ₹10, 000$ Value decreases per year $= ₹800$ (i) Value of phone after 3 years

$= 10, 000 - 800 \times 3 = 10, 000 - 2400 = ₹7600$

(ii)

Time $(t)$ in years	Value $(V)$ in ₹.
0	10,000
1	9,200
2	8,400
3	7,600
4	6,800
5	6,000
6	5,200
7	4,400
8	3,600

(iii) Expression is $V = 10000 - 800 t$ where $V =$ value and $t =$ time. 3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. (i) Find the population of the village after 6 years. (ii) Make a table of values for t varying from 0 to 10 years and show how the population, P , increases every year. (iii) Find an expression that relates P and t and explain why it represents linear growth.

Sol. Initial population of village $= 750$ People move city to village per year $= 50$ (i) Population after 6 years $= 750 + 50 \times 6 = 750 + 300 = 1050$

Sol. Given that $y = ax + b$ Case-I: $x = 10, y = 400$ $10 a + b = 400$ $\Rightarrow b = 400 - 10 a$

Case-II: $x = 14, y = 500$ $14 a + b = 500$

Put value of $b$ from (i) to (ii) $14 a + 400 - 10 a = 500$ $4 a = 100$ $a = 25$ put this value to equation (i) $b = 400 - 10 (25)$ $b = 150$ 2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹ 800 . When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x , according to the relation $y = a x + b$ , find the values of $a$ and $b$ . Sol. Given that $y = ax + b$ Case-I: $x = 10, y = 800$ $10 a + b = 800$ $\Rightarrow b = 800 - 10 a$

Case-II: $x = 15, y = 1100$ $15 a + b = 1100$

Put value of $b$ from (i) to (ii) $15 a + 800 - 10 a = 1100$ $5 a = 300$ $a = 60$ put this value to equation (i) $b = 800 - 10 \times 60$ $b = 800 - 600$ $b = 200$ 3. Consider the relationship between temperature measured in degrees Celsius $(^{\circ} C)$ and degrees Fahrenheit ( $^{\circ} F$ ), which is given by $^{\circ} C = a^{\circ} F + b$ . Find a and b , given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When $^{\circ} C = 0,^{\circ} F = 32$ and when $^{\circ} C = 100,^{\circ} F = 212$ . Use this information to find a and b , and thus, the linear relationship between $^{\circ} C$ and $^{\circ} F$ .)

Sol. Given that $^{\circ} C = a^{\circ} F + b$

Case-I:

$^{\circ} C = 0,^{\circ} F = 32$ $0 = a \times 32 + b$ $32 a + b = 0$ $b = - 32 a$

Case-II: $^{\circ} C = 100,^{\circ} F = 212$ $100 = a \times 212 + b$ $212 a + b = 100$

Put value of $b$ from (i) to (ii) $212 a - 32 a = 100$ $180 a = 100$ $a = \frac{100}{180}$ $a = \frac{5}{9}$ Put this value to equation (i) $b = - 32 \times \frac{5}{9}$ $b = - \frac{160}{9}$

Exercise : 2.6

1.Draw the graphs of the following sets of lines. In each case, reflect on the role of 'a' and 'b'. (i) $y = 4 x, y = 2 x, y = x$ (ii) $y = - 6 x, y = - 3 x, y = - x$ (iii) $y = 5 x, y = - 5 x$ (iv) $y = 3 x - 1, y = 3 x, y = 3 x + 1$ (v) $y = - 2 x - 3, y = - 2 x, y = 2 x + 3$

Sol. (i) $y = 4 x$ Put $x = 0$

$y = 4 \times 0 y = 0$

Put $x = 1$

$y = 4 \times 1 y = 4$

$x$	$0$	$1$
$y$	$0$	$4$

$y = 2 x$ Put $x = 0$

$y = 2 \times 0 y = 0$

Put $x = 1$

$y = 2 \times 1 y = 2$

$X$	0	1
$y$	0	2

$y = x$ Put $x = 0$

$y = 0$

Put $x = 1$

$y = 1$

$X$	0	1
$y$	0	1

\to b = 0

, that's why all three lines passes from origin.

→ As 'a' increases from 1 to 2 to 4 the slope gets steeper. (ii) $y = - 6 x$

Put $x = 0$

$y = - 6 \times 0 y = 0$

Put $x = 1$

$y = - 6 \times 1 y = - 6$

$x$	0	1
$y$	0	-6

$y = - 3 x$ Put $x = 0$

$y = - 3 \times 0 y = 0$

Put $x = 1$

$y = - 3 \times 1 y = - 3$

$X$	0	1
$y$	0	-3

$y = - x Put x = 0 y = 0$

Put $x = 1$

$y = - 1$

$X$	0	1
$y$	0	-1

→ $b = 0$ , that's why all three lines passes from origin.

→ Because 'a' values are negative they travel 'downhill' and because values gets larger the downhill drop becomes much steeper. (iii) To draw the graph, we join points which lie on the line For $y = 5 x$ Put $x = 0$

$y = 5 \times 0 y = 0$

Put $x = 1$

$y = 5 \times 1 y = 5$

$x$	$0$	$1$
$y$	$0$	$5$

$For y Put x y y = - 5 x = 0 = - 5 \times 0 = 0$ $Put x y y = 1 = - 5 \times 1 = - 5$

$x$	0	1
$y$	0	-5

These two lines have the exact same level of steepness because the number 5 is the same.
The only difference is the sign. $y = 5 x$ is a steep uphill climb, while $y = - 5 x$ is a steep downhill drop. (iv) To draw the graph, we join points which lie on the line For $y = 3 x - 1$

Put $x = 0$

$y = 3 \times 0 - 1 y = 0 - 1 y = - 1$

Put $x = 1$

$y = 3 \times 1 - 1 y = 3 - 1 y = 2$

$x$	$0$	$1$
$y$	-1	2

For $y = 3 x$ Put $x = 0$

$y = 3 \times 0 y = 0$

Put $x = 1$

$y = 3 \times 1 y = 3$

$x$	$0$	$1$
$y$	$0$	$3$

For $y = 3 x + 1$ Put $x = 0$

$y = 3 \times 0 + 1 y = 0 + 1 y = 1$

Put $x = 1$

$y = 3 \times 1 + 1 y = 3 + 1 y = 4$

Because the slope ' a ' is exactly the same for all three, they have the exact same steepness.
This means they are parallel lines that will never touch.
Changing the ' b ' value simply shifts the line vertically. $y = 3 x$ goes through the center, $y = 3 x + 1$ is shifted one unit up, and $y = 3 x - 1$ is shifted one unit down. (v) To draw the graph, we join points which lie on the line

For $y = - 2 x - 3$ Put $x = 0$

$y = - 2 \times 0 - 3 y = 0 - 3 y = - 3$

Put $x = 1$

$y = - 2 \times 1 - 3 y = - 2 - 3 y = - 5$

$x$	0	1
$y$	-3	-5

For $y = - 2 x$ Put $x = 0$

$y = - 2 \times 0 y = 0$

Put $x = 1$

$y = - 2 \times 1 y = - 2$

$x$	0	1
$y$	0	-2

For $y = - 2 x + 3$ Put $x = 0$

$y = - 2 \times 0 + 3 y = 0 + 3 y = 3$

Put $x = 1$

$y = - 2 \times 1 + 3 y = - 2 + 3 y = 1$

$x$	$0$	$1$
$y$	$3$	$1$

Because the slope ' a ' is exactly the same for all three lines (-2), they all have the exact same "downhill" steepness.

Because they have the same steepness, they are perfectly parallel lines-they will never, ever intersect.

The 'b' values tell us how these parallel lines are stacked.

The middle line $(y = - 2 x)$ goes straight through the origin $(0, 0)$ .
The line $y = - 2 x + 3$ is that exact same line, just shifted straight up like an elevator by 3 units.
The line $y = - 2 x - 3$ is shifted straight down by 3 units.

End of Chapter Exercise

1.Write a polynomial of degree 3 in the variable $x$ , in which the coefficient of the $x^{2}$ term is -7 . Sol. $2 x^{3} - 7 x^{2} + 5 x + 30$

2.Find the values of the following polynomials at the indicated values of the variables. (i) $5 x^{2} - 3 x + 7$ if $x = 1$ (ii) $4 t^{3} - t^{2} + 6$ if $t = a$

Sol. (i) Let $p (x) = 5 x^{2} - 3 x + 7$ Putting $x = 1$

$p (1) = 5 (1)^{2} - 3 (1) + 7 = 5 - 3 + 7 = 9$

(ii) Let $p (t) = 4 t^{3} - t^{2} + 6$

Putting $t = a$

3.If we multiply a number by $\frac{5}{2}$ and add $\frac{2}{3}$ to the product, we get $\frac{- 7}{12}$ . Find the number. Sol. Let Number be x $\frac{5}{2} \times x + \frac{2}{3} = \frac{- 7}{12}$ $\frac{5 x}{2} + \frac{2}{3} = \frac{- 7}{12}$ $\frac{5 x}{2} = \frac{- 7}{12} - \frac{2}{3}$ $\frac{5 x}{2} = \frac{- 7 - 8}{12}$ $\frac{5 x}{2} = \frac{- 15}{12}$ $x = \frac{- 15}{12} \times \frac{2}{5}$ $x = \frac{- 3}{6}$ $x = \frac{- 1}{2}$ Hence the number is $- \frac{1}{2}$ .

4.A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Sol. Let the smaller number be x Given that a positive number is 5 times another number.

Thus, Positive number $= 5 \times$ Smaller number $= 5 x$

Now, 21 added to both numbers. So, our numbers become: New smaller number $= x + 21$ New bigger number $= 5 x + 21$ New bigger number $= 2 \times$ New smaller number $5 x + 21 = 2 \times (x + 21)$ $5 x + 21 = 2 x + 42$ $5 x - 2 x + 21 = 42$ $3 x + 21 = 42$ $3 x = 42 - 21$ $3 x = 21$ $x = \frac{21}{3}$ $x = 7$ Thus, our numbers are Smaller number $= x = 7$ Bigger number $= 5 x = 5 \times 7 = 35$

5. If you have ₹ 800 and you save ₹ 250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

Sol. (i) Thus, Initial Value = ₹ 800 Value after 1 month $= 800 + 250$ = ₹ 1,050 Value after 2 months $= 800 + 250 \times 2$ $= 800 + 500 =$ ₹ 1,300 Value after 3 months $= 800 + 250 \times 3$ $= 800 + 750 = ₹1, 550$ Similarly, we can write Value after 6 months $= 800 + 250 \times 6$ $= 800 + 1500 = ₹2, 300$ (ii) We also need to find value after 2 years

Since 2 years $= 2 \times 12$ months $= 24$ months We can write Value after 2 years = Value after 24 months $= 800 + 250 \times 24$ $= 800 + 6, 000 = ₹6, 800$ We are also asked to express it as a linear pattern

Let time in months $= t$ And, value of money (in ₹) after t months be $v (t)$ Since value v starts at ₹ 800 and adds ₹250 every month t. Our Expression is $v (t) = 800 + 250 t$

4.0Important Key Concepts of NCERT Solutions for Class 9 Maths Chapter 2

Topic	What Students Learn
Variables, Coefficients & Constants	Basic building blocks of a polynomial expression, and how each part — variable, coefficient, and constant — contributes to its value
Degree of a Polynomial	Meaning and calculation of degree, and how it decides whether a polynomial is linear, quadratic, or cubic
Types of Polynomials	Linear, quadratic, cubic, and constant polynomials, classified by their degree, with examples of each
Linear Polynomials	Degree 1 polynomials written in the form ax + b, and why they represent a constant rate of change
Value of a Polynomial	Substituting a given value for the variable to evaluate a linear polynomial, applied to simple word problems
Linear Patterns	Real-life quantities that change at a constant rate, like fixed fares, recharge plans, or savings that grow steadily
Linear Growth & Decay	Modelling quantities that increase or decrease steadily over time, such as a filling or draining water tank
Linear Relationships (y = ax + b)	Standard form of a linear relationship, where a shows the rate of change and b shows the starting value
Visualising Linear Relationships	Plotting straight-line graphs, reading the slope and y-intercept, and connecting algebra with coordinate geometry
Parallel Lines	Identifying parallel lines by comparing their slopes, and understanding why they never intersect

5.0Exercise-wise NCERT Class 9 Maths Chapter 2 - Introduction to Linear polynomials

Get a quick overview of the concepts covered in each exercise of NCERT Class 9 Maths Chapter 2 – Introduction to Linear Polynomials. This exercise-wise summary helps build conceptual clarity and makes revision more focused and effective.

Exercise	Key Concepts Covered
Exercise 2.1	Understanding polynomials, identifying variables, constants, coefficients, and determining the degree of a polynomial.
Exercise 2.2	Evaluating the value of linear and quadratic polynomials by substituting given values and solving simple word problems.
Exercise 2.3	Understanding linear relationships, identifying dependent and independent variables, and representing real-life situations using linear expressions.
Exercise 2.4	Forming linear equations from practical situations and solving them using algebraic methods.
Exercise 2.5	Understanding slope, rate of change, and the relationship between variables in linear equations.
Exercise 2.6	Representing linear equations graphically, interpreting graphs, and relating equations to straight-line graphs.
End-of-Chapter Exercise	Applying all chapter concepts through mixed problems covering linear polynomials, equations, slope, graphs, and real-life applications

6.0Quick Revision on Class 9 Maths Chapter 2 - Introduction to Linear Polynomials

7.0Related Study Materials Class 9 Maths

Improve your concepts and prepare confidently for school and CBSE exams with ALLEN’s Class 9 Maths study materials designed based on the NCERT Solutions. along with the the latest syllabus, NCERT textbooks, revision notes, sample papers and previous years’ question papers to improve your learning.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 2 Maths Class 9 NCERT Solutions

Identifies Linear Polynomials: Enables students to identify linear polynomials by degree, coefficients and constant terms.
Explains Polynomial Structure: Builds understanding of variables, coefficients, and the degree of a polynomial.
Teaches Evaluation Skills: Shows how to evaluate a linear polynomial by substituting different values for the variable.
Builds Graph Interpretation: It explains how linear polynomials are represented as graphs of straight lines.
Develops Understanding of Linear Relationships Illustrates the use of linear polynomials to express mathematical relationships.
Builds a strong foundation Prepares Students for Linear Equations, Functions, and Higher Algebraic Concepts

NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials

1.0Download NCERT Class 9 Maths Chapter 2 Introduction to Linear Polynomials Solutions

NCERT Solutions Class 9 Maths Chapter 2

2.0Learning Outcomes - NCERT Class 9 Maths Chapter 2 Solutions

Identify the variable, coefficient, constant term, and degree of a polynomial.
Distinguish between constant, linear, quadratic, and cubic polynomials based on their degree.
Understand the characteristics of a linear polynomial.
Find the value of a polynomial by substituting the given value of the variable.
Recognise and represent linear relationships using algebraic expressions.
Understand functions as an input-output process through simple examples.
Model real-life situations using linear growth and linear decay.
Represent and interpret simple linear relationships using tables and graphs.