Class 9 Maths Ganita Manjari Chapter 2 Solutions (NCERT 2026–27) – Introduction to Linear Polynomials

These Class 9 Maths Ganita Manjari Chapter 2 solutions cover Introduction to Linear Polynomials from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step so you can understand each concept and revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 2 Exercises: 2.1–2.6, End-of-Chapter Session: 2026–27
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Chapter 2 Overview

Chapter 2 of Ganita Manjari, Introduction to Linear Polynomials, builds on algebraic expressions to define polynomials in one variable and their degree, then focuses on linear polynomials (degree 1). It shows how linear expressions model real-life linear patterns, linear growth and decay, and linear relationships of the form y = ax + b, and how to plot these as straight lines. The Class 9 Maths Ganita Manjari Chapter 2 solutions below work through every exercise step by step.

Key Concepts & Definitions

Polynomial (in one variable): an algebraic expression like x2 + 5x + 3 made of terms with whole-number powers of the variable.

Degree: the highest power of the variable. Degree 1 = linear, 2 = quadratic, 3 = cubic, 0 = constant.

Linear polynomial: a polynomial of degree 1, e.g. 2x + 3 or 5 − 4y.

Linear pattern: a sequence in which the difference between consecutive terms is constant.

Linear growth / decay: a quantity that increases (growth) or decreases (decay) by a fixed amount over equal intervals.

Linear relationship: y = ax + b, a straight line with slope a and y-intercept b; when b = 0 the line passes through the origin.

Key Facts & Formulas

Value of a polynomial: substitute the given value of the variable and simplify.

Linear relationship: y = ax + b — a = slope (rate of change), b = y-intercept (value when x = 0).

Plotting a line: find any two points (e.g. x = 0 and one more), plot and join them.

Parallel lines: have the same slope a (only b differs). Growth ⇒ a > 0; decay ⇒ a < 0.

Exercise Set 2.1

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

SOLUTION (i) Highest power = 2 → degree 2. (ii) Highest power = 3 → degree 3. (iii) −9 is a constant → degree 0. (iv) Highest power = 1 → degree 1.

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

SOLUTION Degree 1: 2x + 5;   Degree 2: x2 − 3x + 1;   Degree 3: 4x3 + x2 − 2. (Other correct answers are possible.)

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

SOLUTION Coefficient of x2 = 6; coefficient of x3 = −3.

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

SOLUTIONThere is no z term, so the coefficient of z is 0.

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

SOLUTIONThe constant term is −10.
Think and Reflect Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm? Answer. Perimeter = 4 × side: 4, 6, 8, 10, 12 cm. Each time the side increases by 0.5 cm, the perimeter increases by a constant 2 cm (a linear pattern).
Think and Reflect A chess club charges ₹200 joining fee plus ₹50 per match. If a player paid ₹750, how many matches did he play? Answer. 200 + 50m = 750 ⇒ 50m = 550 ⇒ m = 11 matches.

Exercise Set 2.2

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

SOLUTION (i) 5(0) − 3 = −3. (ii) 5(−1) − 3 = −8. (iii) 5(2) − 3 = 7.

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

SOLUTION (i) 7(0) − 4(0) + 6 = 6. (ii) 7(9) − 4(−3) + 6 = 63 + 12 + 6 = 81. (iii) 7(16) − 4(4) + 6 = 112 − 16 + 6 = 102.

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.

SOLUTION Let Salil’s age = x, so mother’s age = 3x. After 5 years: (x + 5) + (3x + 5) = 70 ⇒ 4x + 10 = 70 ⇒ 4x = 60 ⇒ x = 15. ∴ Salil is 15 years and his 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.

SOLUTION Let the integers be 2k and 5k. Then 5k − 2k = 63 ⇒ 3k = 63 ⇒ k = 21. ∴ the 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?

SOLUTION Let five-rupee coins = x, so two-rupee coins = 3x. Value: 5x + 2(3x) = 88 ⇒ 5x + 6x = 88 ⇒ 11x = 88 ⇒ x = 8. ∴ 8 five-rupee coins and 24 two-rupee coins (32 coins in all).

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?

SOLUTION Let the shorter piece = x, so the longer = 4x. Then x + 4x = 300 ⇒ 5x = 300 ⇒ x = 60. ∴ the pieces are 60 feet and 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?

SOLUTION Let width = w, so length = 2w + 3. Perimeter: 2(w + 2w + 3) = 24 ⇒ 2(3w + 3) = 24 ⇒ 3w + 3 = 12 ⇒ w = 3. ∴ width = 3 cm, length = 2(3) + 3 = 9 cm.
Think and Reflect The area of the rectangle, 10x − x2, is a function of x. What value does the expression take when x = 6 cm? Answer. 10(6) − 62 = 60 − 36 = 24 cm2.

Exercise Set 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 nth month.

SOLUTION Amount after n months = 500 + 150n. Month 2 → 500 + 300 = ₹800, month 3 → ₹950, month 4 → ₹1100, … Linear expression: 500 + 150n.

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, … hours? Find a linear expression to represent the number of members at the end of the nth hour.

SOLUTION After 1 hr → 111, 2 hr → 102, 3 hr → 93, … Linear expression: 120 − 9n.

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.

SOLUTION Area = 13 × breadth. (i) 13 × 12 = 156 cm2; (ii) 13 × 10 = 130 cm2; (iii) 13 × 8 = 104 cm2. Linear pattern: area = 13b (it falls by 26 cm2 for every 2 cm fall in 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.

SOLUTION Volume = 7 × 11 × height = 77h. (i) 77 × 5 = 385 cm3; (ii) 77 × 9 = 693 cm3; (iii) 77 × 13 = 1001 cm3. Linear pattern: volume = 77h.

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.

SOLUTION Pages left after n days = 500 − 20n. After 15 days: 500 − 20(15) = 500 − 300 = 200 pages.
Think and Reflect Using 2n − 1, how many tiles are there in the 15th and 26th stages? Which stage contains 21 tiles and 47 tiles? Answer. 15th: 2(15) − 1 = 29; 26th: 2(26) − 1 = 51. 2n − 1 = 21 ⇒ n = 11; 2n − 1 = 47 ⇒ n = 24.

Exercise Set 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. (iii) Find an expression that relates h and t, and explain why it represents linear growth.

SOLUTION (i) h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet. (ii) t: 0,1,2,…,10 → h: 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75. (iii) h = 1.75 + 0.5t. The height increases by a fixed 0.5 ft each month (slope > 0), so it is linear growth.

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

SOLUTION (i) v = 10000 − 800(3) = ₹7600. (ii) t: 0–8 → v: 10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600. (iii) v = 10000 − 800t. The value falls by a fixed ₹800 each year (slope < 0), so it is linear decay.

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 after 6 years. (ii) Make a table of values for t from 0 to 10 years. (iii) Find an expression relating P and t, and explain why it represents linear growth.

SOLUTION (i) P = 750 + 50(6) = 1050. (ii) t: 0–10 → P: 750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200, 1250. (iii) P = 750 + 50t. The population rises by a fixed 50 each year (slope > 0) ⇒ linear growth.

4. A telecom company charges ₹600 for a recharge scheme. The prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation for the remaining balance b(x) after x days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for x from 1 to 10 days.

SOLUTION (i) b(x) = 600 − 15x. The balance falls by a fixed ₹15 per day (slope < 0) ⇒ linear decay. (ii) 600 − 15x = 0 ⇒ x = 40 days. (iii) x: 1–10 → b: 585, 570, 555, 540, 525, 510, 495, 480, 465, 450.
Think and Reflect For the cost C(d) = 100 + 60d: what is the cost for 15 km? For how many km will the cost be ₹700? Answer. C(15) = 100 + 60(15) = ₹1000. 700 = 100 + 60d ⇒ 60d = 600 ⇒ d = 10 km.

Exercise Set 2.5

1. A learning platform charges a fixed monthly fee plus a cost per module. When 10 modules were accessed the bill was ₹400; for 14 modules it was ₹500. If y = ax + b, find a and b.

SOLUTION 400 = 10a + b  and  500 = 14a + b. Subtract: 100 = 4a ⇒ a = 25. b = 400 − 10(25) = 150. ∴ a = 25, b = 150 (y = 25x + 150).

2. A gym charges a fixed monthly fee plus a cost per hour of badminton. For 10 hours the bill was ₹800; for 15 hours it was ₹1100. If y = ax + b, find a and b.

SOLUTION 800 = 10a + b  and  1100 = 15a + b. Subtract: 300 = 5a ⇒ a = 60. b = 800 − 10(60) = 200. ∴ a = 60, b = 200 (y = 60x + 200).

3. The relation between Celsius and Fahrenheit is °C = a·°F + b. Given ice melts at 0°C / 32°F and water boils at 100°C / 212°F, find a and b.

SOLUTION 0 = 32a + b  and  100 = 212a + b. Subtract: 100 = 180a ⇒ a = 100/180 = 5/9. b = −32a = −32(5/9) = −160/9. ∴ °C = (5/9)°F − 160/9 = (5/9)(°F − 32).
Think and Reflect In y = 20x + 150, what do the numbers 20 and 150 represent? Answer. 20 is the slope — the cost per GB of data; 150 is the y-intercept — the fixed monthly fee charged even when x = 0.

Exercise Set 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 = 4x, y = 2x, y = x (ii) y = –6x, y = –3x, y = –x (iii) y = 5x, y = –5x (iv) y = 3x – 1, y = 3x, y = 3x + 1 (v) y = –2x – 3, y = –2x, y = 2x + 3

SOLUTION Plot each using two points, then observe the role of a (slope) and b (y-intercept): (i) y = 4x → (0,0),(1,4); y = 2x → (1,2); y = x → (1,1). All pass through the origin (b = 0); the line gets steeper as a increases. (ii) y = −6x,−3x,−x → (1,−6),(1,−3),(1,−1). All through the origin with negative slope (fall left-to-right); steeper as |a| increases. (iii) y = 5x → (1,5); y = −5x → (1,−5). Same steepness, opposite signs — mirror images in the x-axis. (iv) y = 3x−1, 3x, 3x+1 → same slope 3, y-intercepts −1, 0, 1. They are parallel (a fixed, b changes). (v) y = −2x−3 and y = −2x are parallel (slope −2); y = 2x+3 has slope +2, so it is not parallel to them. Conclusion: a controls slope/direction, b controls the y-intercept; equal a ⇒ parallel lines.
Think and Reflect Complete the table for y = 2x + 1: x = 1, 2, 5, 7, 9, 12, 20. Answer. y = 2x + 1 → x: 1,2,5,7,9,12,20 give y: 3, 5, 11, 15, 19, 25, 41.

Class 9 Maths Ganita Manjari Chapter 2 Solutions — End-of-Chapter Exercises

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

SOLUTIONOne example: x3 − 7x2 + 2x + 1 (any degree-3 polynomial with the x2 coefficient equal to −7).

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

SOLUTION (i) 5(1) − 3(1) + 7 = 5 − 3 + 7 = 9. (ii) Substitute t = a: 4a3 − a2 + 6.

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

SOLUTION Let the number be x: (5/2)x + 2/3 = −7/12. (5/2)x = −7/12 − 2/3 = −7/12 − 8/12 = −15/12 = −5/4. x = (−5/4) × (2/5) = −10/20 = −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?

SOLUTION Let the smaller number = x, larger = 5x. After adding 21: (5x + 21) and (x + 21). 5x + 21 = 2(x + 21) ⇒ 5x + 21 = 2x + 42 ⇒ 3x = 21 ⇒ x = 7. ∴ the numbers are 7 and 35 (check: 56 = 2 × 28).

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.

SOLUTION Amount after n months = 800 + 250n. (i) 800 + 250(6) = ₹2300. (ii) 2 years = 24 months: 800 + 250(24) = ₹6800.

*6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

SOLUTION Let the digits be a (tens) and b (units). Number = 10a + b, reversed = 10b + a. Sum = 11(a + b) = 143 ⇒ a + b = 13. Also |a − b| = 3. Solving: a = 8, b = 5 (or a = 5, b = 8). ∴ the numbers are 85 and 58.

*7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also find where these lines cut the y-axis. (i) y = –3x + 4   (ii) 2y = 4x + 7   (iii) 5y = 6x – 10   (iv) 3y = 6x – 11 Are any of the lines parallel?

SOLUTION Write each as y = ax + b: (i) y = −3x + 4 → slope −3, y-intercept 4, cuts y-axis at (0, 4). (ii) y = 2x + 7/2 → slope 2, y-intercept 3.5, cuts (0, 3.5). (iii) y = (6/5)x − 2 → slope 1.2, y-intercept −2, cuts (0, −2). (iv) y = 2x − 11/3 → slope 2, y-intercept −11/3, cuts (0, −11/3). Lines (ii) and (iv) are parallel (both have slope 2).

*8. The relation between Kelvin (x K) and Fahrenheit (y °F) is y = (9/5)(x – 273) + 32. (i) Find y if x = 313 K.   (ii) If y = 158 °F, find x.

SOLUTION (i) y = (9/5)(313 − 273) + 32 = (9/5)(40) + 32 = 72 + 32 = 104 °F. (ii) 158 = (9/5)(x − 273) + 32 ⇒ 126 = (9/5)(x − 273) ⇒ x − 273 = 70 ⇒ x = 343 K.

*9. Work done = constant force × distance. Express this as a linear equation in two variables (work w and distance d), and draw its graph for a constant force of 3 units. What is the work done when the distance travelled is 2 units?

SOLUTION With force = 3: w = 3d. This is a straight line through the origin with slope 3 (points (0,0), (1,3), (2,6)). When d = 2: w = 3(2) = 6 units.

*10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). (i) Find the polynomial p(x).   (ii) Find where its graph cuts the axes.   (iii) Draw the graph and verify.

SOLUTION (i) Slope a = (11 − 5)/(3 − 1) = 3; b = 5 − 3(1) = 2 ⇒ p(x) = 3x + 2. (ii) y-axis: x = 0 ⇒ (0, 2). x-axis: 3x + 2 = 0 ⇒ x = −2/3 ⇒ (−2/3, 0).

*11. Let p(x) = ax + b and q(x) = cx + d be linear polynomials such that: (i) p(0) = 5; (ii) p(x) – q(x) cuts the x-axis at (3, 0); (iii) p(x) + q(x) = 6x + 4 for all real x. Find p(x) and q(x).

SOLUTION (i) p(0) = b = 5. (iii) a + c = 6 and b + d = 4 ⇒ d = −1. (ii) p(x) − q(x) = (a − c)x + (b − d) = (a − c)x + 6; at x = 3: 3(a − c) + 6 = 0 ⇒ a − c = −2. With a + c = 6: a = 2, c = 4. ∴ p(x) = 2x + 5 and q(x) = 4x − 1.

*12. Hexagons are made with matchsticks; a new hexagon shares a side with the previous one. (i) Draw the next two stages — how many matchsticks at each? (ii) Complete the table. (iii) Find the rule for the nth stage. (iv) Matchsticks for the 15th stage? (v) Can 200 matchsticks form a stage?

SOLUTION Stage 1 needs 6 matchsticks; each new hexagon adds 5 (one shared side). So 1→6, 2→11, 3→16. (i) Stage 4 = 21, Stage 5 = 26 matchsticks. (ii)/(iii) Rule for the nth stage: 5n + 1. (iv) 15th stage: 5(15) + 1 = 76. (v) 5n + 1 = 200 ⇒ 5n = 199 ⇒ n = 39.8, not a whole number ⇒ No, 200 matchsticks cannot form a stage.

*13. p(x) = ax + b and q(x) = cx + d such that: (i) p(x) passes through (2, 3) and (6, 11); (ii) q(x) passes through (4, –1); (iii) q(x) is parallel to p(x). Find p(x) and q(x), and where each meets the x-axis.

SOLUTION (i) Slope of p = (11 − 3)/(6 − 2) = 2; b = 3 − 2(2) = −1 ⇒ p(x) = 2x − 1. (iii) Parallel ⇒ slope of q is also 2. (ii) Through (4, −1): −1 = 2(4) + d ⇒ d = −9 ⇒ q(x) = 2x − 9. x-axis: p(x) = 0 ⇒ x = 1/2, i.e. (1/2, 0); q(x) = 0 ⇒ x = 9/2, i.e. (9/2, 0). (Being parallel, the two lines never meet each other.)

*14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

SOLUTION f(x) = ax + a = a(x + 1). At x = −1, f(−1) = 0 for every a. So all such lines pass through the point (−1, 0) (their common x-intercept), and being a > 0 they all rise from left to right.

Common Mistakes to Avoid

Watch out for these

  • Confusing a coefficient (the number) with the term or the variable.
  • Forgetting that a non-zero constant has degree 0, and that a missing term means coefficient 0.
  • Swapping slope (a) and y-intercept (b) in y = ax + b.
  • Sign slips when transposing terms while solving a linear equation.
  • Substituting negatives wrongly: (−3)2 = 9, not −9.
  • Misreading word problems (which quantity is ‘twice’ or ‘5 times’ the other).

Practice MCQs & Assertion–Reason

1. The degree of the polynomial 7x − 4 is:

(a) 0    (b) 1    (c) 2    (d) 7

2. The value of 3x − 5 at x = 2 is:

(a) 1    (b) −1    (c) 11    (d) 6

3. A polynomial of degree 1 is called a:

(a) constant polynomial    (b) linear polynomial    (c) quadratic polynomial    (d) cubic polynomial

4. In y = ax + b, the letter ‘a’ represents the:

(a) y-intercept    (b) slope    (c) constant term    (d) degree

5. The y-intercept of the line y = 2x + 5 is:

(a) 2    (b) 5    (c) −5    (d) 0

6. The constant term of 4x2 − 3x + 9 is:

(a) 4    (b) −3    (c) 9    (d) 0

7. Two lines are parallel if they have the same:

(a) y-intercept    (b) slope    (c) x-intercept    (d) degree

8. The line y = 3x passes through the point:

(a) (0, 3)    (b) (3, 0)    (c) (0, 0)    (d) (1, 0)

9. If 2x + 1 = 9, then x =

(a) 3    (b) 4    (c) 5    (d) 8

10. The coefficient of x in 5x2 − 7x + 2 is:

(a) 5    (b) −7    (c) 2    (d) 7

Answer key: 1-(b), 2-(a), 3-(b), 4-(b), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The degree of a non-zero constant polynomial is 0.

Reason: A constant c can be written as c·x0.

A-R 2. Assertion: The lines y = 2x + 3 and y = 2x − 4 are parallel.

Reason: Parallel lines have equal slopes.

A-R 3. Assertion: The graph of y = ax (b = 0) passes through the origin.

Reason: When x = 0, y = 0 for the equation y = ax.

A-R 4. Assertion: 3x2 + 2x + 1 is a linear polynomial.

Reason: A linear polynomial has degree 1.

A-R 5. Assertion: In linear decay the slope is negative.

Reason: Linear decay means the quantity decreases by a fixed amount over equal intervals.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • A polynomial’s degree is the highest power of the variable; degree 1 = linear, 2 = quadratic, 3 = cubic, 0 = constant.
  • To find a polynomial’s value, substitute the variable and simplify.
  • A linear pattern has a constant difference between consecutive terms.
  • Linear growth: increases by a fixed amount (slope a > 0); linear decay: decreases by a fixed amount (slope a < 0).
  • A linear relationship is y = ax + b: a = slope, b = y-intercept; if b = 0 the line passes through the origin.
  • Parallel lines share the same slope a (only b differs).

How to score full marks in this chapter

State the degree/coefficient/constant clearly, show each step when solving linear equations, and for word problems always define the variable first (“Let … = x”). For graph questions, write the line as y = ax + b, find two neat points, and label the slope and y-intercept.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 2 about?

It covers polynomials and degree, linear polynomials, linear patterns, linear growth and decay, linear relationships y = ax + b, and plotting straight-line graphs.

What is a linear polynomial?

A linear polynomial is a polynomial of degree 1, such as 2x + 3 or 5 − 4y.

What do a and b mean in y = ax + b?

a is the slope of the line and b is the y-intercept (where the line cuts the y-axis).

Are these Class 9 Maths Ganita Manjari Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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