Introduction to | Class 9 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Introduction to – this guide gives you a concise, exam-ready overview of Introduction to from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
2.1 INTRODUCTION
This section introduces the concept of algebraic expressions and specifically focuses on linear polynomials. Algebraic expressions are combinations of numbers, variables (also called letter-numbers), and operations such as addition, subtraction, and multiplication. The section begins with real-life examples to illustrate how algebraic expressions can be used to represent quantities and solve problems. For instance, in Example 1, the total number of pens and pencils bought by Raju is expressed as 4x + 5y + 3, where x and y represent the number of red and blue boxes respectively, 4 and 5 are coefficients, and 3 is a constant term. This example helps students understand the components of an algebraic expression: terms, variables, coefficients, and constants.
Example 2 extends this idea to a practical problem involving the cost of fencing and decorating a rectangular garden. The total cost is expressed as 200l + 160w + 50lw, where l and w are the length and width of the garden. This example introduces multiplication of variables and constants within algebraic expressions.
Example 3 considers the area of rectangles formed by bending a wire of fixed length. The area is expressed as x(10 - x) or 10x - x², introducing the concept of polynomials involving powers of variables.
The section then defines univariate polynomials—algebraic expressions involving one variable with non-negative integer exponents. It explains the degree of a polynomial as the highest power of the variable present. Examples of polynomials of different degrees are given: cubic (degree 3), quadratic (degree 2), linear (degree 1), and constant (degree 0).
This foundational section sets the stage for deeper exploration of linear polynomials, emphasizing the structure and components of algebraic expressions and polynomials.
📊 Diagram: Fig. 2.1 shows red and blue boxes representing pens and pencils respectively; Fig. 2.2 illustrates a rectangular garden with length l and width w.
🧪 Activity: Think and Reflect questions encourage identifying terms, variables, and coefficients in algebraic expressions and comparing different expressions.
🔗 Connection: This section introduces polynomials and variables, leading to the next section's focus on linear polynomials and their properties.
Frequently asked questions
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.
(i) Initial height = 1.75 feet, growth per month = 0.5 feet Height after 7 months = Initial height + 7 × growth per month = 1.75 + 7 × 0.5 = 1.75 + 3.5 = 5.25 feet
(ii) Table of values:
| t (months) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| h (feet) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
(iii) Expression relating height h and time t:
h = 1.75 + 0.5t
This is a linear expression because h changes by a consta
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, $\nu$, depreciates with time. (iii) Find an expression that relates $\nu$ and $t$, and explain why it represents linear decay.
(i) Initial value = ₹10,000 Depreciation per year = ₹800 Value after 3 years = Initial value - 3 × depreciation = 10000 - 3 × 800 = 10000 - 2400 = ₹7600
(ii) Table of values:
| t (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| ν (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) Expression relating value ν and time t:
ν = 10000 - 800t
This represents linear decay because the value decreases by a fixed amount (₹800) every y
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.
(i) Initial population = 750 Increase per year = 50 Population after 6 years = 750 + 6 × 50 = 750 + 300 = 1050
(ii) Table of values:
| t (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
(iii) Expression relating population P and time t:
P = 750 + 50t
This represents linear growth because the population increases by a fixed number (50) every year. The
4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation that models the remaining balance $b(x)$ after using the scheme for $x$ days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for $x$ varying from 1 to 10 days and show how the balance $b(x)$, reduces with time.
(i) Initial balance = ₹600 Daily reduction = ₹15 Equation for balance after x days:
b(x) = 600 - 15x
This represents linear decay because the balance decreases by a fixed amount (₹15) each day.
(ii) To find when balance runs out, set b(x) = 0:
0 = 600 - 15x 15x = 600 x = 600 / 15 = 40 days
(iii) Table of values:
| x (days) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| b(x) (₹) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
Ready to ace this chapter?
Get the full Introduction to chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.
Study smarter with ConceptScroll
Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.
Start learning freeContinue reading
- Predicting What Comes Next: Exploring Sequences and Progressions | Class 9 Mathematics Notes
Clear NCERT-aligned notes on Predicting What Comes Next: Exploring Sequences and Progressions for Class 9 Mathematics.
- Predicting What Comes Next: Exploring Sequences and Progressions | Class 9 Mathematics Notes
Clear NCERT-aligned notes on Predicting What Comes Next: Exploring Sequences and Progressions for Class 9 Mathematics.
- Predicting What Comes Next: Exploring Sequences and Progressions | Class 9 Mathematics Notes
Clear NCERT-aligned notes on Predicting What Comes Next: Exploring Sequences and Progressions for Class 9 Mathematics.