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

2.1 INTRODUCTION

Fig. 2.2: Rectangular garden of length ℓ and width w (Example 2)

Think and Reflect (after Example 2)

Context from the textbook — Example 2: total cost of fencing and sowing a rectangular garden of length $𝑙$ metres and width $𝑤$ metres is given by the algebraic expression $200 𝑙 + 160 𝑤 + 50 𝑙 𝑤$ . Example 1 had the expression $4 𝑥 + 5 𝑦 + 3$ .

Question

1. Can you identify the terms, variables and coefficients of the algebraic expression

200 𝑙 + 160 𝑤 + 50 𝑙 𝑤

Solution

The algebraic expression from Example 2 is

200 𝑙 + 160 𝑤 + 50 𝑙 𝑤 .

Terms: A term is each part of the expression that is added or subtracted. Here there are three terms: $200 𝑙, 160 𝑤, 50 𝑙 𝑤 .$
Variables: The letter-numbers that stand for unknown quantities are $𝑙$ (length) and $𝑤$ (width).
Coefficients: The numerical factor multiplying each variable part is its coefficient:
- Coefficient of $𝑙$ is $200$ (from the term $200 𝑙$ ).
- Coefficient of $𝑤$ is $160$ (from the term $160 𝑤$ ).
- Coefficient of the product $𝑙 𝑤$ is $50$ (from the term $50 𝑙 𝑤$ ).

Question

2. How is

200 𝑙 + 160 𝑤 + 50 𝑙 𝑤

different from the algebraic expression in Example 1 (

4 𝑥 + 5 𝑦 + 3

Solution

Compare Example 2 ( $200 𝑙 + 160 𝑤 + 50 𝑙 𝑤$ ) with Example 1 ( $4 𝑥 + 5 𝑦 + 3$ ):

Product term: Example 2 has a product term $50 𝑙 𝑤$ , in which two different variables are multiplied together ( $𝑙 \times 𝑤$ ). In Example 1, the variables $𝑥$ and $𝑦$ appear only separately in linear terms $4 𝑥$ and $5 𝑦$ (there is no $𝑥 𝑦$ term).
Constant term: Example 2 has no constant term (no standalone number without a variable). Example 1 has a constant term $3$ .
Variables used: Example 2 uses $𝑙$ and $𝑤$ ; Example 1 uses $𝑥$ and $𝑦$ . Both use two variables, but the structure of the terms is different as noted above.

Think and Reflect (after Example 3)

Context from the textbook — Example 3: a wire of length $20 c m$ is bent to form rectangles. If the length is $𝑥 c m$ , then the width is $(10 - 𝑥) c m$ . The area of such a rectangle is $𝑥 (10 - 𝑥)$ , which expands to the algebraic expression $10 𝑥 - 𝑥 2$ . Example 1 had the expression $4 𝑥 + 5 𝑦 + 3$ .

Question

1. Can you identify the terms, variables and coefficients of the algebraic expression

10 𝑥 - 𝑥 2

Solution

The algebraic expression from Example 3 (area of the rectangle) is

10 𝑥 - 𝑥 2

(which is the same as $𝑥 (10 - 𝑥)$ , or as a sum of terms $10 𝑥 + (- 𝑥 2)$ ).

Terms: There are two terms: $10 𝑥 a n d - 𝑥 2 .$
Variables: Only one variable is used: $𝑥$ .
Coefficients:
- The term $10 𝑥$ is $10 \cdot 𝑥 1$ , so the coefficient of $𝑥$ is $10$ .
- The term $- 𝑥 2$ is $(- 1) \cdot 𝑥 2$ , so the coefficient of $𝑥 2$ is $- 1$ .
Constant term: There is no constant term (no standalone number without $𝑥$ ).

Question

2. Can you point out any similarity or difference between the algebraic expressions obtained in Example 1 (

4 𝑥 + 5 𝑦 + 3

) and Example 3 (

10 𝑥 - 𝑥 2

Solution

Compare Example 1 ( $4 𝑥 + 5 𝑦 + 3$ ) with Example 3 ( $10 𝑥 - 𝑥 2$ ):

Feature	Example 1: $4 𝑥 + 5 𝑦 + 3$	Example 3: $10 𝑥 - 𝑥 2$
Similarity	Both are algebraic expressions with numerical coefficients and variables
Number of variables	Two variables: $𝑥$ and $𝑦$	One variable: $𝑥$ only
Highest power (degree)	Highest power is $1$ (linear)	Highest power is $2$ (quadratic), because of $- 𝑥 2$
Constant term	Has constant term $3$	No constant term
Product of different variables	No product term like $𝑥 𝑦$	No product of two different variables (only powers of the same $𝑥$ )

So: both are algebraic expressions, but Example 3 is a one-variable quadratic polynomial, while Example 1 is a two-variable linear expression with a constant term.

EXERCISE SET 2.1

Question

1. Find the degrees of the following polynomials:

(i) $2 𝑥 2 - 5 𝑥 + 3$

(ii) $𝑦 3 + 2 𝑦 - 1$

(iii) $- 9$

(iv) $4 𝑧 - 3$

Solution

The degree of a polynomial is the highest exponent (power) of the variable in the expression.

(i) Degree is $2$ : In $2 𝑥 2 - 5 𝑥 + 3$ , the highest power of the variable $𝑥$ is $2$ .
(ii) Degree is $3$ : In $𝑦 3 + 2 𝑦 - 1$ , the highest power of the variable $𝑦$ is $3$ .
(iii) Degree is $0$ : The constant $- 9$ can be written as $- 9 𝑥 0$ . Since the variable's power is $0$ , the degree of any non-zero constant polynomial is $0$ .
(iv) Degree is $1$ : In $4 𝑧 - 3$ , the variable $𝑧$ has an implicit exponent of $1$ ( $4 𝑧 1 - 3$ ). Therefore, its degree is $1$ .

Question

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

Solution

(Note: Answers may vary; any polynomial with the correct highest exponent is correct.)

Degree 1 (Linear Polynomial): $3 𝑥 + 5$
Degree 2 (Quadratic Polynomial): $𝑥 2 - 4 𝑥 + 7$
Degree 3 (Cubic Polynomial): $2 𝑦 3 + 6 𝑦 2 - 𝑦 - 8$

Question

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

Solution

Look at the term containing $𝑥 2$ : it is $+ 6 𝑥 2$ . Therefore, the coefficient of $𝑥 2$ is $6$ .
Look at the term containing $𝑥 3$ : it is $- 3 𝑥 3$ . Therefore, the coefficient of $𝑥 3$ is $- 3$ .

Question

4. What is the coefficient of

𝑧

in the polynomial

4 𝑧 3 + 5 𝑧 2 - 11

Solution

The polynomial does not contain a term with the variable $𝑧$ to the first power ( $𝑧 1$ ). We can rewrite the polynomial by inserting a zero term: $4 𝑧 3 + 5 𝑧 2 + 0 𝑧 - 11$ . Therefore, the coefficient of $𝑧$ is $0$ .

Question

5. What is the constant term of the polynomial

9 𝑥 3 + 5 𝑥 2 - 8 𝑥 - 10

Solution

The constant term is the number that stands alone without any variable multiplying it. In this polynomial, that term is $- 10$ .

2.2 LINEAR POLYNOMIALS

Think and Reflect (after Example 4)

Question

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?

Solution

The formula for the perimeter of a square with side length $𝑥$ is $𝑃 = 4 𝑥$ . Substituting each value of $𝑥$ :

For $𝑥 = 1 c m$ : $𝑃 = 4 \times 1 = 𝟒 c m$
For $𝑥 = 1.5 c m$ : $𝑃 = 4 \times 1.5 = 𝟔 c m$
For $𝑥 = 2 c m$ : $𝑃 = 4 \times 2 = 𝟖 c m$
For $𝑥 = 2.5 c m$ : $𝑃 = 4 \times 2.5 = 𝟏 𝟎 c m$
For $𝑥 = 3 c m$ : $𝑃 = 4 \times 3 = 𝟏 𝟐 c m$

Question

What happens when the side increases by

0.5 c m

Every time the side length increases by $0.5 c m$ , the perimeter increases by a constant amount of $2 c m$ (since $4 \times 0.5 = 2$ ). This constant rate of increase is the defining feature of a linear relationship.

Think and Reflect (after Example 5)

Question

If a player paid ₹750, how many matches did he play?

Solution

From Example 5, the linear polynomial representing the total cost for playing $𝑚$ matches is $200 + 50 𝑚$ . Set this expression equal to the amount paid ( $750$ ) and solve for $𝑚$ :

200 + 50 𝑚 = 750

50 𝑚 = 750 - 200

50 𝑚 = 550

𝑚 = 55050 = 𝟏 𝟏

The player played $11$ matches.

Think and Reflect (after Fig. 2.3 / Input-Output Process)

Fig. 2.3: A linear expression as an input–output process

Question

We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle,

10 𝑥 - 𝑥 2

is a function of

𝑥

. Can you interpret this as an input-output process? What value does the expression take when

𝑥 = 6 c m

Solution

Input-Output Interpretation: Yes. In an input-output machine, the side length $𝑥$ (in cm) acts as the input. The machine processes this number using the rule "multiply the input by $10$ , then subtract the square of the input" ( $10 𝑥 - 𝑥 2$ ). The resulting number is the output, which represents the area of the rectangle (in $c m 2$ ).
Value when $𝑥 = 6 c m$ : Substitute $6$ into the function:

A r e a = 10 (6) - (6) 2 = 60 - 36 = 𝟐 𝟒 c m 𝟐

EXERCISE SET 2.2

Question

1. Find the value of the linear polynomial

5 𝑥 - 3

if:

(i) $𝑥 = 0$

(ii) $𝑥 = - 1$

(iii) $𝑥 = 2$

Solution

(i) For $𝑥 = 0$ : Substitute $0$ for $𝑥$ : $5 (0) - 3 = 0 - 3 = - 𝟑$
(ii) For $𝑥 = - 1$ : Substitute $- 1$ for $𝑥$ : $5 (- 1) - 3 = - 5 - 3 = - 𝟖$
(iii) For $𝑥 = 2$ : Substitute $2$ for $𝑥$ : $5 (2) - 3 = 10 - 3 = 𝟕$

Question

2. Find the value of the quadratic polynomial

7 𝑠 2 - 4 𝑠 + 6

if:

(i) $𝑠 = 0$

(ii) $𝑠 = - 3$

(iii) $𝑠 = 4$

Solution

(i) For $𝑠 = 0$ : Substitute $0$ for $𝑠$ : $7 (0) 2 - 4 (0) + 6 = 0 - 0 + 6 = 𝟔$
(ii) For $𝑠 = - 3$ : Substitute $- 3$ for $𝑠$ :

7 (- 3) 2 - 4 (- 3) + 6 = 7 (9) + 12 + 6 = 63 + 12 + 6 = 𝟖 𝟏

(iii) For $𝑠 = 4$ : Substitute $4$ for $𝑠$ :

7 (4) 2 - 4 (4) + 6 = 7 (16) - 16 + 6 = 112 - 16 + 6 = 𝟏 𝟎 𝟐

Question

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 present age be $𝑥$ years.
Then, his mother's present age is $3 𝑥$ years.
After $5$ years:

Salil's age will be $(𝑥 + 5)$ years.
His mother's age will be $(3 𝑥 + 5)$ years.

Set the sum of their future ages equal to $70$ :

(𝑥 + 5) + (3 𝑥 + 5) = 70

4 𝑥 + 10 = 70

4 𝑥 = 60

𝑥 = 𝟏 𝟓

Calculate their present ages:

Salil's age $= 𝑥 = 𝟏 𝟓 y e a r s$
Mother's age $= 3 𝑥 = 3 \times 15 = 𝟒 𝟓 y e a r s$

Question

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

Solution

Since the ratio is $2 : 5$ , let the smaller integer be $2 𝑥$ and the larger integer be $5 𝑥$ .
Set up an equation using their difference:

5 𝑥 - 2 𝑥 = 63

3 𝑥 = 63

𝑥 = 𝟐 𝟏

Find the actual integers by multiplying by $𝑥$ :

Smaller integer $= 2 𝑥 = 2 \times 21 = 𝟒 𝟐$
Larger integer $= 5 𝑥 = 5 \times 21 = 𝟏 𝟎 𝟓$

Question

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

Solution

Let the number of ₹5 coins Ruby has be $𝑥$ .
Then, the number of ₹2 coins she has is $3 𝑥$ .
Calculate the total monetary value of the coins:

Value from ₹5 coins $= 5 \times 𝑥 = 5 𝑥$ rupees
Value from ₹2 coins $= 2 \times (3 𝑥) = 6 𝑥$ rupees

Set the total monetary value equal to ₹88:

5 𝑥 + 6 𝑥 = 88

11 𝑥 = 88

𝑥 = 𝟖

Find the coin counts:

Number of ₹5 coins $= 𝑥 = 𝟖 c o i n s$
Number of ₹2 coins $= 3 𝑥 = 3 \times 8 = 𝟐 𝟒 c o i n s$

Question

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 length of the shorter piece be $𝑥$ feet.
Then, the length of the longer piece is $4 𝑥$ feet.
The sum of the lengths of both pieces equals the total length of the fence:

𝑥 + 4 𝑥 = 300

5 𝑥 = 300

𝑥 = 𝟔 𝟎

Calculate the two lengths:

Shorter piece $= 𝑥 = 𝟔 𝟎 f e e t$
Longer piece $= 4 𝑥 = 4 \times 60 = 𝟐 𝟒 𝟎 f e e t$

Question

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 the width of the rectangle be $𝑤 c m$ .
The length $𝑙$ is three more than twice the width: $𝑙 = 2 𝑤 + 3$ .
The formula for the perimeter of a rectangle is $𝑃 = 2 (𝑙 + 𝑤)$ . Substitute the expressions and set equal to $24$ :

2 ((2 𝑤 + 3) + 𝑤) = 24

2 (3 𝑤 + 3) = 24

6 𝑤 + 6 = 24

6 𝑤 = 18

𝑤 = 𝟑

Calculate the dimensions:

Width $= 𝑤 = 𝟑 c m$
Length $= 𝑙 = 2 (3) + 3 = 6 + 3 = 𝟗 c m$

2.3 EXPLORING LINEAR PATTERNS

Think and Reflect (after Fig. 2.4 - Growing Pattern of Square Tiles)

Fig. 2.4: A growing pattern of square tiles

Question

Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.

Solution

Looking at Fig. 2.4 in the book: Stage 1 has $1$ tile, Stage 2 has $3$ tiles, Stage 3 has $5$ tiles, and Stage 4 has $7$ tiles.
Notice that at each stage, $2$ tiles are added to the previous stage.
Next three stages (Stages 5, 6, and 7):
Stage 5: $7 + 2 = 𝟗 t i l e s$
Stage 6: $9 + 2 = 𝟏 𝟏 t i l e s$
Stage 7: $11 + 2 = 𝟏 𝟑 t i l e s$
Sequence up to Stage 7: $𝟏, 𝟑, 𝟓, 𝟕, 𝟗, 𝟏 𝟏, 𝟏 𝟑$

Think and Reflect (after deriving $2 𝑛 - 1$ )

Question

Using the expression

2 𝑛 - 1

, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?

Solution

Tiles in the 15th stage ( $𝑛 = 15$ ):

N u m b e r o f t i l e s = 2 (15) - 1 = 30 - 1 = 𝟐 𝟗 t i l e s

Tiles in the 26th stage ( $𝑛 = 26$ ):

N u m b e r o f t i l e s = 2 (26) - 1 = 52 - 1 = 𝟓 𝟏 t i l e s

Which stage contains 21 tiles? Set the formula equal to $21$ and solve for $𝑛$ :

2 𝑛 - 1 = 21 ⟹ 2 𝑛 = 22 ⟹ 𝑛 = 𝟏 𝟏 t h s t a g e

Which stage contains 47 tiles? Set the formula equal to $47$ and solve for $𝑛$ :

2 𝑛 - 1 = 47 ⟹ 2 𝑛 = 48 ⟹ 𝑛 = 𝟐 𝟒 t h s t a g e

Think and Reflect (after Example 7)

Question

What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?

Solution

From Example 7, Bela starts with ₹100 and spends ₹5 daily. The linear polynomial for the amount left after $𝑛$ days is $100 - 5 𝑛$ .

Amount left on the 15th day ( $𝑛 = 15$ ):

A m o u n t l e f t = 100 - 5 (15) = 100 - 75 = ₹ 𝟐 𝟓

Days to spend the entire amount: Set the amount left to $0$ and solve for $𝑛$ :

100 - 5 𝑛 = 0 ⟹ 5 𝑛 = 100 ⟹ 𝑛 = 𝟐 𝟎 d a y s

Think and Reflect (after Example 8)

Question

For how many km will the fare be ₹130?

Solution

From Example 8, the auto-rickshaw fare formula for a distance of $𝑛 k m$ ( $𝑛 \geq 2$ ) is $15 𝑛 - 5$ . Set this formula equal to ₹130 and solve for $𝑛$ :

15 𝑛 - 5 = 130

15 𝑛 = 135

𝑛 = 13515 = 𝟗 k m

A travel distance of $9 k m$ costs ₹130.

EXERCISE SET 2.3

Question

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.

Solution

Understanding the sequence:
At the start (Month 0): She has ₹500.
At the end of Month 1: $500 + 150 = ₹ 650$ .
At the end of Month 2: $650 + 150 = ₹ 𝟖 𝟎 𝟎$ .
At the end of Month 3: $800 + 150 = ₹ 𝟗 𝟓 𝟎$ .
At the end of Month 4: $950 + 150 = ₹ 𝟏, 𝟏 𝟎 𝟎$ , and so on.
Linear Expression: Since she adds ₹150 every month to her initial ₹500 balance, the amount of money at the end of the $𝑛$ th month is given by the linear expression:

𝟓 𝟎 𝟎 + 𝟏 𝟓 𝟎 𝐧

Question

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

𝑛

th hour.

Solution

Calculating members remaining hour by hour:
After 1 hour ( $𝑛 = 1$ ): $120 - 9 (1) = 𝟏 𝟏 𝟏 m e m b e r s$
After 2 hours ( $𝑛 = 2$ ): $120 - 9 (2) = 𝟏 𝟎 𝟐 m e m b e r s$
After 3 hours ( $𝑛 = 3$ ): $120 - 9 (3) = 𝟗 𝟑 m e m b e r s$
After 4 hours ( $𝑛 = 4$ ): $120 - 9 (4) = 𝟖 𝟒 m e m b e r s$
Linear Expression: Because 9 members leave each hour, the number of members remaining at the end of the $𝑛$ th hour decreases linearly:

𝟏 𝟐 𝟎 - 𝟗 𝐧

Question

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

The area of a rectangle is calculated as $A r e a = L e n g t h \times B r e a d t h$ . With a fixed length of $13 c m$ :

(i) For breadth $= 12 c m$ : $A r e a = 13 \times 12 = 𝟏 𝟓 𝟔 c m 𝟐$
(ii) For breadth $= 10 c m$ : $A r e a = 13 \times 10 = 𝟏 𝟑 𝟎 c m 𝟐$
(iii) For breadth $= 8 c m$ : $A r e a = 13 \times 8 = 𝟏 𝟎 𝟒 c m 𝟐$
Linear Pattern: Notice that every time the breadth decreases by $2 c m$ , the area decreases by a constant $26 c m 2$ ( $156 \to 130 \to 104$ ). If we let $𝑏$ represent the breadth in cm, the linear expression for the area is:

𝟏 𝟑 𝐛

Question

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

The volume of a rectangular box is $V o l u m e = L e n g t h \times B r e a d t h \times H e i g h t$ . Here, the base area is constant: $L e n g t h \times B r e a d t h = 7 \times 11 = 77 c m 2$ .

(i) For height $= 5 c m$ : $V o l u m e = 77 \times 5 = 𝟑 𝟖 𝟓 c m 𝟑$
(ii) For height $= 9 c m$ : $V o l u m e = 77 \times 9 = 𝟔 𝟗 𝟑 c m 𝟑$
(iii) For height $= 13 c m$ : $V o l u m e = 77 \times 13 = 𝟏, 𝟎 𝟎 𝟏 c m 𝟑$
Linear Pattern: Notice that as the height increases in equal intervals of $4 c m$ ( $5 \to 9 \to 13$ ), the volume increases by a constant interval of $308 c m 3$ ( $385 \to 693 \to 1001$ ). If $ℎ$ is the height in cm, the linear pattern for volume is:

𝟕 𝟕 𝐡

Question

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 15 days:

In $15$ days, Sarita reads a total of $20 \times 15 = 300$ pages.

P a g e s l e f t = 500 - 300 = 𝟐 𝟎 𝟎 p a g e s

Linear Pattern: Since the number of unread pages decreases by a constant $20$ pages each day, let $𝑛$ represent the number of days spent reading. The linear polynomial for pages left is:

𝟓 𝟎 𝟎 - 𝟐 𝟎 𝐧

2.4 LINEAR GROWTH AND LINEAR DECAY

Think and Reflect (after Example 9)

Question

What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?

Solution

From Example 9, the journey cost function is $𝐶 (𝑑) = 100 + 60 𝑑$ , where $𝑑$ is distance in km.

Cost for travelling $15 k m$ ( $𝑑 = 15$ ):

𝐶 (15) = 100 + 60 (15) = 100 + 900 = ₹ 𝟏, 𝟎 𝟎 𝟎

Distance when the cost is ₹700: Set $𝐶 (𝑑) = 700$ and solve for $𝑑$ :

100 + 60 𝑑 = 700

60 𝑑 = 600

𝑑 = 60060 = 𝟏 𝟎 k m

Think and Reflect (after Example 10)

Question

What will be the height of the water at the end of 5 months?

Solution

From Example 10, the water tank height function is $ℎ (𝑡) = 3 - 0.5 𝑡$ , where $𝑡$ is the number of months.

Height at the end of 5 months ( $𝑡 = 5$ ):

ℎ (5) = 3 - 0.5 (5) = 3 - 2.5 = 𝟎 . 𝟓 m e t e r s

EXERCISE SET 2.4

Question

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.

Solution

(i) Height after 7 months ( $𝑡 = 7$ ):

H e i g h t = 1.75 + 0.5 (7) = 1.75 + 3.5 = 𝟓 . 𝟐 𝟓 f e e t

(ii) Table of values (for $𝑡 = 0$ to $10$ ):

Month, $𝑡$	0	1	2	3	4	5	6	7	8	9	10
Height, $ℎ$ (ft)	1.75	2.25	2.75	3.25	3.75	4.25	4.75	5.25	5.75	6.25	6.75

(iii) Expression and Explanation:
Expression: $𝐡 (𝐭) = 𝟏 . 𝟕 𝟓 + 𝟎 . 𝟓 𝐭$
Explanation: This equation models linear growth because as the independent variable (time $𝑡$ ) increases by a steady interval of $1$ month, the dependent variable (height $ℎ$ ) increases by a fixed, constant amount ( $+ 0.5 f t$ ).

Question

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.

Solution

(i) Value after 3 years ( $𝑡 = 3$ ):

V a l u e = 10000 - 800 (3) = 10000 - 2400 = ₹ 𝟕, 𝟔 𝟎 𝟎

(ii) Table of values (for $𝑡 = 0$ to $8$ ):

Year, $𝑡$	0	1	2	3	4	5	6	7	8
Value, $𝑣$ (₹)	10000	9200	8400	7600	6800	6000	5200	4400	3600

(iii) Expression and Explanation:
Expression: $𝐯 (𝐭) = 𝟏 𝟎 𝟎 𝟎 𝟎 - 𝟖 𝟎 𝟎 𝐭$
Explanation: This equation models linear decay because as time ( $𝑡$ ) advances by equal $1$ -year intervals, the phone's value ( $𝑣$ ) decreases by a constant, unchanging amount ( $- ₹ 800$ ).

Question

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.

Solution

(i) Population after 6 years ( $𝑡 = 6$ ):

P o p u l a t i o n = 750 + 50 (6) = 750 + 300 = 𝟏, 𝟎 𝟓 𝟎 p e o p l e

(ii) Table of values (for $𝑡 = 0$ to $10$ ):

Year, $𝑡$	0	1	2	3	4	5	6	7	8	9	10
Population, $𝑃$	750	800	850	900	950	1000	1050	1100	1150	1200	1250

(iii) Expression and Explanation:
Expression: $𝐏 (𝐭) = 𝟕 𝟓 𝟎 + 𝟓 𝟎 𝐭$
Explanation: This equation models linear growth because the population size ( $𝑃$ ) increases by the same fixed number ( $+ 50 p e o p l e$ ) over each equal $1$ -year time interval.

Question

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.

Solution

(i) Equation and Explanation:
Equation: $𝐛 (𝐱) = 𝟔 𝟎 𝟎 - 𝟏 𝟓 𝐱$
Explanation: This models linear decay because the account balance decreases by a steady, unchanging amount ( $- ₹ 15$ ) for every single day ( $𝑥$ ) that passes.

(ii) When the balance runs out ( $𝑏 (𝑥) = 0$ ):

600 - 15 𝑥 = 0 ⟹ 15 𝑥 = 600 ⟹ 𝑥 = 60015 = 𝟒 𝟎 d a y s

The prepaid balance will be completely gone after $40$ days.

(iii) Table of values (for $𝑥 = 1$ to $10$ days):

Day, $𝑥$	1	2	3	4	5	6	7	8	9	10
Balance, $𝑏 (𝑥)$ (₹)	585	570	555	540	525	510	495	480	465	450

2.5 LINEAR RELATIONSHIPS

Think and Reflect (after Example 11)

Question

Can you guess what the numbers 20 and 150 in the equation

𝑦 = 20 𝑥 + 150

represent?

Solution

In the telecom data equation $𝑦 = 20 𝑥 + 150$ :

$20$ represents the rate per GB: It is the additional variable cost in rupees for every $1 G B$ of internet data used ( $𝑎$ ).
$150$ represents the fixed monthly fee: It is the base charge in rupees that a customer must pay every month even if zero internet data ( $𝑥 = 0$ ) is consumed ( $𝑏$ ).

EXERCISE SET 2.5

Question

1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill

𝑦

depends on the number of modules accessed,

𝑥

, according to the relation

𝑦 = 𝑎 𝑥 + 𝑏

, find the values of

𝑎

and

𝑏

Solution

Use the given data points $(𝑥, 𝑦)$ to form two linear equations:

For $𝑥 = 10$ , $𝑦 = 400 ⟹ 𝟒 𝟎 𝟎 = 𝟏 𝟎 𝐚 + 𝐛$ (Equation 1)
For $𝑥 = 14$ , $𝑦 = 500 ⟹ 𝟓 𝟎 𝟎 = 𝟏 𝟒 𝐚 + 𝐛$ (Equation 2)

Subtract Equation 1 from Equation 2 to eliminate $𝑏$ :

(500 - 400) = (14 𝑎 + 𝑏) - (10 𝑎 + 𝑏)

100 = 4 𝑎

𝑎 = 1004 = 𝟐 𝟓

Substitute $𝑎 = 25$ back into Equation 1 to find $𝑏$ :

400 = 10 (25) + 𝑏

400 = 250 + 𝑏

𝑏 = 400 - 250 = 𝟏 𝟓 𝟎

Conclusion: $𝐚 = 𝟐 𝟓$ (cost per module is ₹25) and $𝐛 = 𝟏 𝟓 𝟎$ (fixed monthly fee is ₹150). The relationship is $𝑦 = 25 𝑥 + 150$ .

Question

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

Solution

Form two equations from the given information:

For $𝑥 = 10$ , $𝑦 = 800 ⟹ 𝟖 𝟎 𝟎 = 𝟏 𝟎 𝐚 + 𝐛$ (Equation 1)
For $𝑥 = 15$ , $𝑦 = 1100 ⟹ 𝟏 𝟏 𝟎 𝟎 = 𝟏 𝟓 𝐚 + 𝐛$ (Equation 2)

Subtract Equation 1 from Equation 2:

1100 - 800 = 15 𝑎 - 10 𝑎

300 = 5 𝑎

𝑎 = 3005 = 𝟔 𝟎

Substitute $𝑎 = 60$ into Equation 1:

800 = 10 (60) + 𝑏

800 = 600 + 𝑏

𝑏 = 800 - 600 = 𝟐 𝟎 𝟎

Conclusion: $𝐚 = 𝟔 𝟎$ (court cost is ₹60/hour) and $𝐛 = 𝟐 𝟎 𝟎$ (fixed gym fee is ₹200). The relationship is $𝑦 = 60 𝑥 + 200$ .

Question

3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°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 °C and °F.)

Solution

Let $𝑦 = \circ C$ and $𝑥 = \circ F$ so that the equation is $𝑦 = 𝑎 𝑥 + 𝑏$ .
Substitute the melting point of ice ( $0 \circ C = 32 \circ F$ ):

0 = 32 𝑎 + 𝑏 ⟹ 𝐛 = - 𝟑 𝟐 𝐚

(Equation 1)

Substitute the boiling point of water ( $100 \circ C = 212 \circ F$ ):

𝟏 𝟎 𝟎 = 𝟐 𝟏 𝟐 𝐚 + 𝐛

(Equation 2)

Substitute $𝑏 = - 32 𝑎$ into Equation 2:

100 = 212 𝑎 - 32 𝑎

100 = 180 𝑎

𝑎 = 100180 = 𝟓 𝟗

Now, calculate $𝑏$ :

𝑏 = - 32 (59) = - 𝟏 𝟔 𝟎 𝟗

Conclusion: $𝐚 = 𝟓 𝟗$ and $𝐛 = - 𝟏 𝟔 𝟎 𝟗$ . The linear relationship is:

\circ C = 59 \circ F - 1609 o r s i m p l y \circ C = 𝟓 𝟗 (\circ F - 𝟑 𝟐)

2.6 VISUALISING LINEAR RELATIONSHIPS

Think and Reflect (after Fig. 2.5 / Line $𝑦 = 2 𝑥 + 1$ )

Fig. 2.5: The straight line y = 2x + 1

Question

Identify other points on the line by completing the following table:

$𝑥$	5	2	1	7	9	12	20
$𝑦$			3	15

Solution

Use the equation of the line, $𝑦 = 2 𝑥 + 1$ , to find the missing $𝑦$ -coordinates for each $𝑥$ -value:

For $𝑥 = 5$ : $𝑦 = 2 (5) + 1 = 10 + 1 = 𝟏 𝟏$
For $𝑥 = 2$ : $𝑦 = 2 (2) + 1 = 4 + 1 = 𝟓$
For $𝑥 = 1$ : $𝑦 = 2 (1) + 1 = 𝟑$ (given)
For $𝑥 = 7$ : $𝑦 = 2 (7) + 1 = 𝟏 𝟓$ (given)
For $𝑥 = 9$ : $𝑦 = 2 (9) + 1 = 18 + 1 = 𝟏 𝟗$
For $𝑥 = 12$ : $𝑦 = 2 (12) + 1 = 24 + 1 = 𝟐 𝟓$
For $𝑥 = 20$ : $𝑦 = 2 (20) + 1 = 40 + 1 = 𝟒 𝟏$

Question

Completed Table:

$𝑥$	5	2	1	7	9	12	20
$𝑦$	11	5	3	15	19	25	41

Think and Reflect (after Fig. 2.9)

Fig. 2.9: Graphs of y = ½x, y = x and y = 2x

Question

Does this help you to conclude anything about the linear equation

𝑦 = 𝑎 𝑥

𝑎 > 0

𝑎

varies? What happens when

𝑎 > 1

and when

𝑎 < 1

Solution

Looking at Fig. 2.9 (which plots $𝑦 = 12 𝑥$ , $𝑦 = 𝑥$ , and $𝑦 = 2 𝑥$ together):

Passes through origin: Regardless of the positive value of $𝑎$ , every line of the form $𝑦 = 𝑎 𝑥$ always passes directly through the origin point $(0, 0)$ .
When $𝑎 > 1$ (e.g., $𝑦 = 2 𝑥$ ): The line becomes steeper than the baseline $𝑦 = 𝑥$ (it tilts closer to the vertical $𝑦$ -axis). Larger values of $𝑎$ mean faster growth.
When $0 < 𝑎 < 1$ (e.g., $𝑦 = 12 𝑥$ ): The line becomes less steep than $𝑦 = 𝑥$ (it tilts closer to the horizontal $𝑥$ -axis). Fractional values of $𝑎$ represent slower growth.

Think and Reflect (after Fig. 2.11)

Fig. 2.11: Graphs of y = 3x + 1 and y = −3x + 1

Question

Differentiate between the graphs of the equations

𝑦 = 3 𝑥 + 1

and

𝑦 = - 3 𝑥 + 1

Solution

Similarity ( $𝑦$ -intercept): Both lines have the exact same constant term ( $𝑏 = + 1$ ). This means both lines cut across the vertical $𝑦$ -axis at the exact same point: $(0, 1)$ .
Difference (Direction/Slope):
The graph of $𝑦 = 3 𝑥 + 1$ has a positive coefficient ( $𝑎 = + 3$ ), so it slants upwards from left to right (representing linear growth).
The graph of $𝑦 = - 3 𝑥 + 1$ has a negative coefficient ( $𝑎 = - 3$ ), so it slants downwards from left to right (representing linear decay).
Geometric Relation: The two straight lines are mirror reflections of each other across the vertical $𝑦$ -axis, crossing each other at $(0, 1)$ .

Think and Reflect (after Fig. 2.13)

Fig. 2.13: Parallel lines y = 2x − 1, y = 2x + 1, y = 2x + 5

Question

Does this help you to conclude anything about the linear equation

𝑦 = 𝑎 𝑥 + 𝑏

when

𝑎

is fixed but

𝑏

varies?

Solution

Looking at Fig. 2.13 (which plots $𝑦 = 2 𝑥 - 1$ , $𝑦 = 2 𝑥 + 1$ , and $𝑦 = 2 𝑥 + 5$ ):

When the slope $𝑎$ is fixed ( $𝑎 = 2$ ), the steepness and angle of the lines never change. Therefore, all these lines are strictly parallel to each other and will never intersect.
When you vary the value of $𝑏$ , you simply shift the entire straight line vertically up or down along the coordinate plane:
A larger positive $𝑏$ (like $+ 5$ ) shifts the line upward to cut the $𝑦$ -axis at $(0, 5)$ .
A negative $𝑏$ (like $- 1$ ) shifts the line downward to cut the $𝑦$ -axis at $(0, - 1)$ .

EXERCISE SET 2.6

Sample graphs for Exercise Set 2.6 (i): y = 3x − 1, y = 3x, y = 3x + 1

Question

1. Draw the graphs of the following sets of lines. In each case, reflect on the role of '

𝑎

' and '

𝑏

(i) $𝑦 = 3 𝑥 - 1, 𝑦 = 3 𝑥, 𝑦 = 3 𝑥 + 1$

(ii) $𝑦 = 4 𝑥, 𝑦 = 2 𝑥, 𝑦 = 𝑥$

(iii) $𝑦 = - 6 𝑥, 𝑦 = - 3 𝑥, 𝑦 = - 𝑥$

(iv) $𝑦 = 5 𝑥, 𝑦 = - 5 𝑥$

(v) $𝑦 = - 2 𝑥 - 3, 𝑦 = - 2 𝑥, 𝑦 = 2 𝑥 + 3$

Solution

(To draw any linear graph on graph paper, pick two simple $𝑥$ -values—like $𝑥 = 0$ and $𝑥 = 1$ —calculate the corresponding $𝑦$ -values, plot the two points, and draw a straight line through them with a ruler.)

(i) Reflection on $𝑦 = 3 𝑥 - 1$ , $𝑦 = 3 𝑥$ , and $𝑦 = 3 𝑥 + 1$ :
Role of $𝑎$ : Here, $𝑎 = 3$ is fixed for all three equations. Because their slopes are identical, all three lines are parallel.
Role of $𝑏$ : The values of $𝑏$ are $- 1, 0,$ and $+ 1$ . This determines where each line cuts the $𝑦$ -axis: at $(0, - 1)$ , $(0, 0)$ , and $(0, 1)$ respectively. Varying $𝑏$ shifts the parallel line up or down.

(ii) Reflection on $𝑦 = 4 𝑥$ , $𝑦 = 2 𝑥$ , and $𝑦 = 𝑥$ :
Role of $𝑏$ : Here, $𝑏 = 0$ for all three equations. Therefore, all three straight lines pass through the origin $(0, 0)$ .
Role of $𝑎$ : As the positive slope $𝑎$ increases ( $1 \to 2 \to 4$ ), the lines become progressively steeper, climbing faster away from the $𝑥$ -axis toward the vertical $𝑦$ -axis.

(iii) Reflection on $𝑦 = - 6 𝑥$ , $𝑦 = - 3 𝑥$ , and $𝑦 = - 𝑥$ :
Role of $𝑏$ : Since $𝑏 = 0$ , all three lines pass directly through the origin $(0, 0)$ .
Role of $𝑎$ : Because $𝑎$ is negative ( $- 1, - 3, - 6$ ), all three lines slant downwards from left to right (linear decay). As the number becomes more negative (moving from $- 1$ to $- 6$ ), the downward slope becomes much steeper.

(iv) Reflection on $𝑦 = 5 𝑥$ and $𝑦 = - 5 𝑥$ :
Role of $𝑏$ : Both lines pass through the origin $(0, 0)$ because $𝑏 = 0$ .
Role of $𝑎$ : The slopes are equal in steepness ( $5$ ) but opposite in sign ( $+ 5$ and $- 5$ ). This creates two lines that form an X-shape at the origin; they are exact mirror images of one another across the vertical $𝑦$ -axis.

(v) Reflection on $𝑦 = - 2 𝑥 - 3$ , $𝑦 = - 2 𝑥$ , and $𝑦 = 2 𝑥 + 3$ :
Comparing the first two ( $𝑦 = - 2 𝑥 - 3$ and $𝑦 = - 2 𝑥$ ): Both share the same negative slope ( $𝑎 = - 2$ ), so they are parallel lines sloping downwards. They cross the $𝑦$ -axis at $(0, - 3)$ and $(0, 0)$ respectively.
Comparing with the third ( $𝑦 = 2 𝑥 + 3$ ): This line has a positive slope ( $𝑎 = + 2$ ), so it slants upwards and intersects the other two lines. It crosses the $𝑦$ -axis at $(0, 3)$ .

END-OF-CHAPTER EXERCISES

Question

1. Write a polynomial of degree 3 in the variable

𝑥

, in which the coefficient of the

𝑥 2

term is

- 7

Solution

(Note: Many answers are possible as long as the highest exponent of $𝑥$ is $3$ and the term with $𝑥 2$ is multiplied by $- 7$ .) An example is:

𝐱 𝟑 - 𝟕 𝐱 𝟐 + 𝟒 𝐱 - 𝟗

Question

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 $𝑡 = 𝑎$

Solution

(i) Substitute $𝑥 = 1$ :

5 (1) 2 - 3 (1) + 7 = 5 - 3 + 7 = 𝟗

(ii) Substitute $𝑡 = 𝑎$ :

Replace every $𝑡$ with the algebraic letter $𝑎$ :

𝟒 𝐚 𝟑 - 𝐚 𝟐 + 𝟔

Question

3. If we multiply a number by

52

and add

23

to the product, we get

- 7 12

. Find the number.

Solution

Let the unknown number be $𝑥$ .
Write the word problem as a linear equation:

52 𝑥 + 23 = - 712

Subtract $23$ from both sides. To do this, express $23$ with a denominator of $12$ ( $2 \times 4 3 \times 4 = 812$ ):

52 𝑥 = - 712 - 812

52 𝑥 = - 1512

Reduce $- 1512$ by dividing numerator and denominator by $3$ :

52 𝑥 = - 54

Multiply both sides by the reciprocal $25$ to isolate $𝑥$ :

𝑥 = - 54 \times 25 = - 24 = - 𝟏 𝟐

The required number is $- 12$ (or $- 0.5$ ).

Question

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 positive number be $𝑥$ .
Then, the larger positive number is $5 𝑥$ .
After adding $21$ to both numbers, the new numbers are:

Smaller new number $= (𝑥 + 21)$
Larger new number $= (5 𝑥 + 21)$

Since $𝑥$ is positive, $(5 𝑥 + 21)$ is clearly bigger than $(𝑥 + 21)$ . Therefore, the larger new number must be twice the smaller new number:

5 𝑥 + 21 = 2 (𝑥 + 21)

Expand and solve for $𝑥$ :

5 𝑥 + 21 = 2 𝑥 + 42

5 𝑥 - 2 𝑥 = 42 - 21

3 𝑥 = 21

𝑥 = 𝟕

Find both original numbers:

Smaller number $= 𝑥 = 𝟕$
Larger number $= 5 𝑥 = 5 \times 7 = 𝟑 𝟓$

Question

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

(i) Amount after 6 months:

A m o u n t = 800 + 250 (6) = 800 + 1500 = ₹ 𝟐, 𝟑 𝟎 𝟎

(ii) Amount after 2 years (24 months):

A m o u n t = 800 + 250 (24) = 800 + 6000 = ₹ 𝟔, 𝟖 𝟎 𝟎

Linear Pattern Expression:

Let $𝑚$ represent the number of months passed. The total amount saved follows the linear polynomial:

𝟖 𝟎 𝟎 + 𝟐 𝟓 𝟎 𝐦

Question

\*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 tens digit of the original number be $𝑡$ and the units digit be $𝑢$ .
The value of the original two-digit number is $(10 𝑡 + 𝑢)$ .
When the digits are interchanged, the new number's value is $(10 𝑢 + 𝑡)$ .
The sum of the original number and the interchanged number is $143$ :

(10 𝑡 + 𝑢) + (10 𝑢 + 𝑡) = 143

11 𝑡 + 11 𝑢 = 143

11 (𝑡 + 𝑢) = 143

𝑡 + 𝑢 = 14311 ⟹ 𝐭 + 𝐮 = 𝟏 𝟑

(Equation 1)

We are given that the digits differ by $3$ . This gives two possible cases:

Case 1: The tens digit is larger ( $𝑡 - 𝑢 = 3$ )

Add this equation to Equation 1:

(𝑡 + 𝑢) + (𝑡 - 𝑢) = 13 + 3 ⟹ 2 𝑡 = 16 ⟹ 𝐭 = 𝟖

Since $𝑡 = 8$ , substitute into Equation 1: $8 + 𝑢 = 13 ⟹ 𝐮 = 𝟓$ . This gives the number $85$ .

Case 2: The units digit is larger ( $𝑢 - 𝑡 = 3 ⟹ - 𝑡 + 𝑢 = 3$ )

Add this equation to Equation 1:

(𝑡 + 𝑢) + (- 𝑡 + 𝑢) = 13 + 3 ⟹ 2 𝑢 = 16 ⟹ 𝐮 = 𝟖

Since $𝑢 = 8$ , substitute into Equation 1: $𝑡 + 8 = 13 ⟹ 𝐭 = 𝟓$ . This gives the number $58$ .

Conclusion: Both two-digit numbers satisfying the conditions are $58$ and $85$ .

Question

\*7. Draw the graph of the following equations, and identify their slopes and

𝑦

-intercepts. Also, find the coordinates of the points where these lines cut the

𝑦

-axis.

(i) $𝑦 = - 3 𝑥 + 4$

(ii) $2 𝑦 = 4 𝑥 + 7$

(iii) $5 𝑦 = 6 𝑥 - 10$

(iv) $3 𝑦 = 6 𝑥 - 11$

Question

Are any of the lines parallel?

Solution

To find the slope ( $𝑎$ ) and $𝑦$ -intercept ( $𝑏$ ), rewrite each equation into the standard slope-intercept form: $𝑦 = 𝑎 𝑥 + 𝑏$ . A line cuts the $𝑦$ -axis at the coordinate $(0, 𝑏)$ .

(i) $𝑦 = - 3 𝑥 + 4$ :
Already in standard form.
Slope ( $𝑎$ ): $- 𝟑$
$𝑦$ -intercept ( $𝑏$ ): $𝟒$
Cuts $𝑦$ -axis at coordinate: $(𝟎, 𝟒)$

(ii) $2 𝑦 = 4 𝑥 + 7$ :
Divide all terms by $2$ : $𝑦 = 2 𝑥 + 72$ (or $𝑦 = 2 𝑥 + 3.5$ )
Slope ( $𝑎$ ): $𝟐$
$𝑦$ -intercept ( $𝑏$ ): $𝟕 𝟐$ (or $3.5$ )
Cuts $𝑦$ -axis at coordinate: $(𝟎, 𝟕 𝟐)$ or $(0, 3.5)$

(iii) $5 𝑦 = 6 𝑥 - 10$ :
Divide all terms by $5$ : $𝑦 = 65 𝑥 - 2$ (or $𝑦 = 1.2 𝑥 - 2$ )
Slope ( $𝑎$ ): $𝟔 𝟓$ (or $1.2$ )
$𝑦$ -intercept ( $𝑏$ ): $- 𝟐$
Cuts $𝑦$ -axis at coordinate: $(𝟎, - 𝟐)$

(iv) $3 𝑦 = 6 𝑥 - 11$ :
Divide all terms by $3$ : $𝑦 = 2 𝑥 - 113$
Slope ( $𝑎$ ): $𝟐$
$𝑦$ -intercept ( $𝑏$ ): $- 𝟏 𝟏 𝟑$ (or $\approx - 3.67$ )
Cuts $𝑦$ -axis at coordinate: $(𝟎, - 𝟏 𝟏 𝟑)$

Are any of the lines parallel?

Yes. Two lines are parallel if their slopes ( $𝑎$ ) are equal. Here, line (ii) and line (iv) both share the exact same slope of $2$ . Therefore, lines (ii) and (iv) are parallel.

(To draw these on graph paper: use the $𝑦$ -intercept $(0, 𝑏)$ as your starting point for each line, choose another $𝑥$ -value like $𝑥 = 1$ or $𝑥 = 5$ to compute a second point, and connect them with a ruler.)

Question

\*8. If the temperature of a liquid can be measured in Kelvin units as

𝑥 K

and in Fahrenheit units as

𝑦 \circ F

, the relation between the two systems of measurement of temperature is given by the linear equation

𝑦 = 95 (𝑥 - 273) + 32

(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313 K$ .

(ii) If the temperature is $158 \circ F$ then find the temperature in Kelvin.

Solution

(i) Given $𝑥 = 313 K$ , find $𝑦$ :

𝑦 = 95 (313 - 273) + 32

𝑦 = 95 (40) + 32

𝑦 = 9 (8) + 32 = 72 + 32 = 𝟏 𝟎 𝟒 \circ F

The temperature is $104 \circ F$ .

(ii) Given $𝑦 = 158 \circ F$ , find $𝑥$ :

158 = 95 (𝑥 - 273) + 32

Subtract $32$ from both sides:

158 - 32 = 95 (𝑥 - 273)

126 = 95 (𝑥 - 273)

Multiply both sides by the reciprocal $59$ :

126 \times 59 = 𝑥 - 273

14 \times 5 = 𝑥 - 273

70 = 𝑥 - 273

𝑥 = 70 + 273 = 𝟑 𝟒 𝟑 K

The temperature is $343 K$ .

Question

\*9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work

𝑤

and distance

𝑑

), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.

Graph of w = 3d (work vs distance, force = 3)

Solution

Linear Equation:

We are given that $W o r k = F o r c e \times D i s t a n c e$ . Let $𝑤$ be the work done and $𝑑$ be the distance travelled. With a constant force of $3 u n i t s$ , the linear equation is:

𝐰 = 𝟑 𝐝

Work done when distance $𝑑 = 2 u n i t s$ :

𝑤 = 3 (2) = 𝟔 u n i t s o f w o r k

Verification by Graphing:

Table of points to plot:

For $𝑑 = 0$ : $𝑤 = 3 (0) = 0 ⟹ (𝟎, 𝟎)$
For $𝑑 = 1$ : $𝑤 = 3 (1) = 3 ⟹ (𝟏, 𝟑)$
For $𝑑 = 2$ : $𝑤 = 3 (2) = 6 ⟹ (𝟐, 𝟔)$

Drawing the Graph: On a graph paper, label the horizontal axis as Distance ( $𝑑$ ) and the vertical axis as Work ( $𝑤$ ). Plot the points $(0, 0)$ , $(1, 3)$ , and $(2, 6)$ and draw a straight line passing through them from the origin.
Visual Verification: Locate $𝑑 = 2$ on the horizontal axis, move vertically upward until you touch the drawn straight line, and look across to the vertical axis. The value on the vertical axis will clearly read $6$ , verifying our calculation!

Question

\*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 the coordinates where the graph of $𝑝 (𝑥)$ cuts the axes.

(iii) Draw the graph of $𝑝 (𝑥)$ and verify your answers.

Graph of p(x) = 3x + 2

Solution

(i) Find the polynomial $𝑝 (𝑥)$ :

Let the general linear polynomial be $𝑝 (𝑥) = 𝑎 𝑥 + 𝑏$ .

Since it passes through $(1, 5)$ , substitute $𝑥 = 1$ and $𝑝 (1) = 5$ :

𝐚 + 𝐛 = 𝟓

(Equation 1)

Since it passes through $(3, 11)$ , substitute $𝑥 = 3$ and $𝑝 (3) = 11$ :

𝟑 𝐚 + 𝐛 = 𝟏 𝟏

(Equation 2)

Subtract Equation 1 from Equation 2:

(3 𝑎 + 𝑏) - (𝑎 + 𝑏) = 11 - 5 ⟹ 2 𝑎 = 6 ⟹ 𝐚 = 𝟑

Substitute $𝑎 = 3$ into Equation 1:

3 + 𝑏 = 5 ⟹ 𝐛 = 𝟐

Polynomial: $𝐩 (𝐱) = 𝟑 𝐱 + 𝟐$

(ii) Coordinates where graph cuts the axes:
Cutting the $𝑦$ -axis (set $𝑥 = 0$ ):

𝑦 = 3 (0) + 2 = 2 ⟹ C o o r d i n a t e i s (𝟎, 𝟐)

Cutting the $𝑥$ -axis (set $𝑦 = 𝑝 (𝑥) = 0$ ):

3 𝑥 + 2 = 0 ⟹ 3 𝑥 = - 2 ⟹ 𝑥 = - 23 ⟹ C o o r d i n a t e i s (- 𝟐 𝟑, 𝟎)

(or $\approx (- 0.67, 0)$ )

(iii) Graph and Verification:
Plot the known points $(0, 2)$ , $(1, 5)$ , and $(3, 11)$ on graph paper and draw a straight line through them.
Look at where the line crosses the axes: it clearly intersects the vertical $𝑦$ -axis at $+ 2$ , and the horizontal $𝑥$ -axis just to the left of zero at $- 23$ , verifying our answers!

Question

\*11. Let

𝑝 (𝑥) = 𝑎 𝑥 + 𝑏

and

𝑞 (𝑥) = 𝑐 𝑥 + 𝑑

be two linear polynomials such that:

(i) $𝑝 (0) = 5$ .

(ii) The polynomial $𝑝 (𝑥) - 𝑞 (𝑥)$ cuts the $𝑥$ -axis at $(3, 0)$ .

(iii) The sum $𝑝 (𝑥) + 𝑞 (𝑥)$ is equal to $6 𝑥 + 4$ for all real $𝑥$ .

Question

Find the polynomials

𝑝 (𝑥)

and

𝑞 (𝑥)

Solution

From condition (i):

𝑝 (0) = 𝑎 (0) + 𝑏 = 5 ⟹ 𝐛 = 𝟓

So, we know $𝑝 (𝑥) = 𝑎 𝑥 + 5$ .

From condition (iii):

𝑝 (𝑥) + 𝑞 (𝑥) = (𝑎 𝑥 + 5) + (𝑐 𝑥 + 𝑑) = (𝑎 + 𝑐) 𝑥 + (5 + 𝑑)

We are told this sum equals $6 𝑥 + 4$ . By comparing the coefficients of $𝑥$ and the constant terms:

Constant terms: $5 + 𝑑 = 4 ⟹ 𝐝 = - 𝟏$
$𝑥$ -coefficients: $𝑎 + 𝑐 = 6 ⟹ 𝐜 = 𝟔 - 𝐚$

So, we know $𝑞 (𝑥) = (6 - 𝑎) 𝑥 - 1$ .

From condition (ii):

First, find the difference polynomial $𝐷 (𝑥) = 𝑝 (𝑥) - 𝑞 (𝑥)$ :

𝐷 (𝑥) = (𝑎 𝑥 + 5) - [(6 - 𝑎) 𝑥 - 1]

𝐷 (𝑥) = 𝑎 𝑥 + 5 - (6 - 𝑎) 𝑥 + 1

𝐷 (𝑥) = (𝑎 - 6 + 𝑎) 𝑥 + 6 = (2 𝑎 - 6) 𝑥 + 6

We are told this polynomial cuts the $𝑥$ -axis at $(3, 0)$ . This means when $𝑥 = 3$ , $𝐷 (3) = 0$ :

(2 𝑎 - 6) (3) + 6 = 0

6 𝑎 - 18 + 6 = 0

6 𝑎 - 12 = 0 ⟹ 6 𝑎 = 12 ⟹ 𝐚 = 𝟐

Find $𝑐$ and write the final polynomials:

Since $𝑎 = 2$ , calculate $𝑐 = 6 - 𝑎 = 6 - 2 = 𝟒$ .

$𝐩 (𝐱) = 𝟐 𝐱 + 𝟓$
$𝐪 (𝐱) = 𝟒 𝐱 - 𝟏$

Question

\*12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(Stage 1: 1 hexagon, Stage 2: 2 hexagons, Stage 3: 3 hexagons)

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

(ii) Complete the following table:

Matchstick hexagon pattern (Stages 1–3)

Stage	1	2	3	4	5	$𝑛$
Number of matchsticks

(iii) Find a rule to determine the number of matchsticks required for the $𝑛$ th stage.

(iv) How many matchsticks will be required for the 15th stage of the pattern?

(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Solution

Understanding the matchstick count:
Stage 1 (1 standalone hexagon) requires $6$ matchsticks.
Stage 2 (2 attached hexagons sharing 1 side) requires $6 + 5 = 𝟏 𝟏 m a t c h s t i c k s$ .
Stage 3 (3 attached hexagons sharing 2 sides) requires $11 + 5 = 𝟏 𝟔 m a t c h s t i c k s$ .
Notice that every new attached hexagon requires $5$ additional matchsticks (since one side is already shared).

(i) Next two stages (Stage 4 and Stage 5):
Stage 4: Draw a horizontal chain of $4$ connected hexagons.

M a t c h s t i c k s r e q u i r e d = 16 + 5 = 𝟐 𝟏 m a t c h s t i c k s

Stage 5: Draw a horizontal chain of $5$ connected hexagons.

M a t c h s t i c k s r e q u i r e d = 21 + 5 = 𝟐 𝟔 m a t c h s t i c k s

(ii) Completed Table:

Stage	1	2	3	4	5	$𝑛$
Number of matchsticks	6	11	16	21	26	$5 𝑛 + 1$

(iii) Rule for the $𝑛$ th stage:

Since the count increases by $5$ at each stage starting from $6$ at Stage 1, the formula for $𝑛$ hexagons is:

M a t c h s t i c k s = 6 + 5 (𝑛 - 1) = 6 + 5 𝑛 - 5 = 𝟓 𝐧 + 𝟏

(iv) Matchsticks for the 15th stage ( $𝑛 = 15$ ):

M a t c h s t i c k s = 5 (15) + 1 = 75 + 1 = 𝟕 𝟔 m a t c h s t i c k s

(v) Can 200 matchsticks form a stage?

Set our rule equal to $200$ and check if $𝑛$ is a whole integer:

5 𝑛 + 1 = 200

5 𝑛 = 199

𝑛 = 1995 = 𝟑 𝟗 . 𝟖

Justification: No, 200 matchsticks cannot form a stage. The stage number $𝑛$ must always be a whole positive integer (you cannot have a fraction like $0.8$ of a stage in this geometric pattern). To form a complete stage, you would need either $196$ matchsticks (for Stage 39) or $201$ matchsticks (for Stage 40).

Question

\*13. Let

𝑝 (𝑥) = 𝑎 𝑥 + 𝑏

and

𝑞 (𝑥) = 𝑐 𝑥 + 𝑑

be two linear polynomials such that:

(i) The graph of $𝑝 (𝑥)$ passes through the points $(2, 3)$ and $(6, 11)$ .

(ii) The graph of $𝑞 (𝑥)$ passes through the point $(4, - 1)$ .

(iii) The graph of $𝑞 (𝑥)$ is parallel to the graph of $𝑝 (𝑥)$ .

Question

Find the polynomials

𝑝 (𝑥)

and

𝑞 (𝑥)

. Also, find the coordinates of the point where these lines meet the

𝑥

-axis.

Solution

Find polynomial $𝑝 (𝑥)$ using points $(2, 3)$ and $(6, 11)$ :

For $(2, 3)$ : $𝟐 𝐚 + 𝐛 = 𝟑$ (Equation 1)
For $(6, 11)$ : $𝟔 𝐚 + 𝐛 = 𝟏 𝟏$ (Equation 2)
Subtract Equation 1 from Equation 2:

4 𝑎 = 8 ⟹ 𝐚 = 𝟐

Substitute $𝑎 = 2$ into Equation 1:

2 (2) + 𝑏 = 3 ⟹ 4 + 𝑏 = 3 ⟹ 𝐛 = - 𝟏

Thus, $𝐩 (𝐱) = 𝟐 𝐱 - 𝟏$ .

Find polynomial $𝑞 (𝑥)$ :

Since $𝑞 (𝑥)$ is parallel to $𝑝 (𝑥)$ , it must have the exact same slope ( $𝑐 = 𝑎 = 2$ ). Thus, $𝑞 (𝑥) = 2 𝑥 + 𝑑$ .
We are given that $𝑞 (𝑥)$ passes through $(4, - 1)$ . Substitute $𝑥 = 4$ and $𝑞 (4) = - 1$ :

- 1 = 2 (4) + 𝑑 ⟹ - 1 = 8 + 𝑑 ⟹ 𝐝 = - 𝟗

Thus, $𝐪 (𝐱) = 𝟐 𝐱 - 𝟗$ .

Find coordinates where they meet the $𝑥$ -axis (set $𝑦 = 0$ ):

For $𝑝 (𝑥)$ : Set $2 𝑥 - 1 = 0 ⟹ 2 𝑥 = 1 ⟹ 𝑥 = 12$ .

The line $𝑝 (𝑥)$ meets the $𝑥$ -axis at $(𝟏 𝟐, 𝟎)$ (or $(0.5, 0)$ ).

For $𝑞 (𝑥)$ : Set $2 𝑥 - 9 = 0 ⟹ 2 𝑥 = 9 ⟹ 𝑥 = 92$ .

The line $𝑞 (𝑥)$ meets the $𝑥$ -axis at $(𝟗 𝟐, 𝟎)$ (or $(4.5, 0)$ ).

Question

\*14. What do all linear functions of the form

𝑓 (𝑥) = 𝑎 𝑥 + 𝑎

𝑎 > 0

, have in common?

Solution

Let us factor out the common algebraic factor $𝑎$ from the expression:

𝑓 (𝑥) = 𝑎 (𝑥 + 1)

Now, notice what happens if we calculate the value of the function when $𝑥 = - 1$ :

𝑓 (- 1) = 𝑎 (- 1 + 1) = 𝑎 (0) = 𝟎

Because the result is $0$ regardless of what positive number $𝑎$ is, we can conclude two striking features that all these functions share in common:

Common $𝑥$ -intercept: Every single graph of this form passes through the exact same point on the $𝑥$ -axis: $(- 1, 0)$ . They all pivot around this single anchor point!
Equal Slope and $𝑦$ -intercept: Because the equation is written as $𝑦 = 𝑎 𝑥 + 𝑎$ , the numerical value of the slope ( $𝑎$ ) is always equal to its $𝑦$ -intercept ( $𝑎$ ). Each line cuts across the vertical $𝑦$ -axis at the exact same height as its rate of climb!

2.1 INTRODUCTION

Think and Reflect (after Example 2)

Think and Reflect (after Example 3)

EXERCISE SET 2.1

2.2 LINEAR POLYNOMIALS

Think and Reflect (after Example 4)

Think and Reflect (after Example 5)

Think and Reflect (after Fig. 2.3 / Input-Output Process)

EXERCISE SET 2.2

2.3 EXPLORING LINEAR PATTERNS

Think and Reflect (after Fig. 2.4 - Growing Pattern of Square Tiles)

Think and Reflect (after deriving 2⁢𝑛 −1)

Think and Reflect (after Example 7)

Think and Reflect (after Example 8)

EXERCISE SET 2.3

2.4 LINEAR GROWTH AND LINEAR DECAY

Think and Reflect (after Example 9)

Think and Reflect (after Example 10)

EXERCISE SET 2.4

2.5 LINEAR RELATIONSHIPS

Think and Reflect (after Example 11)

EXERCISE SET 2.5

2.6 VISUALISING LINEAR RELATIONSHIPS

Think and Reflect (after Fig. 2.5 / Line 𝑦 =2⁢𝑥 +1)

Think and Reflect (after Fig. 2.9)

Think and Reflect (after Fig. 2.11)

Think and Reflect (after Fig. 2.13)

EXERCISE SET 2.6

END-OF-CHAPTER EXERCISES

Think and Reflect (after deriving $2 𝑛 - 1$ )

Think and Reflect (after Fig. 2.5 / Line $𝑦 = 2 𝑥 + 1$ )