Chapter 2: Introduction to Linear Polynomials

Think and Reflect — Page 17

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

The algebraic expression is:

200l + 160w + 50lw

Answer: In the expression 200l + 160w + 50lw:

Terms

The terms are: 200l, 160w and 50lw

Variables

The variables are:

l and w

Coefficients

So, the terms are 200l, 160w, 50lw, the variables are l and w, and the coefficients are 200, 160 and 50.

Question 2. How is it different from the algebraic expression in Example 1?

Answer: The algebraic expression in Example 1 is: 4x + 5y + 3

The algebraic expression in Example 2 is: 200l + 160w + 50lw

The difference is:

Therefore, Example 2 is different from Example 1 because it contains the term 50lw, where two variables are multiplied together.

Think and Reflect — Page 17

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

The algebraic expression is: 10x − x²

Answer: In the expression 10x − x²:

Terms

The terms are: 10x and −x²

Variable

The variable is: x

Coefficients

So, the terms are 10x and −x², the variable is x, and the coefficients are 10 and −1.

Question 2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?

Answer: The expression in Example 1 is: 4x + 5y + 3

The expression in Example 3 is: 10x − x²

Similarity

Both are algebraic expressions because both contain variables, coefficients and mathematical operations.

Difference

Therefore, Example 1 has two variables and is linear, while Example 3 has one variable and is quadratic.

Exercise Set 2.1

Question 1. Find the degrees of the following polynomials:

Question 1(i). 2x² − 5x + 3

Answer: The polynomial is: 2x² − 5x + 3

The highest power of x is 2.

Therefore, the degree of the polynomial is: 2

Question 1(ii). y³ + 2y − 1

Answer: The polynomial is:

y³ + 2y − 1

The highest power of y is 3.

Therefore, the degree of the polynomial is: 3

Question 1(iii). −9

Answer: The polynomial is: −9

This is a constant polynomial.

A non-zero constant polynomial has degree 0.

Therefore, the degree of the polynomial is: 0

Question 1(iv). 4z − 3

Answer: The polynomial is: 4z − 3

The highest power of z is 1.

Therefore, the degree of the polynomial is: 1

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

Answer: Examples of polynomials of degrees 1, 2 and 3 are:

So, one possible answer is:

Degree 1: 2x + 5
Degree 2: x² + 3x + 1
Degree 3: x³ − 2x² + 4x − 7

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

Answer: The polynomial is: x⁴ − 3x³ + 6x² − 2x + 7

Now compare the terms:

Therefore: Coefficient of x² = 6

Coefficient of x³ = −3

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

Answer: The polynomial is:

4z³ + 5z² − 11

There is no term containing only z.

We can write the polynomial as: 4z³ + 5z² + 0z − 11

Therefore, the coefficient of z is: 0

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

Answer: The polynomial is: 9x³ + 5x² − 8x − 10

The constant term is the term that does not contain any variable.

Here, the constant term is: −10

Therefore, the answer is: −10

Think and Reflect — Page 19

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?

Answer: Perimeter of a square:

Perimeter = 4 × side

So, the perimeters are: 4 cm, 6 cm, 8 cm, 10 cm and 12 cm

When the side increases by 0.5 cm, the perimeter increases by: 4 × 0.5 = 2 cm

Therefore, the perimeter increases by 2 cm each time.

Think and Reflect — Page 19

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

Answer: The total amount paid is given by: 200 + 50m

where m is the number of matches played.

Given amount paid = ₹750

So,

200 + 50m = 750

Subtract 200 from both sides: 50m = 750 − 200

50m = 550

Divide by 50: m = 550 ÷ 50

m = 11

Therefore, the player played 11 matches.

Think and Reflect — Page 20

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, 10x – x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?

Answer: Yes, the expression 10x − x² can be interpreted as an input-output process.

Here:

• Input = value of x

• Rule = 10x − x²

• Output = area of the rectangle

So, when we put a value of x, the expression gives the area of the rectangle.

Given: x = 6 cm

Now substitute x = 6 in the expression:

10x − x² = 10 × 6 − 6²

= 60 − 36

= 24

Therefore, when x = 6 cm, the value of the expression is 24.

So, the area of the rectangle is 24 cm².

Exercise Set 2.2

Question 1. Find the value of the linear polynomial 5x – 3 if:

Question 1(i). x = 0

Answer: The polynomial is: 5x − 3

Substitute x = 0:

5 × 0 − 3 = 0 − 3 = −3

Therefore, the value is −3.

Question 1(ii). x = −1

Answer: The polynomial is: 5x − 3

Substitute x = −1:

5 × (−1) − 3 = −5 − 3 = −8

Therefore, the value is −8.

Question 1(iii). x = 2

Answer: The polynomial is: 5x − 3

Substitute x = 2:

5 × 2 − 3 = 10 − 3 = 7

Therefore, the value is 7.

Question 2. Find the value of the quadratic polynomial 7s² – 4s + 6 if:

Question 2(i). s = 0

Answer: The polynomial is: 7s² − 4s + 6

Substitute s = 0:

7(0)² − 4(0) + 6 = 0 − 0 + 6 = 6

Therefore, the value is 6.

Question 2(ii). s = −3

Answer: The polynomial is: 7s² − 4s + 6

Substitute s = −3:

7(−3)² − 4(−3) + 6

= 7 × 9 + 12 + 6

= 63 + 12 + 6

= 81

Therefore, the value is 81.

Question 2(iii). s = 4

Answer: The polynomial is: 7s² − 4s + 6

Substitute s = 4:

7(4)² − 4(4) + 6

= 7 × 16 − 16 + 6

= 112 − 16 + 6

= 102

Therefore, the value is 102.

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.

Answer: Let Salil’s present age be x years.

Then his mother’s present age = 3x years

After 5 years: Salil’s age = x + 5

Mother’s age = 3x + 5

According to the question:

(x + 5) + (3x + 5) = 70

4x + 10 = 70

4x = 70 − 10

4x = 60

x = 15

So, Salil’s present age = 15 years

Mother’s present age:

3x = 3 × 15 = 45 years

Therefore, Salil is 15 years old and his mother is 45 years old.

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

Answer: The ratio of the two integers is 2:5.

Let the integers be: 2x and 5x

Their difference is 63.

So, 5x − 2x = 63

3x = 63

x = 21

Now, First integer = 2x = 2 × 21 = 42

Second integer = 5x = 5 × 21 = 105

Therefore, the two integers are 42 and 105.

Question 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?

Answer: Let the number of five-rupee coins be x.

Then the number of two-rupee coins = 3x

Value of five-rupee coins: 5 × x = 5x

Value of two-rupee coins: 2 × 3x = 6x

Total amount = ₹88

So, 5x + 6x = 88

11x = 88

x = 8

Therefore, Number of five-rupee coins = 8

Number of two-rupee coins = 3x = 3 × 8 = 24

Therefore, Ruby has 8 five-rupee coins and 24 two-rupee coins.

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?

Answer: Let the length of the shorter piece be x feet.

Then the length of the longer piece = 4x feet

Total length of fence = 300 feet

So, x + 4x = 300

5x = 300

x = 60

Shorter piece = 60 feet

Longer piece: 4x = 4 × 60 = 240 feet

Therefore, the two pieces are 60 feet and 240 feet.

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?

Answer: Let the width of the rectangle be x cm.

Length is three more than twice its width.

So, Length = 2x + 3 cm

Perimeter of rectangle:

2(length + width) = 24

Substitute the values: 2(2x + 3 + x) = 24

2(3x + 3) = 24

6x + 6 = 24

6x = 24 − 6

6x = 18

x = 3

So, width = 3 cm

Length: 2x + 3 = 2 × 3 + 3 = 6 + 3 = 9 cm

Therefore, the dimensions of the rectangle are:

Length = 9 cm
Width = 3 cm

Think and Reflect — Page 22

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.

Answer: From the given pattern:

Each new stage has 2 more squares than the previous stage.

So, the next three stages are:

Therefore, the sequence up to Stage 7 is:

1, 3, 5, 7, 9, 11, 13

Think and Reflect — Page 22

Question. Using the expression 2n – 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?

Answer: The number of tiles at stage n is given by: 2n − 1

Tiles in the 15th stage

Put n = 15:

2n − 1 = 2 × 15 − 1

= 30 − 1

= 29

So, the 15th stage will have 29 tiles.

Tiles in the 26th stage

Put n = 26:

2n − 1 = 2 × 26 − 1

= 52 − 1

= 51

So, the 26th stage will have 51 tiles.

Stage containing 21 tiles

Given: 2n − 1 = 21

2n = 22

n = 11

So, Stage 11 will contain 21 tiles.

Stage containing 47 tiles

Given: 2n − 1 = 47

2n = 48

n = 24

So, Stage 24 will contain 47 tiles.

Think and Reflect — Page 23

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

Answer: Bela has ₹100 and spends ₹5 every day.

Amount left after n days: 100 − 5n

Amount left on the 15th day

Put n = 15: 100 − 5 × 15

= 100 − 75

= 25

So, ₹25 will be left on the 15th day.

Number of days to spend the entire amount

For the entire amount to be spent, amount left = 0.

So, 100 − 5n = 0

5n = 100

n = 20

Therefore, it will take 20 days for the entire amount to be spent.

Think and Reflect — Page 23

Question. For how many km will the fare be ₹130?

Answer: The fare after the initial 2 km is given by:

15n − 5

where n is the distance travelled in km.

Given fare = ₹130

So, 15n − 5 = 130

Add 5 to both sides:

15n = 135

Divide by 15: n = 9

Therefore, the fare will be ₹130 for 9 km.

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 nth month.

Answer: Initial amount = ₹500

Pocket money received every month = ₹150

Amount at the end of the 1st month: 500 + 150 = ₹650

Amount at the end of the 2nd month: 500 + 2 × 150 = 500 + 300 = ₹800

Amount at the end of the 3rd month: 500 + 3 × 150 = 500 + 450 = ₹950

Amount at the end of the 4th month: 500 + 4 × 150 = 500 + 600 = ₹1100

So, from the second month onwards:

The linear expression for the amount at the end of the nth month is: 500 + 150n

Therefore, the amount at the end of the nth month is ₹(500 + 150n).

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 nth hour.

Answer: Initial number of members = 120

Number of members dropping out every hour = 9

After 1 hour: 120 − 9 = 111

After 2 hours: 120 − 2 × 9 = 120 − 18 = 102

After 3 hours: 120 − 3 × 9 = 120 − 27 = 93

After 4 hours: 120 − 4 × 9 = 120 − 36 = 84

So, the number of members remaining is:

At the end of the nth hour: 120 − 9n

Therefore, the linear expression is: 120 − 9n

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.

Answer: Length of rectangle = 13 cm

Area of rectangle: Area = length × breadth

So,

Area = 13 × breadth

Question 3(i). Breadth = 12 cm

Area = 13 × 12 = 156 cm²

Question 3(ii). Breadth = 10 cm

Area = 13 × 10 = 130 cm²

Question 3(iii). Breadth = 8 cm

Area = 13 × 8 = 104 cm²

If breadth is represented by b, then area is: A = 13b

Therefore, the linear pattern representing the area is: A = 13b

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.

Answer: Length of rectangular box = 7 cm

Breadth of rectangular box = 11 cm

Volume of rectangular box:

Volume = length × breadth × height

So, Volume = 7 × 11 × height

Volume = 77 × height

Question 4(i). Height = 5 cm

Volume = 77 × 5 = 385 cm³

Question 4(ii). Height = 9 cm

Volume = 77 × 9 = 693 cm³

Question 4(iii). Height = 13 cm

Volume = 77 × 13 = 1001 cm³

If height is represented by h, then volume is: V = 77h

Therefore, the linear pattern representing the volume is: V = 77h

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.

Answer: Total number of pages = 500

Pages read every day = 20

Pages read in 15 days: 20 × 15 = 300

Pages left after 15 days: 500 − 300 = 200

Therefore, after 15 days, 200 pages will be left.

If the number of days is represented by n, then pages left after n days: 500 − 20n

Therefore, the linear pattern is: 500 − 20n

Think and Reflect — Page 24

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

Answer: The cost of the journey is given by: C(d) = 100 + 60d

where d is the distance travelled in km.

Cost for travelling 15 km

Put d = 15: C(15) = 100 + 60 × 15

= 100 + 900

= ₹1000

So, the cost for travelling 15 km is ₹1000.

Distance for which cost is ₹700

Given: C(d) = 700

So, 100 + 60d = 700

60d = 700 − 100

60d = 600

d = 10

Therefore, the cost of the journey will be ₹700 for 10 km.

Think and Reflect — Page 25

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

Answer: The height of water is given by: h(t) = 3 − 0.5t

where t is time in months.

Put t = 5: h(5) = 3 − 0.5 × 5

= 3 − 2.5

= 0.5 m

Therefore, the height of water at the end of 5 months will be 0.5 m.

Exercise Set 2.4

Question 1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

Question 1(i). Find the height after 7 months.

Answer: Initial height of plant = 1.75 feet

Growth per month = 0.5 feet

Height after 7 months: h = 1.75 + 0.5 × 7

= 1.75 + 3.5

= 5.25 feet

Therefore, the height of the plant after 7 months is 5.25 feet.

Question 1(ii). Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.

Answer: The height of the plant after t months is: h = 1.75 + 0.5t

Question 1(iii). Find an expression that relates h and t, and explain why it represents linear growth.

Answer: The expression is: h = 1.75 + 0.5t

This represents linear growth because the height of the plant increases by a fixed amount of 0.5 feet every month.

Question 2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

Question 2(i). Find the value of the phone after 3 years.

Answer: Initial value of phone = ₹10,000

Decrease per year = ₹800

Value after 3 years: v = 10000 − 800 × 3

= 10000 − 2400

= ₹7600

Therefore, the value of the phone after 3 years is ₹7600.

Question 2(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.

Answer: The value of the phone after t years is: v = 10000 − 800t

Question 2(iii). Find an expression that relates v and t, and explain why it represents linear decay.

Answer: The expression is: v = 10000 − 800t

This represents linear decay because the value of the phone decreases by a fixed amount of ₹800 every year.

Question 3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

Question 3(i). Find the population of the village after 6 years.

Answer: Initial population = 750

Increase every year = 50

Population after 6 years: P = 750 + 50 × 6

= 750 + 300

= 1050

Therefore, the population of the village after 6 years is 1050.

Question 3(ii). Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.

Answer: The population after t years is: P = 750 + 50t

Question 3(iii). Find an expression that relates P and t, and explain why it represents linear growth.

Answer: The expression is: P = 750 + 50t

This represents linear growth because the population increases by a fixed number of 50 people every year.

Question 4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

Question 4(i). Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.

Answer: Initial recharge balance = ₹600

Balance reduced each day = ₹15

Remaining balance after x days: b(x) = 600 − 15x

This represents linear decay because the balance decreases by a fixed amount of ₹15 every day.

Question 4(ii). After how many days will the balance run out?

Answer: The balance will run out when: b(x) = 0

So, 600 − 15x = 0

15x = 600

x = 600 ÷ 15

x = 40

Therefore, the balance will run out after 40 days.

Question 4(iii). Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.

Answer: Using: b(x) = 600 − 15x

Think and Reflect — Page 27

Question. Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?

Answer: The equation is: y = 20x + 150

Here:

• x represents the amount of internet data used in GB.

• y represents the total monthly bill in rupees.

In this equation: 20 represents the additional cost per GB of internet data.

150 represents the fixed monthly fee.

Therefore, the company charges ₹150 as fixed monthly fee and ₹20 per GB of data used.

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 = ax + b, find the values of a and b.

Answer: The relation is: y = ax + b

Given: When x = 10, y = 400

So, 400 = 10a + b — equation (1)

When x = 14, y = 500

So, 500 = 14a + b — equation (2)

Subtract equation (1) from equation (2):

500 − 400 = 14a − 10a

100 = 4a

a = 25

Now substitute a = 25 in equation (1): 400 = 10 × 25 + b

400 = 250 + b

b = 150

Therefore: a = 25 and b = 150

So, the linear relationship is: y = 25x + 150

Here, ₹25 is charged per module and ₹150 is the fixed monthly fee.

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 = ax + b, find the values of a and b.

Answer: The relation is: y = ax + b

Given: When x = 10, y = 800

So,

800 = 10a + b — equation (1)

When x = 15, y = 1100

So,

1100 = 15a + b — equation (2)

Subtract equation (1) from equation (2):

1100 − 800 = 15a − 10a

300 = 5a

a = 60

Now substitute a = 60 in equation (1): 800 = 10 × 60 + b

800 = 600 + b

b = 200

Therefore: a = 60 and b = 200

So, the linear relationship is: y = 60x + 200

Here, ₹60 is charged per hour and ₹200 is the fixed monthly fee.

Question 3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °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.

Answer: The relation is:

°C = a °F + b

Given: When °F = 32, °C = 0

So, 0 = 32a + b — equation (1)

When °F = 212, °C = 100

So, 100 = 212a + b — equation (2)

Subtract equation (1) from equation (2): 100 − 0 = 212a − 32a

100 = 180a

a = 100/180

a = 5/9

Now substitute a = 5/9 in equation (1): 0 = 32 × 5/9 + b

0 = 160/9 + b

b = −160/9

Therefore: a = 5/9 and b = −160/9

So, the linear relationship is: °C = 5/9 °F − 160/9

or

°C = 5/9 (°F − 32)

Think and Reflect — Page 28

Question. Identify other points on the line by completing the following table.

Given line: y = 2x + 1

Answer: We use: y = 2x + 1

For x = 1: y = 2 × 1 + 1 = 3

For x = 2: y = 2 × 2 + 1 = 5

For x = 5: y = 2 × 5 + 1 = 11

For x = 7: y = 2 × 7 + 1 = 15

For x = 9: y = 2 × 9 + 1 = 19

For x = 12: y = 2 × 12 + 1 = 25

For x = 20: y = 2 × 20 + 1 = 41

Completed table:

Therefore, the points are:

(1, 3), (2, 5), (5, 11), (7, 15), (9, 19), (12, 25), (20, 41)

Think and Reflect — Page 33

Question. Differentiate between the graphs of the equations y = 3x + 1, and y = −3x + 1.

Answer: The two equations are: y = 3x + 1

and

y = −3x + 1

Both equations are of the form: y = ax + b

For y = 3x + 1

Here: a = 3, b = 1

So, the slope is positive.

This means the line rises from left to right.

For y = −3x + 1

Here: a = −3, b = 1

So, the slope is negative.

This means the line falls from left to right.

Difference

Both lines cut the y-axis at the same point (0, 1), but their directions are different.

Think and Reflect — Page 35

Question. Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?

Answer: Yes.

In the equation: y = ax + b

• a represents the slope of the line.

• b represents the y-intercept.

When a is fixed but b varies:

1. The slope of the lines remains the same.

2. The lines remain parallel to each other.

3. The lines cut the y-axis at different points.

4. The graph shifts upward or downward depending on the value of b.

Example: y = 2x − 1, y = 2x + 1, y = 2x + 5

All these lines have the same slope 2, so they are parallel.

But their y-intercepts are different:

• y = 2x − 1 cuts y-axis at −1

• y = 2x + 1 cuts y-axis at 1

• y = 2x + 5 cuts y-axis at 5

Therefore, when a is fixed and b varies, the lines are parallel and shift up or down.

Exercise Set 2.6

Question 1(i). Draw the graphs of y = 4x, y = 2x, y = x. In each case, reflect on the role of ‘a’ and ‘b’.

Answer: All equations are of the form: y = ax + b

Here, b = 0 in all three equations.
So, all lines pass through the origin (0, 0).

Table of points

Reflection

Since all slopes are positive, all lines rise from left to right.

The larger the value of a, the steeper the line.

Question 1(ii). Draw the graphs of y = −6x, y = −3x, y = −x. In each case, reflect on the role of ‘a’ and ‘b’.

Answer: All equations are of the form: y = ax + b

Here, b = 0 in all equations.
So, all lines pass through the origin (0, 0).

Table of points

Reflection

Since all slopes are negative, all lines fall from left to right.

The greater the numerical value of negative slope, the steeper the downward line.

Question 1(iii). Draw the graphs of y = 5x, y = −5x. In each case, reflect on the role of ‘a’ and ‘b’.

Answer: Both equations are of the form: y = ax + b

Here, b = 0 in both equations.
So, both lines pass through the origin (0, 0).

Table of points

Reflection

Both lines have the same steepness because the numerical value of slope is 5.

But their directions are opposite because one slope is positive and the other is negative.

Question 1(iv). Draw the graphs of y = 3x − 1, y = 3x, y = 3x + 1. In each case, reflect on the role of ‘a’ and ‘b’.

Answer: All equations are of the form: y = ax + b

Here, a = 3 in all equations.
So, all lines have the same slope and are parallel.

Table of points

Reflection

Since a is the same, the lines are parallel.

Since b is different, the lines cut the y-axis at different points.

Question 1(v). Draw the graphs of y = −2x − 3, y = −2x, y = 2x + 3. In each case, reflect on the role of ‘a’ and ‘b’.

Answer: The equations are: y = −2x − 3

y = −2x

y = 2x + 3

Table of points

Reflection

Conclusion

• y = −2x − 3 and y = −2x have the same slope −2, so they are parallel.

• y = 2x + 3 has positive slope 2, so it rises from left to right.

• The value of b tells where the line cuts the y-axis.

• The value of a tells the direction and steepness of the line.

End-of-Chapter Exercises

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

Answer: A polynomial of degree 3 must have highest power 3.

One possible polynomial is: x³ − 7x² + 2x + 5

Here, the coefficient of x² is −7.

Therefore, one answer is: x³ − 7x² + 2x + 5

Question 2. Find the values of the following polynomials at the indicated values of the variables.

Question 2(i). 5x² – 3x + 7 if x = 1

Answer: Given polynomial: 5x² − 3x + 7

Put x = 1:

5(1)² − 3(1) + 7

= 5 − 3 + 7

= 9

Therefore, the value of the polynomial is 9.

Question 2(ii). 4t³ – t² + 6 if t = a

Answer: Given polynomial: 4t³ − t² + 6

Put t = a:

4a³ − a² + 6

Therefore, the value of the polynomial is: 4a³ − a² + 6

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

Answer: Let the number be x.

According to the question: 5x/2 + 2/3 = −7/12

Subtract 2/3 from both sides: 5x/2 = −7/12 − 2/3

LCM of 12 and 3 is 12.

5x/2 = −7/12 − 8/12

5x/2 = −15/12

5x/2 = −5/4

Now multiply both sides by 2/5:

x = −5/4 × 2/5

x = −10/20

x = −1/2

Therefore, the number is:

−1/2

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?

Answer: Let the smaller number be x.

Then the larger number is 5x.

After adding 21 to both numbers:

Smaller new number = x + 21

Larger new number = 5x + 21

According to the question: 5x + 21 = 2(x + 21)

5x + 21 = 2x + 42

5x − 2x = 42 − 21

3x = 21

x = 7

So, smaller number = 7

Larger number: 5x = 5 × 7 = 35

Therefore, the numbers are: 7 and 35

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.

Answer: Initial amount = ₹800

Monthly saving = ₹250

Amount after n months: 800 + 250n

So, the linear pattern is: A = 800 + 250n

where A is the total amount after n months.

Question 5(i). Amount after 6 months

Put n = 6: A = 800 + 250 × 6

= 800 + 1500

= ₹2300

Therefore, after 6 months, the amount is ₹2300.

Question 5(ii). Amount after 2 years

2 years = 24 months

Put n = 24: A = 800 + 250 × 24

= 800 + 6000

= ₹6800

Therefore, after 2 years, the amount is ₹6800.

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.

Answer: Let the tens digit be a and the ones digit be b.

Original number = 10a + b

Number after interchanging digits = 10b + a

According to the question: (10a + b) + (10b + a) = 143

11a + 11b = 143

11(a + b) = 143

a + b = 13

Also, the digits differ by 3.

So, a − b = 3 or b − a = 3

The two digits whose sum is 13 and difference is 3 are: 8 and 5

Therefore, the possible two-digit numbers are: 85 and 58

Check: 85 + 58 = 143

Hence, the required numbers are 85 and 58.

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. Are any of the lines parallel?

Answer: A line in the form: y = ax + b

has:

• slope = a

• y-intercept = b

• point where it cuts the y-axis = (0, b)

Question 7(i). y = −3x + 4

This is already in the form y = ax + b.

Here: a = −3, b = 4

So, Slope = −3

y-intercept = 4

Point where it cuts y-axis = (0, 4)

For drawing the graph, take two points:

Points: (0, 4) and (1, 1)

Question 7(ii). 2y = 4x + 7

Given: 2y = 4x + 7

Divide by 2: y = 2x + 7/2

Here: a = 2, b = 7/2

So,

Slope = 2

y-intercept = 7/2

Point where it cuts y-axis = (0, 7/2)

For drawing the graph, take two points:

Points: (0, 7/2) and (1, 11/2)

Question 7(iii). 5y = 6x − 10

Given: 5y = 6x − 10

Divide by 5: y = 6x/5 − 2

Here: a = 6/5, b = −2

So, Slope = 6/5

y-intercept = −2

Point where it cuts y-axis = (0, −2)

For drawing the graph, take two points:

Points: (0, −2) and (5, 4)

Question 7(iv). 3y = 6x − 11

Given: 3y = 6x − 11

Divide by 3: y = 2x − 11/3

Here: a = 2, b = −11/3

So, Slope = 2

y-intercept = −11/3

Point where it cuts y-axis = (0, −11/3)

For drawing the graph, take two points:

Points: (0, −11/3) and (1, −5/3)

Summary Table

Lines with the same slope are parallel.

Here, equations (ii) and (iv) have the same slope 2.

Therefore, the lines: 2y = 4x + 7 and 3y = 6x – 11 are parallel.

Question 8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9/5 (x – 273) + 32.

Question 8(i). Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

Answer: Given relation: y = 9/5 (x − 273) + 32

Here, x = 313

Substitute: y = 9/5 (313 − 273) + 32

y = 9/5 × 40 + 32

y = 9 × 8 + 32

y = 72 + 32

y = 104

Therefore, the temperature is: 104 °F

Question 8(ii). If the temperature is 158 °F, then find the temperature in Kelvin.

Answer: Given: y = 158

Using the formula: y = 9/5 (x − 273) + 32

Substitute y = 158: 158 = 9/5 (x − 273) + 32

Subtract 32 from both sides: 158 − 32 = 9/5 (x − 273)

126 = 9/5 (x − 273)

Multiply both sides by 5/9: 126 × 5/9 = x − 273

14 × 5 = x − 273

70 = x − 273

x = 343

Therefore, 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.

Answer: Work done = Force × Distance

Let work done be w and distance be d.

Given force = 3 units

So, w = 3d

This is the required linear equation.

Table for graph

Plot the points: (0, 0), (1, 3), (2, 6), (3, 9)

Join these points to get a straight line.

Work done when distance is 2 units

Put d = 2: w = 3 × 2

w = 6

Therefore, the work done is 6 units.

The point (2, 6) lies on the graph, so the answer is verified.

Question 10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).

Question 10(i). Find the polynomial p(x).

Answer: Let the linear polynomial be: p(x) = ax + b

The graph passes through (1, 5).

So, 5 = a(1) + b

a + b = 5 — equation (1)

The graph also passes through (3, 11).

So, 11 = 3a + b — equation (2)

Subtract equation (1) from equation (2): 11 − 5 = 3a − a

6 = 2a

a = 3

Now substitute a = 3 in equation (1): 3 + b = 5

b = 2

Therefore, p(x) = 3x + 2

Question 10(ii). Find the coordinates where the graph of p(x) cuts the axes.

Answer: The polynomial is: p(x) = 3x + 2

This means: y = 3x + 2

Point where graph cuts y-axis

On the y-axis, x = 0.

So, y = 3(0) + 2 = 2

Therefore, the graph cuts the y-axis at: (0, 2)

Point where graph cuts x-axis

On the x-axis, y = 0.

So, 0 = 3x + 2

3x = −2

x = −2/3

Therefore, the graph cuts the x-axis at:

(−2/3, 0)

Question 10(iii). Draw the graph of p(x) and verify your answers.

Answer: The graph is of: y = 3x + 2

Take some points:

Plot the points: (−2/3, 0), (0, 2), (1, 5), (3, 11)

Join them to get a straight line.

The graph passes through (1, 5) and (3, 11), and cuts the axes at: (0, 2) and (−2/3, 0)

Hence, the answers are verified.

Question 11. Let p(x) = ax + b and q(x) = cx + 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 6x + 4 for all real x.
Find the polynomials p(x) and q(x).

Answer: Given:

p(x) = ax + b

q(x) = cx + d

From condition (i)

p(0) = 5

So, a(0) + b = 5

b = 5

From condition (iii)

p(x) + q(x) = 6x + 4

So, (ax + b) + (cx + d) = 6x + 4

(a + c)x + (b + d) = 6x + 4

Comparing coefficients: a + c = 6 — equation (1)

b + d = 4

But b = 5.

So, 5 + d = 4

d = −1

From condition (ii)

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

This means when x = 3, p(x) − q(x) = 0.

Now, p(x) − q(x) = (ax + b) − (cx + d)

= (a − c)x + (b − d)

Since b = 5 and d = −1:

b − d = 5 − (−1) = 6

So, p(x) − q(x) = (a − c)x + 6

At x = 3, value is 0:

(a − c)3 + 6 = 0

3(a − c) = −6

a − c = −2 — equation (2)

Now solve: a + c = 6

a − c = −2

Adding both equations: 2a = 4

a = 2

Now, a + c = 6

2 + c = 6

c = 4

Therefore: p(x) = 2x + 5

q(x) = 4x − 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.

Question 12(i). Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

Answer: For one hexagon, matchsticks required = 6

When a new hexagon is added, it shares one side with the previous hexagon.

So, each new hexagon adds only 5 matchsticks.

Stage 1: 6 matchsticks

Stage 2: 6 + 5 = 11 matchsticks

Stage 3: 11 + 5 = 16 matchsticks

Stage 4: 16 + 5 = 21 matchsticks

Stage 5: 21 + 5 = 26 matchsticks

Therefore:

Stage 4 requires 21 matchsticks.

Stage 5 requires 26 matchsticks.

Question 12(ii). Complete the following table.

Answer:

Question 12(iii). Find a rule to determine the number of matchsticks required for the nth stage.

Answer: At Stage 1, matchsticks = 6

Each new stage adds 5 matchsticks.

So, number of matchsticks at Stage n: 6 + 5(n − 1)

= 6 + 5n − 5

= 5n + 1

Therefore, the rule is: 5n + 1

Question 12(iv). How many matchsticks will be required for the 15th stage of the pattern?

Answer: Using the rule: 5n + 1

Put n = 15: 5 × 15 + 1

= 75 + 1

= 76

Therefore, 76 matchsticks will be required for the 15th stage.

Question 12(v). Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer: The number of matchsticks at Stage n is: 5n + 1

Check whether 200 matchsticks can form a stage: 5n + 1 = 200

5n = 199

n = 199/5

n = 39.8

Since the stage number must be a whole number, 39.8 is not possible.

Therefore, 200 matchsticks cannot form a stage in this pattern.

Question 13. Let p(x) = ax + b and q(x) = cx + 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).
Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

Answer: Let:

p(x) = ax + b

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

Slope of p(x): a = (11 − 3) / (6 − 2)

a = 8/4

a = 2

So, p(x) = 2x + b

Since it passes through (2, 3): 3 = 2(2) + b

3 = 4 + b

b = −1

Therefore, p(x) = 2x − 1

Since the graph of q(x) is parallel to the graph of p(x), both have the same slope.

So, q(x) = 2x + d

The graph of q(x) passes through (4, −1).

So, −1 = 2(4) + d

−1 = 8 + d

d = −9

Therefore, q(x) = 2x − 9

Points where the lines meet the x-axis

For the x-axis, y = 0.

For p(x) = 2x − 1

0 = 2x − 1

2x = 1

x = 1/2

So, the graph of p(x) meets the x-axis at: (1/2, 0)

For q(x) = 2x − 9

0 = 2x − 9

2x = 9

x = 9/2

So, the graph of q(x) meets the x-axis at: (9/2, 0)

Therefore: p(x) = 2x − 1

q(x) = 2x − 9

They meet the x-axis at: (1/2, 0) and (9/2, 0) respectively.

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

Answer:

Given: f(x) = ax + a

Take a common: f(x) = a(x + 1)

Since a > 0, the slope is positive.

So, all such functions have the following things in common:

1. They are all linear functions.

2. Their graphs are straight lines.

3. Their slopes are positive, so the lines rise from left to right.

4. In f(x) = ax + a, the slope and y-intercept are both equal to a.

5. All these graphs pass through the point (−1, 0) because:

f(−1) = a(−1) + a

= −a + a

= 0

Therefore, all linear functions of the form:

f(x) = ax + a, a > 0

pass through the common point:

(−1, 0)