Introduction to Linear Polynomials: Class 9 Solutions Part 1 Chapter 2

Linear polynomials are algebraic expressions in one variable whose highest power is 1. In Class 9 Maths Part 1 Chapter 2, students learn how variables, coefficients, constants, degree, linear patterns, linear growth, linear decay, slope and y-intercept help represent real-life situations using expressions and graphs.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 2 explain Introduction to Linear Polynomials from the new Ganita Manjari textbook, where students learn how algebraic expressions can represent changing quantities. The chapter introduces terms, variables, coefficients, constants, degree of a polynomial and linear polynomials.

Ganita Manjari Class 9 Chapter 2 Solutions: Introduction to Linear Polynomials

Ex 2.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.1 Solutions

Question 1:
Find the degrees of the following polynomials.
(i) 2x² – 5x + 3

Solution:

The highest power of

$x$

is 2.

Degree of the polynomial = 2.

(ii) y³ + 2y – 1

Solution:

For

$y^3 + 2y - 1$

, the highest power of

$y$

is 3.

Degree of the polynomial = 3.

(iii) – 9

Solution:

For the polynomial

$-9$

, there is no variable term.

Degree of the polynomial = 0.

(iv) 4z – 3

Solution:

For

$4z - 3$

, the highest power of

$z$

is 1.

Degree of the polynomial = 1.

Question 2:
Write the polynomials of degree 1, 2, and 3.

Solution:

• Degree 1 (Linear): 4x + 2
• Degree 2 (Quadratic): 3x² – 4x + 7
• Degree 3 (Cubic): x³ + 2x² – x + 10

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

Solution:

The term containing x² is 6x², so the coefficient of x² is 6.
The term containing x³ is -3x³, so the coefficient of x³ is -3.

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

Solution:

Since we can rewrite 4z³ + 5z² – 11 as 4z³ + 5z² + 0z – 11,
∴ The coefficient of z in the given polynomial is 0.

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

Solution:

In the polynomial

 

$$9x^3 + 5x^2 - 8x - 10$$

 

the term without a variable is

$-10$

.

Constant term =

$-10$

.

Ex 2.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.2 Solutions

Question 1:
Find the value of the linear polynomial 5x – 3 if:
(i) x = 0

Solution:

Given the polynomial:

 

$$5x - 3$$

 

For

$x = 0$

:

 

$$5(0) - 3 = 0 - 3 = -3$$

 

Answer: $-3$

(ii) x = -1

Solution:

5x - 3 = 5(-1) - 3
= -5 - 3
= -8
Answer: -8

(iii) x = 2

Solution:

5x - 3 = 5(2) - 3
= 10 - 3
= 7
Answer: 7

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

Solution:

7(0)² - 4(0) + 6
= 0 - 0 + 6
= 6

(ii) s = -3

Solution:

7(-3)² - 4(-3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81

(iii) s = 4

Solution:

7(4)² - 4(4) + 6
= 7(16) - 16 + 6
= 112 - 16 + 6
= 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.

Solution:

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 = 60
x = 15
Therefore:
Salil’s present age = 15 years
Mother’s present age = 3 × 15 = 45 years
Answer: 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.

Solution:

Let the two integers be 2x and 5x.
According to the question:
5x - 2x = 63
3x = 63
x = 21
Therefore:
First integer = 2 × 21 = 42
Second integer = 5 × 21 = 105
Answer: 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 of 88, how many coins does she have of each type?

Solution:

Let the number of five-rupee coins be x.
Then, the number of two-rupee coins = 3x.
Value of five-rupee coins = 5x
Value of two-rupee coins = 2 × 3x = 6x
According to the question:
5x + 6x = 88
11x = 88
x = 8
Therefore,
Number of five-rupee coins = 8
Number of two-rupee coins = 3 × 8 = 24
Answer: Ruby has 8 five-rupee coins and 24 two-rupee coins.

Question 6.
A farmer cuts a 300-foot 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 = 4x feet.
According to the question:
x + 4x = 300
5x = 300
x = 60
Therefore,
Shorter piece = 60 feet
Longer piece = 4 × 60 = 240 feet
Answer: The two pieces are 60 feet and 240 feet long.

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 x cm.
Then, the length = 2x + 3 cm.
Perimeter of a rectangle = 2 × (Length + Width)
According to the question:
2[(2x + 3) + x] = 24
2(3x + 3) = 24
6x + 6 = 24
6x = 18
x = 3
Therefore,
Width = 3 cm
Length = 2(3) + 3
= 6 + 3
= 9 cm
Answer: The dimensions of the rectangle are 9 cm × 3 cm.

Ex 2.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.3 Solutions

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.

Solution:

The student initially has ₹500 in her savings account.
She gets ₹150 every month as pocket money.
At the end of the 1st month:
₹500 + ₹150 = ₹650
At the end of the 2nd month:
₹500 + 2 × ₹150 = ₹800
At the end of the 3rd month:
₹500 + 3 × ₹150 = ₹950
Similarly, at the end of the nth month:
Amount = 500 + 150n
Therefore, the required linear expression is:
150n + 500
Answer: The amount in the nth month is ₹(150n + 500).

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.

Solution:

The rally starts with 120 members.
Each hour, 9 members leave the group.
After 1 hour:
120 - 9 = 111 members
After 2 hours:
120 - 2 × 9 = 102 members
After 3 hours:
120 - 3 × 9 = 93 members
Therefore, after n hours:
Number of members = 120 - 9n
Answer: 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.

Solution:

Length of the rectangle = 13 cm
Area of a rectangle = Length × Breadth
(i) When breadth = 12 cm:
Area = 13 × 12
= 156 cm²
(ii) When breadth = 10 cm:
Area = 13 × 10
= 130 cm²
(iii) When breadth = 8 cm:
Area = 13 × 8
= 104 cm²
If the breadth is represented by b cm, then:
Area = 13b
Answer: The linear expression representing the area of the rectangle is 13b.

Question 4.
Suppose the length of a rectangular box is 7 cm and the 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:

Length of the rectangular box = 7 cm
Breadth of the rectangular box = 11 cm
Volume of a rectangular box = Length × Breadth × Height
(i) When height = 5 cm:
Volume = 7 × 11 × 5
= 385 cm³
(ii) When height = 9 cm:
Volume = 7 × 11 × 9
= 693 cm³
(iii) When height = 13 cm:
Volume = 7 × 11 × 13
= 1001 cm³
If the height is represented by h cm, then:
Volume = 7 × 11 × h
= 77h
Answer: The linear expression representing the volume of the rectangular box is 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.

Solution:

Sarita has a book of 500 pages.
She reads 20 pages every day.
Pages read in 15 days:
20 × 15 = 300 pages
Pages left:
500 - 300 = 200 pages
If the number of days is represented by n, then:
Pages left = 500 - 20n
Answer: After 15 days, 200 pages will be left. The linear expression is 500 - 20n.

Ex 2.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.4 Solutions

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

Solution:

Initial height of the plant = 1.75 feet
Growth each month = 0.5 feet
Growth in 7 months:
0.5 × 7 = 3.5 feet
Height after 7 months:
1.75 + 3.5 = 5.25 feet
Answer: The height of the plant after 7 months will be 5.25 feet.

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

Solution:

The initial height of the plant is 1.75 feet, and it grows by 0.5 feet every month.

The height after t months is:

h = 1.75 + 0.5t

Answer: The height increases by 0.5 feet every month. The linear expression is h = 1.75 + 0.5t.

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

Solution:

The height h after t months can be expressed as: h(t)= 1.75 + 0.5;
This expression represents linear growth because the height increases by a constant amount (0.5 feet) for each unit increase in time (months).

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.

Solution:

Find the value of the phone after 3 years.
Initial value of the phone = ₹10,000
Decrease in value every year = ₹800
Decrease in 3 years:
₹800 × 3 = ₹2,400
Value after 3 years:
₹10,000 - ₹2,400 = ₹7,600
Answer: The value of the phone after 3 years is ₹7,600.

(ii) Make a table of values for varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.

Solution:

The initial value of the phone is ₹10,000, and it decreases by ₹800 every year.

The value decreases by ₹800 every year.

So, the linear pattern is:

v = 10,000 - 800t

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

Solution:

The initial value of the phone is ₹10,000, and its value decreases by ₹800 every year.
Therefore, the expression relating v and t is:
v = 10,000 - 800t
where:
v = value of the phone in rupees
t = time in years
This represents linear decay because the value decreases by the same amount, ₹800, every year.
Answer: v = 10,000 - 800t. It represents linear decay because the rate of decrease is constant.

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.

Solution:

Initial population = 750
Increase every year = 50 people
Increase in 6 years:
50 × 6 = 300
Population after 6 years:
750 + 300 = 1,050
Answer: The population of the village after 6 years will be 1,050.

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

Solution:

The initial population of the village is 750, and it increases by 50 people every year.

The population increases by 50 people every year.

The linear expression is:

P = 750 + 50t

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

Solution:

The initial population of the village is 750, and it increases by 50 people every year.
Therefore, the expression relating P and t is:
P = 750 + 50t
where:
P = population of the village
t = time in years
It represents linear growth because the population increases by the same amount, 50 people, every year.
Answer: P = 750 + 50t. It represents linear growth because the rate of increase is constant.

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 forx days. Explain why it represents linear decay.

Solution:

The initial prepaid balance is ₹600.
The balance decreases by ₹15 each day.
So, after x days:
b(x) = 600 - 15x
where:
b(x) = remaining balance after x days
x = number of days
It represents linear decay because the balance decreases by the same amount, ₹15, every day.
Answer: b(x) = 600 - 15x

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

Solution:

The balance runs out when:
b(x) = 0
Using:
600 - 15x = 0
15x = 600
x = 600 ÷ 15
x = 40
Answer: The balance will run out after 40 days.

(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:

Using the equation:

b(x) = 600 - 15x

we get the following values:

Thus, the balance decreases by ₹15 every day.

Ex 2.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.5 Solutions

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.

Solution:

Given:
y = ax + b
When x = 10, y = 400:
400 = 10a + b ...(1)
When x = 14, y = 500:
500 = 14a + b ...(2)
Subtract equation (1) from equation (2):
500 - 400 = 14a - 10a
100 = 4a
a = 25
Substitute a = 25 in equation (1):
400 = 10(25) + b
400 = 250 + b
b = 150
Answer:
a = 25
b = 150
Therefore, the relation is:
y = 25x + 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 billy 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.

Solution:

Given:
y = ax + b
When x = 10, y = 800:
800 = 10a + b ...(1)
When x = 15, y = 1100:
1100 = 15a + b ...(2)
Subtract equation (1) from equation (2):
1100 - 800 = 15a - 10a
300 = 5a
a = 60
Substitute a = 60 in equation (1):
800 = 10(60) + b
800 = 600 + b
b = 200
Answer:
a = 60
b = 200
Therefore, the relation is:
y = 60x + 200

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. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)

Solution:

Given:
°C = a°F + b
When °C = 0 and °F = 32:
0 = 32a + b ...(1)
When °C = 100 and °F = 212:
100 = 212a + b ...(2)
Subtract equation (1) from equation (2):
100 = 180a
a = 100/180
a = 5/9
Substitute a = 5/9 in equation (1):
0 = 32(5/9) + b
b = -160/9
Therefore:
a = 5/9
b = -160/9
Hence, the linear relationship between Celsius and Fahrenheit is:
°C = (5/9)°F - 160/9
or,
°C = (5/9)(°F - 32)
Answer: a = 5/9 and b = -160/9.

Ex 2.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.6 Solutions

Question 1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘6’.
(i) y = 4x,y = 2x,y = x

Solution:

y = 4x, y = 2x, y = x
All lines are of the form y = ax (b = 0).
Observation:
All lines pass through the origin (0, 0).
The value of ‘a’ (slope) determines steepness.
Larger ‘a’ ⇒ steeper line.

(ii) y = – 6x, y = – 3x, y = – x

Solution:

y = – 6x, y = -3x, y =-x
All lines are of the form y = -ax (b = 0).
Observation:
All lines pass through the origin.
Negative ‘a’ means lines slope downward.
Larger magnitude of‘a’ ⇒ steeper downward slope.

(iii) y = 5x, y = -5x

Solution:

y = 5x, y = -5x
All lines are of the form y = ax (b = 0).
Observation: Here, both lines pass through the origin because the y-intercept is 0. y = 5x slopes upward, y = -5x slopes downward; same magnitude of a, same steepness but in opposite directions.

(iv) y = 3x – 1, y = 3x, y = 3x + 1

Solution:

y = 3x – 1, y = 3x, y = 3x + 1
All lines have same slope (a = 3).
Observation: Lines are parallel as slopes are the same (3).
Different values of ‘b’ shift the line. b = -1 line below origin, b = 0 passes through origin b = 1 line above origin.

(v) y = -2x – 3, y = -2x, y = 2x + 3

Solution:

For the lines:

y = -2x - 3
y = -2x
y = 2x + 3

Use the following values to plot the graphs:

For a linear equation:

y = ax + b

For y = -2x - 3:
a = -2, b = -3

For y = -2x:
a = -2, b = 0

For y = 2x + 3:
a = 2, b = 3

Role of a: The value of a determines the slope and direction of the line. When a is negative, the line slopes downward from left to right. When a is positive, the line slopes upward.

Role of b: The value of b determines where the line cuts the y-axis. Thus, the three lines cut the y-axis at -3, 0, and 3 respectively.

Ganita Manjari Class 9 Maths Chapter 2 End of Chapter Exercise Solutions

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

Solution:

A degree 3 polynomial is of the form:
ax³ + bx² + cx + d
The coefficient of the x term must be -7.
So, one possible polynomial is:
P(x) = x³ + 2x² - 7x + 1
Answer: P(x) = x³ + 2x² - 7x + 1
(Answer may vary.)

Question 2.
Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² – 3x + 7 if x = 1

Solution:

Given polynomial:
5x² - 3x + 7
For x = 1:
5(1)² - 3(1) + 7
= 5 - 3 + 7
= 9
Answer: 9

(ii) 4t³ – t² + 6 if t = a

Solution:

Given polynomial:
4t³ - t² + 6
For t = a:
4(a)³ - (a)² + 6
= 4a³ - a² + 6
Answer: 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. 12

Solution:

Let the number be x.
According to the question:
(5/2)x + 2/3 = -7/12
Subtract 2/3 from both sides:
(5/2)x = -7/12 - 2/3
(5/2)x = -7/12 - 8/12
(5/2)x = -15/12
(5/2)x = -5/4
Multiply both sides by 2/5:
x = (-5/4) × (2/5)
x = -1/2
Answer: 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?

Solution:

Let the smaller number be x.
Then, the larger number = 5x.
After adding 21 to both numbers:
Smaller number = x + 21
Larger number = 5x + 21
According to the question:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
5x - 2x = 42 - 21
3x = 21
x = 7
Therefore:
Smaller number = 7
Larger number = 5 × 7 = 35
Answer: The two 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.

Solution:

Initial amount = ₹800
Amount saved every month = ₹250
(i) After 6 months:
Amount saved in 6 months:
₹250 × 6 = ₹1,500
Total amount:
₹800 + ₹1,500 = ₹2,300
Answer: After 6 months, the amount is ₹2,300.
(ii) After 2 years:
2 years = 24 months
Amount saved in 24 months:
₹250 × 24 = ₹6,000
Total amount:
₹800 + ₹6,000 = ₹6,800
Answer: After 2 years, the amount is ₹6,800.
If the number of months is represented by n, then the linear pattern is:
A = 800 + 250n
where A represents the total amount after n months.

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 be x and the units digit be y.
Since the digits differ by 3:
x - y = 3 ...(1)
The original number = 10x + y
The number obtained by interchanging the digits = 10y + x
According to the question:
(10x + y) + (10y + x) = 143
11x + 11y = 143
x + y = 13 ...(2)
Adding equations (1) and (2):
2x = 16
x = 8
Then:
y = 13 - 8 = 5
So, the original number is 85 and the number obtained by interchanging the digits is 58.
Answer: The two numbers are 85 and 58.

Question 7.
Draw the graph of the following equations, and identify their slopes andy-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.

Solution:

(i) y = -3x + 4

Solution:

For the equation:

y = -3x + 4

Comparing it with the standard form:

y = ax + b

we get:

a = -3
b = 4

So:

Slope = -3

y-intercept = 4

The line cuts the y-axis when x = 0.

For x = 0:

y = -3(0) + 4
= 4

Therefore, the line cuts the y-axis at:

(0, 4)

A few points for plotting are:

(ii) 2y = 4x + 7

Solution:

For the equation:

2y = 4x + 7

Divide both sides by 2:

y = 2x + 7/2

Comparing with the standard form:

y = ax + b

we get:

a = 2
b = 7/2

So:

Slope = 2

y-intercept = 7/2 or 3.5

The line cuts the y-axis when x = 0.

For x = 0:

y = 2(0) + 7/2
= 7/2

Therefore, the line cuts the y-axis at:

(0, 7/2)

A few points for plotting are:

(iii) 5y = 6x – 10

Solution:

For the equation:

5y = 6x - 10

Divide both sides by 5:

y = (6/5)x - 2

Comparing with the standard form:

y = ax + b

we get:

a = 6/5
b = -2

So:

Slope = 6/5

y-intercept = -2

The line cuts the y-axis when x = 0.

For x = 0:

y = (6/5)(0) - 2
= -2

Therefore, the line cuts the y-axis at:

(0, -2)

A few points for plotting are:

(iv) 3y = 6x – 11

Solution:

For the equation:

3y = 6x - 11

Divide both sides by 3:

y = 2x - 11/3

Comparing with the standard form:

y = ax + b

we get:

a = 2
b = -11/3

So:

Slope = 2

y-intercept = -11/3

The line cuts the y-axis when x = 0.

For x = 0:

y = 2(0) - 11/3
= -11/3

Therefore, the line cuts the y-axis at:

(0, -11/3)

A few points for plotting are:

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.

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

Solution:

Given:
y = (9/5)(x - 273) + 32
For x = 313 K:
y = (9/5)(313 - 273) + 32
= (9/5)(40) + 32
= 72 + 32
= 104
Answer: The temperature of the liquid is 104°F.

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

Solution:

Given:
y = (9/5)(x - 273) + 32
Here, y = 158°F.
So:
158 = (9/5)(x - 273) + 32
158 - 32 = (9/5)(x - 273)
126 = (9/5)(x - 273)
126 × 5/9 = x - 273
70 = x - 273
x = 343
Answer: 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.

Solution:

Work done is given by:

Work = Force × Distance

Let:

w = work done
d = distance travelled

Since the constant force is 3 units:

w = 3d

So, the required linear equation is:

w = 3d

To draw the graph, take d on the x-axis and w on the y-axis.

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).

Solution:

Let the linear polynomial be:
p(x) = ax + b
The graph passes through (1, 5) and (3, 11).
So:
For x = 1, p(x) = 5:
a + b = 5 ...(1)
For x = 3, p(x) = 11:
3a + b = 11 ...(2)
Subtract equation (1) from equation (2):
3a + b - (a + b) = 11 - 5
2a = 6
a = 3
Substitute a = 3 in equation (1):
3 + b = 5
b = 2
Therefore:
p(x) = 3x + 2
Answer: p(x) = 3x + 2

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

Solution:

From part (i):
p(x) = 3x + 2
To find where the graph cuts the y-axis, put x = 0:
p(0) = 3(0) + 2
= 2
So, the graph cuts the y-axis at:
(0, 2)
To find where the graph cuts the x-axis, put p(x) = 0:
3x + 2 = 0
3x = -2
x = -2/3
So, the graph cuts the x-axis at:
(-2/3, 0)

Answer:
x-axis: (-2/3, 0)
y-axis: (0, 2)

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

Solution:

From part (i), the polynomial is:

p(x) = 3x + 2

To draw the graph, take a few values of x:

Question 11.
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) m = 5.

Solution:

here appears to be a typo in the question. Condition (i) should be:

p(0) = 5

Given:

p(x) = ax + b

Substitute x = 0:

p(0) = a(0) + b

5 = b

Therefore:

b = 5

Hence,

p(x) = ax + 5

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

Solution:

Given:
p(x) = ax + b
q(x) = cx + d
From part (i):
b = 5
Therefore:
p(x) = ax + 5
Now,
p(x) - q(x) = ax + 5 - (cx + d)
= (a - c)x + (5 - d)
Since the graph of p(x) - q(x) cuts the x-axis at (3, 0), put x = 3 and p(x) - q(x) = 0:
3(a - c) + 5 - d = 0
Therefore:
3a - 3c + 5 - d = 0
Answer: 3(a - c) + 5 - d = 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).

Solution:

Given:
p(x) = ax + b
q(x) = cx + d
From part (i):
b = 5
So:
p(x) = ax + 5
From part (iii):
p(x) + q(x) = 6x + 4
Therefore:
(a + c)x + (5 + d) = 6x + 4
Comparing coefficients:
a + c = 6 ...(1)
5 + d = 4
d = -1
So:
q(x) = cx - 1
From part (ii), p(x) - q(x) cuts the x-axis at (3, 0).
Therefore:
p(3) - q(3) = 0
3a + 5 - (3c - 1) = 0
3a - 3c + 6 = 0
a - c = -2 ...(2)
Now, solving equations (1) and (2):
a + c = 6
a - c = -2
Adding both equations:
2a = 4
a = 2
Therefore:
c = 4
Hence,
p(x) = 2x + 5
q(x) = 4x - 1
Answer: p(x) = 2x + 5 and 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.

Solution:

Each hexagon has 6 sides. When a new hexagon is added, it shares one side with the previous hexagon. Therefore, each new hexagon requires only 5 additional matchsticks.

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

Solution:

Stage 4 will contain 4 hexagons.
Number of matchsticks = 16 + 5 = 21
Stage 5 will contain 5 hexagons.
Number of matchsticks = 21 + 5 = 26
Answer:
Stage 4 = 21 matchsticks
Stage 5 = 26 matchsticks

(ii) Complete the following table.

Solution:

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

(iii) Find a rule to determine the number of matchsticks required for the «,h stage.

Solution:

Rule for the nth stage
At Stage 1, 6 matchsticks are required.
Each new stage requires 5 additional matchsticks.
Therefore:
Number of matchsticks = 6 + 5(n - 1)
= 6 + 5n - 5
= 5n + 1
Answer: The number of matchsticks required for the nth stage is 5n + 1.

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

Solution:

Matchsticks required for the 15th stage
Using the rule:
5n + 1
For n = 15:
5(15) + 1
= 75 + 1
= 76
Answer: 76 matchsticks are required for the 15th stage.

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

Solution:

Can 200 matchsticks form a stage in this pattern?
For a stage to have 200 matchsticks:
5n + 1 = 200
5n = 199
n = 199/5
n = 39.8
Since the stage number must be a whole number, 39.8 cannot represent a stage.
Answer: No, 200 matchsticks cannot form a complete stage in this pattern.

Question 13.
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:

Solution:

Given:
p(x) = ax + b
q(x) = cx + d

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

Solution:

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

Since (2, 3) lies on p(x):

3 = 2a + b ...(1)

Since (6, 11) lies on p(x):

11 = 6a + b ...(2)

Subtract equation (1) from equation (2):

11 - 3 = 6a - 2a

8 = 4a

a = 2

Substitute a = 2 in equation (1):

3 = 2(2) + b

3 = 4 + b

b = -1

Therefore:

p(x) = 2x - 1

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

Solution:

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

Since:

q(x) = cx + d

Substitute x = 4 and q(x) = -1:

-1 = 4c + d

Therefore:

4c + d = -1

or,

d = -1 - 4c

So, from this condition alone, q(x) cannot be uniquely determined. Another condition is required to find the exact values of c and d.

Answer:
p(x) = 2x - 1
For q(x): 4c + d = -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.

Solution:

From part (i):
p(x) = 2x - 1
Since q(x) is parallel to p(x), both lines have the same slope.
Therefore:
c = 2
So:
q(x) = 2x + d
The graph of q(x) passes through (4, -1).
Substitute x = 4 and q(x) = -1:
-1 = 2(4) + d
-1 = 8 + d
d = -9
Therefore:
q(x) = 2x - 9
Now, find where each line cuts the x-axis.
For p(x) = 2x - 1:
Set p(x) = 0:
2x - 1 = 0
2x = 1
x = 1/2
So, p(x) cuts the x-axis at:
(1/2, 0)
For q(x) = 2x - 9:
Set q(x) = 0:
2x - 9 = 0
2x = 9
x = 9/2
So, q(x) cuts the x-axis at:
(9/2, 0)
Answer:
p(x) = 2x - 1
q(x) = 2x - 9
p(x) meets the x-axis at (1/2, 0).
q(x) meets the x-axis at (9/2, 0).

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

Solution:

Given:
f(x) = ax + a, where a > 0
Factor out a:
f(x) = a(x + 1)
For every such function, when x = -1:
f(-1) = a(-1 + 1)
= 0
So, all these linear functions pass through the common point:
(-1, 0)
Also, since a > 0, all the lines have a positive slope and rise from left to right.
Answer: All linear functions of the form f(x) = ax + a, a > 0, pass through the point (-1, 0) and have positive slopes.

Related Resource:

• Exercise 2.1 – Degree, Coefficients and Constants
• Exercise 2.2 – Polynomial Values and Linear Equations
• Exercise 2.3 – Linear Patterns
• Exercise 2.4 – Linear Growth and Linear Decay
• Exercise 2.5 – Linear Relationships
• Exercise 2.6 – Graphs of Linear Relationships
• End-of-Chapter Exercises – Introduction to Linear Polynomials