Mastering Chapter 2 of the Class 9 Ganita Manjari syllabus is essential for building a strong foundation in algebra. Our comprehensive guide provides detailed NCERT Solutions Ch 2 Introduction To Linear Polynomials Class 9 to help you excel.

This chapter simplifies the world of variables, constants, and degrees of polynomials. Whether you are finding the value of a polynomial or solving real-world linear growth problems, these step-by-step solutions make learning intuitive.

Exercise 2.1: Understand degrees and coefficients.
Exercise 2.2: Solve linear and quadratic polynomial values.
Exercise 2.3 & 2.4: Explore linear patterns and decay.
Exercise 2.5 & 2.6: Master linear equations and their graphs.

Achieve clarity and confidence in your mathematics exams with our expert-verified answers. Use these NCERT Solutions Ch 2 Introduction To Linear Polynomials Class 9 to study smarter and score better.

Table of Contents

Exercise Set 2.1

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

Solution to Question 1

The degree of a polynomial is the highest power of the variable present in the expression.

(i) The expression is 2x² – 5x + 3. The highest power of x is 2. Therefore, Degree = 2.
(ii) The expression is y³ + 2y – 1. The highest power of y is 3. Therefore, Degree = 3.
(iii) The expression is –9. A constant non-zero number can be written as -9x⁰. Therefore, Degree = 0.
(iv) The expression is 4z – 3. The highest power of z is 1. Therefore, Degree = 1.

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

Solution to Question 2

Here are examples of polynomials of degrees 1, 2, and 3. (Note: Many other valid examples exist)

Degree 1 (Linear Polynomial): 3x + 5
(The highest power of the variable x is 1)
Degree 2 (Quadratic Polynomial): 4y² – 2y + 7
(The highest power of the variable y is 2)
Degree 3 (Cubic Polynomial): 2z³ + z² – 8z + 1
(The highest power of the variable z is 3)

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

Solution to Question 3

The given polynomial is: x⁴ – 3x³ + 6x² – 2x + 7

The coefficient of a term is the numerical multiplier associated with the variable.

The term with x² is +6x². Therefore, the coefficient of x² is 6.
The term with x³ is -3x³. Therefore, the coefficient of x³ is -3.

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

Solution to Question 4

The given polynomial is: 4z³ + 5z² – 11

In this polynomial, the term containing the variable z (with a power of 1) is completely missing.

This means it can be mathematically represented as + 0z. Therefore, the coefficient of z is exactly 0.

What is the constant term of the polynomial 9×3 + 5×2 – 8x –10?
Recall that polynomials of degree 1 are called linear polynomials.
In this chapter, we shall study linear polynomials.

Solution to Question 5

The given polynomial is: 9x³ + 5x² – 8x – 10

The constant term in a polynomial is the numerical term that does not have any variable attached to it (or technically, it’s multiplied by the variable to the power of 0).

Looking at the expression, the term without a variable is -10. Therefore, the constant term is -10.

Also Read | Orienting Yourself: The Use of Coordinates NCERT Solutions Ch 1 Class 9

Exercise Set 2.2

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

Solution to Question 1

We substitute the given values of x into the linear polynomial 5x – 3:

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

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

Solution to Question 2

We substitute the given values of s into the quadratic polynomial 7s² – 4s + 6:

(i) For s = 0:
7(0)² – 4(0) + 6 = 0 – 0 + 6 = 6
(ii) For s = -3:
7(-3)² – 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81
(iii) For s = 4:
7(4)² – 4(4) + 6 = 7(16) – 16 + 6 = 112 – 16 + 6 = 102

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 to Question 3

Step 1: Assign variables

Let Salil’s present age be x years.
Therefore, his mother’s present age is 3x years.

Step 2: Determine ages after 5 years

Salil’s age will be x + 5.
Mother’s age will be 3x + 5.

Step 3: Form and solve the equation

(x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15

Final Answer:

Salil’s present age is 15 years and his mother’s present age is (3 × 15) = 45 years.

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

Solution to Question 4

Step 1: Assign variables using the given ratio

Since the ratio of the two positive integers is 2:5, let the smaller integer be 2x and the larger integer be 5x.

Step 2: Form and solve the equation

We are given that their difference is 63:

5x – 2x = 63
3x = 63
x = 21

Final Answer:

Smaller integer = 2 × 21 = 42
Larger integer = 5 × 21 = 105

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

Solution to Question 5

Step 1: Define the quantities

Let the number of ₹5 coins be x.
Since Ruby has 3 times as many ₹2 coins, the number of ₹2 coins is 3x.

Step 2: Calculate total value and form equation

The total value of the coins is ₹88:

(Value of ₹5 coins) + (Value of ₹2 coins) = 88
5(x) + 2(3x) = 88
5x + 6x = 88
11x = 88
x = 8

Final Answer:

Ruby has 8 five-rupee coins and (3 × 8) = 24 two-rupee coins.

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 to Question 6

Step 1: Define the lengths

Let the length of the shorter piece be x feet.
The longer piece is four times as long, so its length is 4x feet.

Step 2: Form and solve the equation

The total length of the fence is 300 feet:

x + 4x = 300
5x = 300
x = 60

Final Answer:

The length of the shorter piece is 60 feet and the length of the longer piece is (4 × 60) = 240 feet.

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 to Question 7

Step 1: Define the dimensions

Let the width of the rectangle be w cm.
The length is three more than twice the width, so length = 2w + 3 cm.

Step 2: Form and solve the equation using perimeter

The formula for the perimeter of a rectangle is P = 2(length + width):

2 × ((2w + 3) + w) = 24
2 × (3w + 3) = 24
6w + 6 = 24
6w = 18
w = 3

Final Answer:

The width of the rectangle is 3 cm and the length is (2(3) + 3) = 9 cm.

Exercise Set 2.3

A student has ₹500 in her savings 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 to Question 1

Step 1: Calculate the amount month by month

Initial Amount (Month 1): ₹500
End of Month 2: 500 + 150 = ₹650
End of Month 3: 650 + 150 = ₹800
End of Month 4: 800 + 150 = ₹950

Step 2: Form the linear expression

The pattern shows an initial amount of 500, with an addition of 150 for every month after the first one.

Amount in nth month = 500 + 150(n – 1)
= 500 + 150n – 150
= 150n + 350

Final Answer:

The linear expression representing the amount she will have in the nth month is 150n + 350.

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 to Question 2

Step 1: Calculate the members remaining hour by hour

Initial Members (Start): 120
After 1 hour: 120 – 9 = 111
After 2 hours: 111 – 9 = 102
After 3 hours: 102 – 9 = 93

Step 2: Form the linear expression

The pattern shows a constant subtraction of 9 members for every hour that passes.

Members after n hours = Initial – 9 × (hours)
= 120 – 9n

Final Answer:

The linear expression representing the number of members at the end of the nth hour is 120 – 9n.

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 to Question 3

Step 1: Calculate the area for the given breadths

The formula for the area of a rectangle is Area = Length × Breadth. Here, the length is fixed at 13 cm.

(i) For breadth = 12 cm: Area = 13 × 12 = 156 cm²
(ii) For breadth = 10 cm: Area = 13 × 10 = 130 cm²
(iii) For breadth = 8 cm: Area = 13 × 8 = 104 cm²

Step 2: Identify the linear pattern

Notice the sequence of areas: 156, 130, 104…
The difference between consecutive areas is -26 cm² (which corresponds to a decrease of 2 cm in breadth, resulting in 13 × 2 = 26 less area). The linear pattern representing the area for any breadth b is simply 13b.

Final Answer:

The areas are 156 cm², 130 cm², and 104 cm². The linear pattern representing the area is Area = 13b, where b is the breadth.

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 to Question 4

Step 1: Calculate the volume for the given heights

The formula for the volume of a rectangular box is Volume = Length × Breadth × Height. Here, Length = 7 cm and Breadth = 11 cm. So, the base area is 7 × 11 = 77 cm².

(i) For height = 5 cm: Volume = 77 × 5 = 385 cm³
(ii) For height = 9 cm: Volume = 77 × 9 = 693 cm³
(iii) For height = 13 cm: Volume = 77 × 13 = 1001 cm³

Step 2: Identify the linear pattern

The sequence of volumes is 385, 693, 1001…
The difference between consecutive volumes is 308 cm³ (which corresponds to an increase of 4 cm in height, resulting in 77 × 4 = 308 more volume). The linear pattern representing the volume for any height h is 77h.

Final Answer:

The volumes are 385 cm³, 693 cm³, and 1001 cm³. The linear pattern representing the volume is Volume = 77h, where h is the height.

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 to Question 5

Step 1: Calculate the pages left after 15 days

The book has 500 pages. Sarita reads 20 pages every day. In 15 days, she will have read:

Pages read = 20 × 15 = 300 pages

The number of pages left will be:

Pages left = 500 – 300 = 200 pages

Step 2: Form the linear pattern

The number of pages left decreases by 20 every single day. If d is the number of days, the number of pages left can be modeled by the linear expression: 500 – 20d.

Final Answer:

There will be 200 pages left after 15 days. The linear pattern is P = 500 – 20d (where P is pages left and d is the number of days).

Exercise Set 2.4

Suppose a plant has a height of 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 to Question 1

(i) Height after 7 months:

Height = 1.75 + (0.5 × 7)
= 1.75 + 3.5
= 5.25 feet

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

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

(iii) Expression relating h and t:

h = 1.75 + 0.5t

Reason for linear growth: This represents linear growth because the height (h) increases by a strictly constant amount (0.5 feet) over equal, constant intervals of time (each month).

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 to Question 2

(i) Value of the phone after 3 years:

Value = 10,000 – (800 × 3)
= 10,000 – 2400
= ₹7,600

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

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

(iii) Expression relating v and t:

v = 10000 – 800t

Reason for linear decay: This represents linear decay because the value (v) decreases by a strictly constant amount (₹800) over equal, constant intervals of time (each year).

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 to Question 3

(i) Population of the village after 6 years:

Population = 750 + (50 × 6)
= 750 + 300
= 1,050 people

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

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 relating P and t:

P = 750 + 50t

Reason for linear growth: This represents linear growth because the population (P) increases by a constant, non-changing amount (50 people) over equal intervals of time (each year).

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 to Question 4

(i) Equation modeling the balance b(x):

The initial balance is ₹600 and it decreases by ₹15 every day (x).

b(x) = 600 – 15x

Reason for linear decay: This represents linear decay because the balance strictly decreases at a constant rate (₹15) per unit of time (each day).

(ii) Days until balance runs out:

Set the remaining balance to zero and solve for x:

0 = 600 – 15x
15x = 600
x = 600 / 15
x = 40 days

(iii) Table of values (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

Exercise Set 2.5

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 :

A gym charges a fixed monthly fee and an additional hourly rate 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 use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Solution:

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. — A gym charges a fixed monthly fee and an additional hourly rate 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.

Consider the relationship between the 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:

Exercise Set 2.6

Page 36

1. Draw the graphs of the following sets of lines. In each case, reflect
on the role of ‘a ’ and ‘b.’

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

(i) y = 4x, y = 2x, y = x

Solution:

(ii) y = – 6x, y = – 3x, y = – x

Solution:

(iii) y = 5x, y = –5x

Solution:

(iv) y = 3x – 1, y = 3x, y = 3x + 1

Solution:

(v) y = –2x – 3, y = –2x, y = 2x + 3

End-of-Chapter Exercises

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

Question 1

Solution:

A polynomial of degree 3 has the highest power of the variable as 3. Its general form is ax³ + bx² + cx + d (where a ≠ 0).

The coefficient of the x² term is given as −7. So, we can set b = −7.

We can choose any real numbers for a (non-zero), c, and d. Let’s choose a = 1, c = 0, and d = 0 for simplicity.

An example of such a polynomial is: x³ − 7x²

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

Question 2

Solution:

Part (i)

Polynomial: 5x² − 3x + 7

Substitute x = 1 into the polynomial:

= 5(1)² − 3(1) + 7

= 5 − 3 + 7

= 9

Part (ii)

Polynomial: 4t³ − t² + 6

Substitute t = a into the polynomial:

= 4(a)³ − (a)² + 6

= 4a³ − a² + 6

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

Question 3

Solution:

Let the unknown number be x.

Multiply the number by 5/2: (5/2)x

Add 2/3: (5/2)x + 2/3

This equals −7/12

(5/2)x + 2/3 = −7/12

Subtract 2/3 from both sides:

(5/2)x = −7/12 − 2/3

Convert to common denominator (12):

(5/2)x = −7/12 − 8/12

(5/2)x = −15/12 = −5/4

Multiply by 2/5:

x = (−5/4) × (2/5)

x = −2/4 = −1/2

The number is: −1/2

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?

Question 4

Solution:

Let the smaller positive number be x.

Then the other positive number is 5x.

If 21 is added to both, the new numbers become (x + 21) and (5x + 21).

Since we are dealing with positive numbers, 5x + 21 will be the larger number. We are given that one of the new numbers becomes twice the other new number. Therefore, the larger new number must be twice the smaller new number:

5x + 21 = 2(x + 21)

Expand the right side:

5x + 21 = 2x + 42

Subtract 2x from both sides:

3x + 21 = 42

Subtract 21 from both sides:

3x = 21

Divide by 3:

x = 7

The smaller number is 7. The larger number is 5x = 5(7) = 35.

The numbers are 7 and 35.

If you have 800 and you save Rs 250 every month, find the amount you have after (i) 6 months, (ii) 2 years. Express this as
a linear pattern.

Question 5

Solution:

Initial amount = ₹800.

Monthly savings = ₹250.

Let y be the total amount after x months. The linear pattern representing the total amount is:

y = 250x + 800

Part (i) Amount after 6 months

Substitute x = 6:

y = 250(6) + 800

y = 1500 + 800 = 2300

The amount after 6 months is ₹2300.

Part (ii) Amount after 2 years

Convert 2 years into months: x = 24 months.

Substitute x = 24:

y = 250(24) + 800

y = 6000 + 800 = 6800

The amount after 2 years is ₹6800.

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.

Question 6

Solution:

Let the tens digit be t and the units digit be u.

The original number is 10t + u.

When the digits are interchanged, the resulting number is 10u + t.

We are given that the sum of the original number and the new number is 143:

(10t + u) + (10u + t) = 143

11t + 11u = 143

Divide the entire equation by 11:

t + u = 13

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

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

System of equations: t + u = 13 and t − u = 3.

Add the two equations:

(t + u) + (t − u) = 13 + 3 ⇒ 2t = 16 ⇒ t = 8

Substitute t = 8 into t + u = 13:

8 + u = 13 ⇒ u = 5

The number for Case 1 is 85.

Case 2: The units digit is larger (u − t = 3)

System of equations: t + u = 13 and u − t = 3.

Add the two equations:

(t + u) + (u − t) = 13 + 3 ⇒ 2u = 16 ⇒ u = 8

Substitute u = 8 into t + u = 13:

t + 8 = 13 ⇒ t = 5

The number for Case 2 is 58.

Both the possible numbers are: 58 and 85.

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 = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?

Solution:

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 = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel? — 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 = –3x + 4(ii) 2y = 4x + 7(iii) 5y = 6x – 10(iv) 3y = 6x – 11
Are any of the lines parallel?

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 are given by the linear
equation y = 95
(x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the The
temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in
Kelvin.

Question 8

Solution:

The linear equation relating Kelvin (x) and Fahrenheit (y) is:

y = ⁹⁄₅ (x − 273) + 32

Part (i) Find Fahrenheit when Kelvin is 313 K

Substitute x = 313 into the equation:

y = ⁹⁄₅ (313 − 273) + 32

y = ⁹⁄₅ (40) + 32

y = 9 × 8 + 32

y = 72 + 32 = 104

The temperature is 104 °F.

Part (ii) Find Kelvin when Fahrenheit is 158 °F

Substitute y = 158 into the equation:

158 = ⁹⁄₅ (x − 273) + 32

Subtract 32 from both sides:

126 = ⁹⁄₅ (x − 273)

Multiply both sides by ⁵⁄₉:

126 × (⁵⁄₉) = x − 273

14 × 5 = x − 273

70 = x − 273

Add 273 to both sides:

x = 70 + 273 = 343

The temperature is 343 K.

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:

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. — 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.

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.

Solution:

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.

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).

Question 11

Solution:

Given two linear polynomials: p(x) = ax + b and q(x) = cx + d.

From condition (i): p(0) = 5

p(0) = a(0) + b = 5 ⇒ b = 5

So, p(x) = ax + 5.

From condition (iii): p(x) + q(x) = 6x + 4

(ax + 5) + (cx + d) = 6x + 4

(a + c)x + (5 + d) = 6x + 4

By comparing the coefficients on both sides:

a + c = 6

5 + d = 4 ⇒ d = −1

So, q(x) = cx − 1.

From condition (ii): p(x) − q(x) cuts the x-axis at (3, 0)

This means when x = 3, the value is 0. That is, p(3) − q(3) = 0, or p(3) = q(3).

Calculate p(3) and q(3):

p(3) = a(3) + 5 = 3a + 5

q(3) = c(3) − 1 = 3c − 1

Set them equal:

3a + 5 = 3c − 1

3a − 3c = −6

Divide by 3:

a − c = −2

Solving for a and c:

We have a system of linear equations:

a + c = 6

a − c = −2

Add the two equations:

2a = 4 ⇒ a = 2

Substitute a = 2 into a + c = 6:

2 + c = 6 ⇒ c = 4

The polynomials are:
p(x) = 2x + 5
q(x) = 4x − 1

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.

(i) Draw the next two stages of the pattern. How many
Will matchsticks be required at these stages?
(ii) Complete the following table.

Stage Number	1	2	4	5	….	n
Number of matchsticks

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

Solutions:

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.(i) Draw the next two stages of the pattern. How many Will matchsticks be required at these stages? (ii) Complete the following table.(iii) Find a rule to determine the number of matchsticks requ — 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. (i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages? (ii) Complete the following table. (iii) Find a rule to determine the number of matchsticks required

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

Solutions:

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

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

Question 14

Solution:

The function is given by f(x) = ax + a, where a > 0.

Let’s find the x-intercept of this function by setting f(x) = 0:

ax + a = 0

a(x + 1) = 0

Since we are given that a > 0, we can divide both sides by a:

x + 1 = 0

x = −1

This shows that the x-intercept for all such functions is the point (−1, 0), regardless of the specific value of a.

What they have in common:

All the lines pass through the exact same point (−1, 0) on the coordinate plane.
Since a > 0, all of these lines have a positive slope, meaning they are all strictly increasing.
The y-intercept and the slope of each line are always numerically equal to the same value (a).