NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 02 Introduction to Linear Polynomials

 

Exercise Set 2.1

 

Question 1. Find the degrees of the following polynomials:
(i) 2x² - 5x + 3 (ii) y³ + 2y - 1 (iii) -9 (iv) 4z - 3
Answer: The degree of a polynomial is the highest power of the variable present in it.
(i) 2x² - 5x + 3 The highest power of x is 2, so degree = 2
(ii) y³ + 2y - 1 The highest power of y is 3, so degree = 3
(iii) -9 This is a constant polynomial with no variable, so degree = 0
(iv) 4z - 3 The highest power of z is 1, so degree = 1
In simple words: Look at each term and find which variable has the biggest exponent (power). That biggest power number is the degree.

Exam Tip: Always identify the term with the highest exponent - constant terms (numbers with no variable) have degree 0. This is a straightforward concept but easy to overlook if you rush.

 

Question 2. Write polynomials of degrees 1, 2 and 3.
Answer: A polynomial with degree 1 (linear polynomial): Example - 2x + 5
A polynomial with degree 2 (quadratic polynomial): Example - x² + 3x + 1
A polynomial with degree 3 (cubic polynomial): Example - x³ - 2x² + x + 4
In simple words: Degree 1 has x to the first power, degree 2 has x squared as the highest power, and degree 3 has x cubed as the highest power.

Exam Tip: Write any polynomial with the correct highest power - there are infinite correct answers. Make sure your highest exponent matches the degree asked for.

 

Question 3. What are the coefficients of x² and x³ in the polynomial x⁴ - 3x³ + 6x² - 2x + 7?
Answer: Looking at the polynomial x⁴ - 3x³ + 6x² - 2x + 7, we identify the number multiplying each variable.
The coefficient of x³ is -3
The coefficient of x² is 6
In simple words: The coefficient is the number in front of the variable. For x³, it is -3. For x², it is 6.

Exam Tip: Be careful with the signs - negative coefficients are still coefficients. Always state the number that multiplies the variable, including its sign.

 

Question 4. What is the coefficient of z in the polynomial 4z³ + 5z² - 11?
Answer: In the polynomial 4z³ + 5z² - 11, we look for any term containing z to the first power. There is no z¹ term (no term with just z) in this polynomial. Therefore, the coefficient of z is 0.
In simple words: If a variable does not appear in the polynomial at all, its coefficient is 0.

Exam Tip: When a variable is absent, its coefficient is zero, not undefined. This applies to any missing power in a polynomial.

 

Question 5. What is the constant term of the polynomial 9x³ + 5x² - 8x - 10?
Answer: The constant term is the number that stands alone, without any variable attached. In the polynomial 9x³ + 5x² - 8x - 10, the constant term is -10.
In simple words: The constant term is just a plain number with no variable - in this case, -10.

Exam Tip: Constant terms have degree 0. Look at the end of the polynomial expression - it usually appears at the end but check the entire expression carefully.

 

Exercise Set 2.2

 

Question 1. Find the value of the linear polynomial 5x - 3 if:
(i) x = 0 (ii) x = -1 (iii) x = 2
Answer:
(i) When x = 0: 5(0) - 3 = -3
(ii) When x = -1: 5(-1) - 3 = -5 - 3 = -8
(iii) When x = 2: 5(2) - 3 = 10 - 3 = 7
In simple words: Put the number in place of x, then calculate following order of operations - multiply first, then subtract.

Exam Tip: Substitute the exact value for x and follow the operations in order. Be careful with negative numbers - use brackets to avoid sign errors.

 

Question 2. Find the value of the quadratic polynomial 7s² - 4s + 6 if:
(i) s = 0 (ii) s = -3 (iii) s = 4
Answer:
(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
In simple words: Square the number first, multiply it by its coefficient, then handle the rest term by term.

Exam Tip: With quadratic polynomials, always square the variable value before multiplying by the coefficient. Work through each term separately to avoid calculation mistakes.

 

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 current age be x years. Then his mother's current age is 3x years.
After 5 years: Salil will be x + 5 years old and his mother will be 3x + 5 years old.
According to the given information: (x + 5) + (3x + 5) = 70
Simplifying: 4x + 10 = 70
4x = 60
x = 15
Therefore, Salil's current age is 15 years and his mother's current age is 45 years.
In simple words: Set up an equation showing both their ages after 5 years add to 70, then solve for x.

Exam Tip: Always define your variable clearly at the start. Set up equations based on the condition given (in this case, their combined age after 5 years), then solve step by step.

 

Question 4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Answer: Let the two integers be 2x and 5x based on their ratio.
Their difference is: 5x - 2x = 63
Simplifying: 3x = 63
x = 21
Therefore, the two integers are 2x = 42 and 5x = 105.
In simple words: When numbers are in ratio 2:5, write them as 2x and 5x, then use the difference to find x.

Exam Tip: Using the ratio form (2x and 5x) is cleaner than setting up two separate unknowns. Always verify by checking both the difference and the ratio.

 

Question 5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total Rs 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 is 3x.
The total value is: 5x + 2(3x) = 88
Simplifying: 5x + 6x = 88
11x = 88
x = 8
Therefore, Ruby has 8 five-rupee coins and 24 two-rupee coins.
In simple words: Multiply each coin type by its value, add them up, and set equal to the total amount to find how many of each.

Exam Tip: Always multiply the number of coins by their individual value. The total value equation must include both coin types with their correct values.

 

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 shorter piece be x feet. Then the longer piece is 4x feet.
Their sum equals the total length: x + 4x = 300
Simplifying: 5x = 300
x = 60
Therefore, the shorter piece is 60 feet and the longer piece is 240 feet.
In simple words: If one piece is 4 times another, write them as x and 4x, add them up to the total, then solve.

Exam Tip: Always check that both pieces add up to the total length given. In ratio problems, using the variable to represent one piece and expressing the other in terms of that variable keeps things simple.

 

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 be x cm. Then the length is 2x + 3 cm.
The perimeter formula gives: 2(length + width) = 24
Substituting: 2[(2x + 3) + x] = 24
Simplifying: 2(3x + 3) = 24
6x + 6 = 24
6x = 18
x = 3
Therefore, the width is 3 cm and the length is 2(3) + 3 = 9 cm.
In simple words: Use the perimeter formula with the width as your variable, express length in terms of width, then solve.

Exam Tip: Always use the perimeter formula correctly - it is 2(length + width). Check your answer by calculating the perimeter with your found dimensions.

 

Exercise Set 2.3

 

Question 1. A student has Rs 500 in her savings bank account. She gets Rs 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: Starting amount in the bank = Rs 500
Monthly pocket money received = Rs 150
At the end of: 2nd month = 500 + 2(150) = Rs 800
3rd month = 500 + 3(150) = Rs 950
4th month = 500 + 4(150) = Rs 1100
And so on...
Let An denote the amount in the nth month.
Then, An = 500 + 150n
The linear expression showing the total amount is: An = 150n + 500
In simple words: Each month, she gets Rs 150 added to her starting amount of Rs 500. So at month n, she has the starting amount plus n times the monthly addition.

Exam Tip: Write the linear pattern clearly showing both the constant (starting amount) and the coefficient of n (monthly addition). This pattern must hold for all values of n starting from month 1.

 

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 members in the rally = 120
Members dropping out each hour = 9
After: 1 hour = 120 - 9 = 111 members
2 hours = 120 - 18 = 102 members
3 hours = 120 - 27 = 93 members
Let Mn represent the number of members after n hours.
Then, Mn = 120 - 9n
The linear expression is: Mn = 120 - 9n
In simple words: Start with 120, and subtract 9 times the number of hours that have passed.

Exam Tip: This is a linear decay pattern (number decreasing over time). Always check that your expression gives reasonable values - members should decrease, not increase.

 

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 = 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
Let breadth = x cm
Area = 13x
The linear pattern is: A = 13x
In simple words: Since length is fixed at 13 cm, the area increases or decreases directly with the breadth - multiply breadth by 13.

Exam Tip: In a linear pattern with one fixed dimension, the area is simply the fixed dimension multiplied by the variable dimension.

 

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 = 7 cm, Breadth = 11 cm
Volume = Length × Breadth × Height = 7 × 11 × h = 77h
(i) When height = 5 cm, volume = 77 × 5 = 385 cm³
(ii) When height = 9 cm, volume = 77 × 9 = 693 cm³
(iii) When height = 13 cm, volume = 77 × 13 = 1001 cm³
Let height = h cm
Volume = 77h
The linear pattern is: V = 77h
In simple words: With length and breadth fixed, multiply their product (77) by the varying height to find volume.

Exam Tip: For volume, find the product of the two fixed dimensions first (7 × 11 = 77), then multiply by the variable dimension.

 

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 pages in the book = 500
Pages read per day = 20
Pages read in 15 days = 20 × 15 = 300
Pages remaining = 500 - 300 = 200
Let Pn represent the pages left after n days.
Then, Pn = 500 - 20n
The linear pattern is: Pn = 500 - 20n
In simple words: Start with 500 pages, and subtract 20 times the number of days that have passed.

Exam Tip: This is a decreasing linear pattern. Verify that after 25 days, there would be zero pages left (500 - 20(25) = 0), which makes sense for a 500-page book read at 20 pages per day.

 

Exercise Set 2.4

 

Question 1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months. (ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month. (iii) Find an expression that relates h and t, and explain why it represents linear growth.
Answer:
(i) The plant grows 0.5 feet each month. After 7 months: h = 1.75 + (0.5 × 7) = 1.75 + 3.5 = 5.25 feet
(ii) Table of values:

(iii) Expression: Let h represent the height after t months.
h = 1.75 + 0.5t
This represents linear growth because the height increases by a fixed amount (0.5 feet) every month, creating a constant rate of change.
In simple words: The plant starts at 1.75 feet and adds 0.5 feet each month. The height increases at a steady, constant rate.

Exam Tip: Linear growth means equal amounts increase each time period - here, 0.5 feet every month. The rate of change (slope) must be constant throughout.

 

Question 2. A mobile phone is bought for Rs 10,000. Its value decreases by Rs 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.
Answer:
(i) The phone loses Rs 800 in value each year. After 3 years: v = 10000 - (800 × 3) = 10000 - 2400 = Rs 7600
(ii) Table of values:

(iii) Expression: Let v represent the value after t years.
v = 10000 - 800t
This represents linear decay because the value decreases by a fixed amount (Rs 800) every year, showing a constant, negative rate of change.
In simple words: The phone starts at Rs 10,000 and loses Rs 800 each year at a steady, unchanging rate.

Exam Tip: Linear decay involves a constant rate of decrease. The negative coefficient (-800) shows the value is dropping - always check that the final value remains positive and realistic.

 

Question 3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years. (ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year. (iii) Find an expression that relates P and t, and explain why it represents linear growth.
Answer:
(i) The village gains 50 people each year. After 6 years: P = 750 + (50 × 6) = 750 + 300 = 1050
(ii) Table of values:

(iii) Expression: Let P represent the population after t years.
P = 750 + 50t
This represents linear growth because the population increases by a fixed number (50 people) every year, showing a constant rate of change.
In simple words: The village starts with 750 people and gains exactly 50 more each year at a steady rate.

Exam Tip: Verify your linear pattern by checking a few values from the table. The difference between consecutive values should always equal the fixed change (here, 50 people per year).

 

Question 4. A telecom company charges Rs 600 for a certain recharge scheme. This prepaid balance is reduced by Rs 15 each day after 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.
Answer:
(i) Expression: Let b(x) represent the remaining balance after x days.
b(x) = 600 - 15x
This represents linear decay because the balance decreases by a fixed amount (Rs 15) every single day, showing a constant negative rate of change.
(ii) When the balance becomes zero: b(x) = 0
600 - 15x = 0
15x = 600
x = 40
The balance will run out after 40 days.
(iii) Table of values:

In simple words: The account starts with Rs 600 and loses Rs 15 every day until it reaches zero after 40 days.

Exam Tip: When finding when the balance runs out, set the expression equal to zero and solve for the variable. Always verify that your answer makes sense - here, 40 days is reasonable for a balance of Rs 600 decreasing by Rs 15 daily.

 

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 observed that when she accessed 10 modules, her bill was Rs 400. When she accessed 14 modules, her bill was Rs 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: Given conditions: When x = 10, y = 400 and when x = 14, y = 500
Using the relation y = ax + b:
For x = 10: 400 = 10a + b ... (1)
For x = 14: 500 = 14a + b ... (2)
Subtracting equation (1) from equation (2):
500 - 400 = 14a - 10a
100 = 4a
a = 25
Substituting a = 25 back into equation (1):
400 = 10(25) + b
400 = 250 + b
b = 150
Therefore, a = 25 and b = 150.
In simple words: Write two equations using the two data points, then subtract them to find a (the cost per module), then use a to find b (the fixed monthly fee).

Exam Tip: Always set up two equations using the given (x, y) pairs. Subtracting them eliminates b, making it easy to find a first. Then substitute back to find b.

 

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 Rs 800. When she used it for 15 hours, her bill was Rs 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.
Answer: Given conditions: When x = 10, y = 800 and when x = 15, y = 1100
Using the relation y = ax + b:
For x = 10: 800 = 10a + b ... (1)
For x = 15: 1100 = 15a + b ... (2)
Subtracting equation (1) from equation (2):
1100 - 800 = 15a - 10a
300 = 5a
a = 60
Substituting a = 60 back into equation (1):
800 = 10(60) + b
800 = 600 + b
b = 200
Therefore, a = 60 and b = 200.
In simple words: The extra charge per hour is a, and the fixed monthly fee is b. Set up two equations and solve by subtraction and substitution.

Exam Tip: The coefficient a always represents the variable cost per unit (per hour here), while b represents the fixed cost that does not change with usage.

 

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: Given relation: °C = a°F + b
Using the first reference point (ice melts at 0°C and 32°F):
0 = 32a + b ... (1)
Using the second reference point (water boils at 100°C and 212°F):
100 = 212a + b ... (2)
Subtracting equation (1) from equation (2):
100 - 0 = 212a - 32a
100 = 180a
a = 100/180 = 5/9
Substituting a = 5/9 back into equation (1):
0 = 32(5/9) + b
0 = 160/9 + b
b = -160/9
Therefore, a = 5/9 and b = -160/9
The linear relationship is: °C = (5/9)°F - 160/9
This can also be written as: °C = (5/9)(°F - 32)
In simple words: Use two known temperature pairs to set up equations, solve for a and b, then you have a formula to convert from Fahrenheit to Celsius.

Exam Tip: This is a classic temperature conversion problem. Always use the exact freezing and boiling points as your reference pairs - they are most reliable. The alternative form °C = (5/9)(°F - 32) is easier to use in practice.

 

Exercise Set 2.6

 

Question 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
Answer: All three lines are of the form y = ax where b = 0.
Observations:
- All three lines pass through the origin (0, 0) because there is no constant term.
- The coefficient a is called the slope and it determines how steep the line is.
- A larger value of a produces a steeper line.
- y = 4x is the steepest, y = 2x is medium, and y = x is the least steep.
In simple words: When there is no constant b, all lines start at the origin. The larger the number multiplying x (which is a), the steeper the line climbs.

Exam Tip: Always identify the slope (a) and y-intercept (b) separately. When b = 0, the line must pass through the origin - this is a quick check for accuracy.

 

Question 1(ii). Draw the graphs of the following sets of lines. In each case, reflect on the role of 'a' and 'b': y = -6x, y = -3x, y = -x
Answer: All three lines are of the form y = ax where b = 0.
Observations:
- All three lines pass through the origin (0, 0) because there is no constant term.
- The negative values of a mean the lines slope downward from left to right.
- A larger magnitude (absolute value) of a produces a steeper downward slope.
- y = -6x has the steepest downward slope, y = -3x is medium, and y = -x is the least steep.
In simple words: Negative slopes make lines go downward. The bigger the negative number, the more steeply the line drops as you move to the right.

Exam Tip: Negative slope means the line falls as x increases. Compare the graphs visually - steeper lines have larger magnitudes, whether positive or negative.

 

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 with degree 3 takes the form ax³ + bx² + cx + d. Since the coefficient of x² must equal - 7, we have b = - 7. One example is x³ - 7x² + 2x + 1. (Any polynomial of degree 3 where the x² coefficient is - 7 is acceptable.)
In simple words: Degree 3 means the highest power is x³. Write any x³ term, then put - 7 times x², then any x term and any number.

Exam Tip: The question accepts ANY valid polynomial meeting the condition - there is no single "correct" answer, so show your working clearly to earn full marks.

 

Question 2. 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
Answer:
(i) Replace x with 1 in 5x² - 3x + 7:
\[ 5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = 9 \]
(ii) Replace t with a in 4t³ - t² + 6:
\[ 4a^3 - a^2 + 6 \]
In simple words: Write the number or letter wherever you see the variable, then work out the answer step by step.

Exam Tip: Always substitute carefully and simplify one step at a time to avoid arithmetic errors.

 

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. From the problem:
\[ \frac{5}{2}x + \frac{2}{3} = -\frac{7}{12} \]
Subtract 2/3 from both sides:
\[ \frac{5}{2}x = -\frac{7}{12} - \frac{2}{3} \]
Convert to a common denominator:
\[ \frac{5}{2}x = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4} \]
Multiply both sides by 2/5:
\[ x = -\frac{5}{4} \times \frac{2}{5} = -\frac{1}{2} \]
The number is - 1/2.
In simple words: Write the sentence as an equation using x for the unknown number. Then move things step by step to get x by itself.

Exam Tip: Show all fraction conversions to a common denominator - this is where marks are often lost.

 

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 = 5x. After adding 21 to each, the new numbers are x + 21 and 5x + 21. According to the condition, one new number is twice the other:
\[ 5x + 21 = 2(x + 21) \]
\[ 5x + 21 = 2x + 42 \]
\[ 3x = 21 \]
\[ x = 7 \]
So the smaller number is 7 and the larger number is 5 × 7 = 35. The two numbers are 7 and 35.
In simple words: One number is 5 times the other. When you add 21 to both, the bigger one is twice the smaller. Use this to find what the numbers are.

Exam Tip: Always define your variable clearly at the start and verify your answer by substituting back into the original condition.

 

Question 5. If you have Rs. 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.
Answer: Starting amount = Rs. 800, Monthly savings = Rs. 250. The linear pattern is:
\[ \text{Amount after } n \text{ months} = 800 + 250n \]
(i) Amount after 6 months:
\[ 800 + 250(6) = 800 + 1500 = \text{Rs. } 2300 \]
(ii) Amount after 2 years: 2 years = 24 months
\[ 800 + 250(24) = 800 + 6000 = \text{Rs. } 6800 \]
The linear equation is A = 800 + 250n, where n is the number of months.
In simple words: Start with 800. Every month add 250. So after 6 months you have 800 + 250 times 6. Convert years to months for the second part.

Exam Tip: Always convert time units consistently (years to months) before substituting into the formula.

 

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 x and the units digit be y. The original number = 10x + y, and the number with interchanged digits = 10y + x. From the first condition, the digits differ by 3:
\[ x - y = 3 \quad \text{...(1)} \]
From the second condition, when we add the original and the interchanged number, we get 143:
\[ (10x + y) + (10y + x) = 143 \]
\[ 11x + 11y = 143 \]
\[ 11(x + y) = 143 \]
\[ x + y = 13 \quad \text{...(2)} \]
Adding equations (1) and (2):
\[ 2x = 16 \implies x = 8 \]
From equation (2): \( y = 13 - 8 = 5 \)
The original number is 85 and the interchanged number is 58. Thus, the two numbers are 85 and 58.
In simple words: Two digits differ by 3. When you write them the other way around and add, you get 143. Figure out what the digits are, then write both numbers.

Exam Tip: Set up two separate equations - one for the digit condition and one for the sum condition - then solve the system.

 

Question 7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = - 3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x - 10
(iv) 3y = 6x - 11
Are any of the lines parallel?
Answer:
(i) y = - 3x + 4
Comparing with y = ax + b: Slope a = - 3, y-intercept b = 4. The line cuts the y-axis at (0, 4).
(ii) 2y = 4x + 7. Divide by 2: y = 2x + 7/2
Slope a = 2, y-intercept b = 7/2. The line cuts the y-axis at (0, 7/2).
(iii) 5y = 6x - 10. Divide by 5: y = (6/5)x - 2
Slope a = 6/5, y-intercept b = - 2. The line cuts the y-axis at (0, - 2).
(iv) 3y = 6x - 11. Divide by 3: y = 2x - 11/3
Slope a = 2, y-intercept b = - 11/3. The line cuts the y-axis at (0, - 11/3).
Parallel lines: Two lines are parallel when they have the same slope. Equations (ii) and (iv) both have slope = 2, so lines (ii) and (iv) are parallel.
In simple words: Rewrite each equation as y = mx + b. The m is the slope and the b is where it crosses the y-axis. Lines with the same slope never meet - those are parallel.

Exam Tip: Always convert to the form y = ax + b first by dividing through - this makes slope and y-intercept obvious.

 

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.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Answer:
(i) When x = 313 K, find y using y = (9/5)(x - 273) + 32:
\[ y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32 = 72 + 32 = 104 \]
The temperature is 104 °F.
(ii) When y = 158 °F, find x:
\[ 158 = \frac{9}{5}(x - 273) + 32 \]
Subtract 32 from both sides:
\[ 126 = \frac{9}{5}(x - 273) \]
Multiply both sides by 5/9:
\[ 70 = x - 273 \]
\[ x = 343 \]
The temperature is 343 K.
In simple words: Plug the given value into the formula. If you have Kelvin, find Fahrenheit. If you have Fahrenheit, work backwards to find Kelvin.

Exam Tip: Be careful with order of operations - subtract inside the brackets before multiplying by 9/5.

 

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: We know that Work = Force × Distance. With work = w, distance = d, and constant force = 3 units:
\[ w = 3d \]
This is the required linear equation. When d = 2 units:
\[ w = 3 \times 2 = 6 \text{ units} \]
Plotting on the graph with d on the x-axis and w on the y-axis, the point (2, 6) lies on the straight line w = 3d, which confirms our answer. The work done is 6 units.
In simple words: Work equals force times distance. With force = 3, the rule is w = 3d. When distance is 2, work is 6.

Exam Tip: Plot at least two or three points to ensure the line is straight, then use the graph to verify specific coordinate values.

 

Question 10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Answer:
(i) Let p(x) = ax + b. Since the graph passes through (1, 5):
\[ a + b = 5 \quad \text{...(1)} \]
Since the graph passes through (3, 11):
\[ 3a + b = 11 \quad \text{...(2)} \]
Subtract (1) from (2):
\[ 2a = 6 \implies a = 3 \]
From (1): \( b = 5 - 3 = 2 \)
Therefore, p(x) = 3x + 2.
(ii) For the y-axis intercept, set x = 0:
\[ y = 3(0) + 2 = 2 \]
The graph cuts the y-axis at (0, 2). For the x-axis intercept, set y = 0:
\[ 0 = 3x + 2 \implies x = -\frac{2}{3} \]
The graph cuts the x-axis at (-2/3, 0).
(iii) From the graph, the line passes through the given points (1, 5) and (3, 11), and it cuts the y-axis at (0, 2) and the x-axis at (-2/3, 0), which verifies our answers.
In simple words: Find the rule y = ax + b using two points. Then find where it crosses each axis by setting the other variable to zero.

Exam Tip: Always verify that your polynomial works by substituting the original two points back into your final equation.

 

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: From condition (i), p(0) = 5:
\[ a(0) + b = 5 \implies b = 5 \]
So p(x) = ax + 5. From condition (iii), p(x) + q(x) = 6x + 4:
\[ (ax + 5) + (cx + d) = 6x + 4 \]
Comparing coefficients:
\[ a + c = 6 \quad \text{...(1)} \]
\[ 5 + d = 4 \implies d = -1 \]
So q(x) = cx - 1. From condition (ii), p(x) - q(x) cuts the x-axis at (3, 0), which means p(3) - q(3) = 0:
\[ (3a + 5) - (3c - 1) = 0 \]
\[ 3a + 5 - 3c + 1 = 0 \]
\[ 3a - 3c = -6 \]
\[ a - c = -2 \quad \text{...(2)} \]
From (1) and (2), adding: \( 2a = 4 \implies a = 2 \). Then \( c = 6 - 2 = 4 \). Therefore, p(x) = 2x + 5 and q(x) = 4x - 1.
In simple words: Use the three conditions one at a time to find each unknown. First find b from condition (i), then use condition (iii) to link a and c, then use condition (ii) to separate them.

Exam Tip: When working with multiple conditions, use them in a logical order - start with the simplest to narrow down possibilities.

 

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.
(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 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.
Answer: A single hexagon needs 6 matchsticks. Each new hexagon shares one side with the previous hexagon, so only 5 new matchsticks are added per stage. The sequence is: Stage 1 = 6, Stage 2 = 11, Stage 3 = 16.
(i) Stage 4: 16 + 5 = 21 matchsticks. Stage 5: 21 + 5 = 26 matchsticks.
(ii) Table of matchsticks:

(iii) The matchstick count forms an arithmetic sequence: 6, 11, 16, 21, 26, ... with first term 6 and common difference 5. So the number of matchsticks in the nth stage is:
\[ M_n = 6 + (n - 1) \times 5 = 6 + 5n - 5 = 5n + 1 \]
(iv) For the 15th stage:
\[ M_{15} = 5(15) + 1 = 75 + 1 = 76 \]
76 matchsticks will be required for the 15th stage.
(v) For 200 matchsticks to form a stage, we need \( 5n + 1 = 200 \), which gives \( 5n = 199 \), so \( n = 39.8 \). Since n must be a whole number, 200 matchsticks cannot form any stage in this pattern.
In simple words: The pattern adds 5 sticks each time. First hexagon is 6, second is 11, third is 16. The rule is 5n + 1. For stage 15, work it out. For 200, see if you get a whole number for n.

Exam Tip: Identify the arithmetic pattern by finding the common difference, then derive the general term formula - this saves time on large-stage questions.

 

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: To find p(x): Since p(x) passes through (2, 3) and (6, 11), the slope is:
\[ m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2 \]
So p(x) = 2x + b. Using point (2, 3):
\[ 3 = 2(2) + b \implies b = -1 \]
Therefore, p(x) = 2x - 1. To find q(x): Since q(x) is parallel to p(x), it has slope 2. So q(x) = 2x + d. Using point (4, -1):
\[ -1 = 2(4) + d \implies d = -9 \]
Therefore, q(x) = 2x - 9. To find the x-axis intercepts: For p(x), set y = 0:
\[ 0 = 2x - 1 \implies x = \frac{1}{2} \]
p(x) meets the x-axis at (1/2, 0). For q(x), set y = 0:
\[ 0 = 2x - 9 \implies x = \frac{9}{2} \]
q(x) meets the x-axis at (9/2, 0). Thus p(x) = 2x - 1 and q(x) = 2x - 9, with x-axis intercepts at (1/2, 0) and (9/2, 0) respectively.
In simple words: Find the slope of p using the two points. Since q is parallel, it has the same slope. Use the third point to find the y-intercept of q. Then find where each line crosses the x-axis.

Exam Tip: Parallel lines have equal slopes - use this relationship to link the two polynomials before finding the constants.

 

Question 14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Answer: Consider f(x) = ax + a where a > 0. We can rewrite this as f(x) = a(x + 1). The common properties are:

1. Slope: The slope is a, and since a > 0, all such lines have positive slope. This means all lines rise from left to right.

2. y-intercept: Setting x = 0 gives f(0) = a. So the y-intercept is (0, a). Since a > 0, all these lines cross the y-axis above the origin.

3. x-intercept: Setting f(x) = 0:
\[ ax + a = 0 \implies a(x + 1) = 0 \]
Since a > 0 and a ≠ 0, we have x + 1 = 0, which gives x = - 1. Every line in this family cuts the x-axis at the single fixed point (-1, 0).

Therefore, all linear functions of the form f(x) = ax + a with a > 0 share these three properties:
- All have positive slope
- All cut the y-axis above the origin
- All pass through the fixed point (-1, 0)
In simple words: Rewrite as a(x + 1). The a is the slope - always positive, so lines go up. At x = 0, y = a, which is positive, so it crosses above the middle. At y = 0, x = -1 always, so every line touches the x-axis at the same point.

Exam Tip: Factoring the expression f(x) = a(x + 1) makes the x-intercept obvious without requiring any algebra beyond the factored form.