NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 – Introduction to Linear Polynomials are available here for Session 2026–27. Exercise Sets 2.1, 2.2, 2.3, 2.4, 2.5, 2.6 and End of Chapter Exercises are explained with step-by-step solutions, clear methods and student-friendly language based on the latest NCERT syllabus under NEP 2020 and NCF 2023.
In this chapter, students learn the basic concepts of linear polynomials, variables, coefficients, constants and degree of a polynomial through simple daily-life situations. The chapter also explains linear equations, linear patterns, linear growth and decay, and relationships of the form y = ax + b. Students further understand how linear relationships are represented on graphs using straight lines, slope and y-intercept concepts.
All questions are solved in an easy and exam-oriented format to help students improve algebraic understanding, logical thinking and problem-solving skills. These solutions are useful for homework, revision, school exams and concept clarity for CBSE Class 9 Mathematics Session 2026–27.
Table of Contents (Quick Links):
Exercise Set 2.1
1. Find the degrees of the following polynomials:
(i) 2x² − 5x + 3
(ii) y³ + 2y − 1
(iii) −9
(iv) 4z – 3
[Concept: The degree of a polynomial is the highest power of the variable in the polynomial. How to Solve: Identify the highest power of the variable in the given polynomial. That highest power is the degree of the polynomial.]
(i) 2x² − 5x + 3
Solution:
Power of x in 2x² = 2
Power of x in -5x = 1
Constant term 3 has degree 0
The highest power of the variable x in this expression is 2.
So, the Degree is 2.
(ii) y³ + 2y − 1
Solution:
Power of y in y³ = 3
Power of y in 2y = 1
Constant term -1 has degree 0
The highest power of the variable y in this expression is 3.
So, the Degree is 3.
(iii) −9
Solution
The polynomial -9 is a constant polynomial.
A non-zero constant polynomial has degree 0
Therefore, the Degree is 0.
(iv) 4z − 3
Solution:
Power of z in 4z = 1
Constant term -3 has degree 0
The highest power of the variable z in this expression is 1.
2. Write polynomials of degrees 1, 2 and 3.
[Concept: The degree of a polynomial is determined by the highest power of the variable. A polynomial is called a linear, quadratic, or cubic polynomial when its highest power is 1, 2, or 3, respectively.
How to Solve: To answer this question, write a polynomial whose highest power of the variable is equal to the required degree.]
Solution:
A polynomial of degree 1 (Linear Polynomial)
Example: 3z + 7
The highest power of the variable z is 1. Therefore, it is a polynomial of degree 1.
A polynomial of degree 2 (Quadratic Polynomial)
Example: x² + 5x + 1
The highest power of the variable x is 2. Therefore, it is a polynomial of degree 2.
Polynomial of degree 3 (Cubic Polynomial)
Example: 5y³ + y² + 2y − 1
The highest power of the variable y is 3. Therefore, it is a polynomial of degree 3.
3. What are the coefficients of x² and x³ in the polynomial x⁴ − 3x³ + 6x² − 2x + 7?
[Concept: The coefficient of a term is the numerical factor multiplied by the variable(s). How to solve: To answer this question, identify the given terms in the polynomial and write their numerical factors (coefficients).]
Solution:
In the given polynomial:
x⁴ − 3x³ + 6x² − 2x + 7
The coefficient of x² is 6.
The coefficient of x³ is −3.
Therefore, The coefficient of x² is 6 and The coefficient of x³ is −3.
4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?
[Concept: The coefficient of a term is the numerical factor multiplied by the variable(s). How to Solve: To answer this question, identify the term containing the given variable and write its numerical factor.]
Solution:
The given polynomial is
4z³ + 5z² − 11
There is no z term written in the polynomial, which means it can be thought of as 0z. Therefore, its coefficient is 0.
5. What is the constant term of the polynomial 9x³ + 5x² − 8x − 10?
Recall that polynomials of degree 1 are called linear polynomials. In this chapter, we shall study linear polynomials.
[Concept: The constant term is the term that does not contain any variable. How to Solve: To answer this question, identify the term in the polynomial that has no variable.]
Solution:
In the given polynomial:
9x³ + 5x² − 8x − 10
The term −10 does not contain any variable.
Therefore, the constant term of the polynomial is -10.
Exercise Set 2.2
1. Find the value of the linear polynomial 5x − 3 if:
(i) x = 0
(ii) x = −1
(iii) x = 2
[Concept: The value of a polynomial represents its calculated output when a specific number is placed in site of its variable. How to Solve: Substitute the given numerical value in place of the variable and solve the expression step-by-step using correct arithmetic operations.]
Solution:
(i) In the given polynomial is:
5x – 3
Substitute x = 0
= 5(0) – 3
= 0 – 3
= −3
Therefore, the value of the polynomial is -3.
(ii) The given polynomial is:
5x – 3
Substitute x = −1
= 5(-1) – 3
= −5 – 3
= −8
(iii) The given polynomial is:
5x – 3
Substitute x = 2
= 5(2) – 3
= 10 – 3
= 7
2. Find the value of the quadratic polynomial 7s² − 4s + 6 if:
(i) s = 0
(ii) s = −3
(iii) s = 4
[Concept: Evaluating a quadratic expression requires replacing its variable with a given value and determining the final numerical result. How to Solve: Plug the given number into the quadratic expression, simplify the squared terms first, and then perform the basic addition or subtraction.]
Solution:
(i) For s = 0
The given polynomial is:
7s² – 4s + 6
Substitute s = 0
= 7(0)² − 4(0) + 6
= 7(0) − 0 + 6
= 6
Therefore, the value of the polynomial is 6.
(ii) For s = −3
The given polynomial is:
7s² – 4s + 6
Substitute s = −3
= 7(−3)² − 4(−3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81
(iii) For s = 4
The given polynomial is:
7s² – 4s + 6
Substitute s = 4
= 7(4)² − 4(4) + 6
= 7(16) − 16 + 6
= 112 − 16 + 6
= 102
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.
[Concept: A linear equation can be formed by representing the unknown quantity with a variable and using the given conditions. How to Solve: Assume the unknown age as a variable, write the age of the other person using the given relation, form a linear equation from the given condition, and solve it.]
Solution:
Let Salil’s present age = x years
Mother’s present age = 3x years
After 5 years:
Salil’s age = x + 5
Mother’s age = 3x + 5
According to question:
(x + 5) + (3x + 5) = 70
⇒ 4x + 10 = 70
⇒ 4x = 70 -10
⇒ 4x = 60
⇒ x = 60/4
⇒ x = 15
Therefore, Salil’s age = 15 years
Mother’s present age = 3 x 15
= 45 years
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
[Concept: When the ratio of two quantities is given, they can be represented as multiples of a common variable. How to Solve: Represent the numbers according to the given ratio, form a linear equation using the given condition, solve it, and find the required numbers.]
Solution:
Let the common ratio between the two integers be x.
Therefore, the two integers are 2x and 5x.
According to the question, the difference between them is 63:
5x − 2x = 63
⇒ 3x = 63
⇒ x = 63/3
⇒ x = 21
Now, substituting the value of x:
First integer = 2x
= 2 × 21
= 42
Second integer = 5x
= 5 × 21
= 105
Therefore, the two positive integers are 42 and 105.
| Verification: Difference: 105 – 42 = 63 Ratio 42 : 105 = 2:5 (Both Conditions are satisfied.) |
5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total ₹88, how many coins does she have of each type?
[Concept: A linear equation can be formed by representing the unknown quantity with a variable and using the given conditions. How to Solve: Assume the number of one type of coin as a variable, express the number of the other type using the given relation, form a linear equation based on the total value, and solve it.]
Solution:
Let the number of five-rupee coins = 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, the total amount is ₹88:
5x + 6x = 88
⇒ 11x = 88
⇒ x = 88/11
⇒ x = 8
Now, calculating the number of coins of each type:
Number of five-rupee coins = x = 8
Number of two-rupee coins = 3x = 3 × 8 = 24
Therefore, Ruby has 8 five-rupee coins and 24 two-rupee coins.
| Verification: Total Value: = (8×5) + (24×2) = 40 + 48 = 88 (Hence, verified.) |
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?
[Concept: A linear equation can be formed by representing the unknown quantity with a variable and using the given conditions. How to Solve: Assume the smaller quantity as a variable, express the larger quantity using the given relation, form a linear equation from the total, and solve it.]
Solution:
Suppose the length of the smaller (छोटे टुकड़े की लंबाई) fence portion = x feet
Hence, the length of the larger (बड़े टुकड़े की लंबाई) fence portion = 4x feet
Based on the given problem statement, the total fence length is 300 feet:
x + 4x = 300
⇒ 5x = 300
⇒ x = 300/5
⇒ x = 60
Now, calculating the length of each piece:
Length of the shorter piece = x = 60 feet
Length of the longer piece = 4x = 4 × 60 = 240 feet
Therefore, the two pieces are 60 feet and 240 feet long.
| Verification: Total Length: 60 + 240 = 300 feet (Both conditions are satisfied.) |
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?
[Concept: A linear equation can be formed by representing the unknown quantity with a variable and using the given conditions. How to Solve: Assume the width as a variable, express the length using the given relation, use the perimeter formula to form a linear equation, and solve it.]
Solution:
Let the width of the rectangle = x cm
Then, the length of the rectangle = 2x + 3 cm
According to the question, the perimeter of the rectangle is 24 cm:
Perimeter = 2 (length + width)
24 = 2 × [(2x + 3) + x]
⇒ 24 = 2 × (3x + 3)
⇒ 24 = 6x + 6
⇒ 24 − 6 = 6x
⇒ 18 = 6x
⇒ x = 3
Therefore, the width is 3 cm.
Length = 2(3) + 3
= 9 cm
Therefore, the dimensions of the rectangle are 9 cm (length) and 3 cm (width).
| Verification: Perimeter = 2 (9+3) = 2 x 12 = 24 cm (The given condition is satisfied). |
Exercise Set 2.3
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.
[Concept: A linear expression is formed by using a variable to represent an unknown quantity and expressing the relationship given in the problem. How to Solve: Identify the fixed amount and the amount added every month. Then use a variable to represent the month number and write the required linear expression.]
Solution:
Let n represent the month number.
Initial amount in the account = ₹500
Amount added every month = ₹150
The amount of money she will have at the end of each month is as follows:
End of 1st month = 500 + 150 = ₹650
End of 2nd month = 650 + 150 = ₹800
End of 3rd month = 800 + 150 = ₹950
End of 4th month = 950 + 150 = ₹1100
From the second month onwards, the amounts are ₹800, ₹950, ₹1100, …
To find the linear expression for the amount in the nᵗʰ month:
Amount in the nᵗʰ month = Initial amount + (Amount added every month × Number of months) Therefore, the required linear expression is 500 + 150n
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.
[Concept: A linear expression is formed by using a variable to represent an unknown quantity and expressing how the quantity changes at a constant rate. How to Solve: Identify the initial number of members and the number of members leaving every hour. Then write a linear expression by subtracting the total members leaving after ‘n’ hours from the initial number of members.]
Solution:
Initial number of members = 120 Members dropping out each hour = 9
The number of members remaining after the first few hours:
After 1 hour = 120 − 9 = 111 members
After 2 hours = 111 − 9 = 102 members
After 3 hours = 102 − 9 = 93 members
Therefore, the number of members remaining after 1, 2, 3, … hours is 111, 102, 93, …
To find the linear expression for the number of members at the end of the nth hour:
Members remaining = Initial members − (Members dropping out per hour × n) Linear expression = 120 − 9n
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.
[Concept: The area of a rectangle is equal to the product of its length and breadth. If one dimension is fixed and the other is represented by a variable, the area can be expressed as a linear expression. How to Solve: Find the area for each given breadth using the formula Area = Length × Breadth. Then replace the breadth with a variable to write the required linear pattern.]
Solution:
Length of the rectangle = 13 cm
The area of the rectangle (Area = length × breadth) for each breadth is:
(i) If breadth = 12 cm Area = 13 × 12 = 156 cm²
(ii) If breadth = 10 cm Area = 13 × 10 = 130 cm²
(iii) If breadth = 8 cm Area = 13 × 8 = 104 cm²
To find the linear pattern representing the area: Let the breadth of the rectangle be x cm.
Linear pattern for the area = 13x
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.
[Concept: The volume of a rectangular box is equal to the product of its length, breadth, and height. If the length and breadth are fixed, the volume varies linearly with the height. How to Solve: Calculate the volume for each given height using the formula Volume = Length × Breadth × Height. Then replace the height with a variable to write the required linear pattern.]
Solution:
Length of the rectangular box = 7 cm Breadth of the rectangular box = 11 cm
The volume of a rectangular box is given by the formula:
Volume = length × breadth × height.
Therefore,
Volume = 7 × 11 × height = 77 × height.
The volume for each given height is:
(i) If height = 5 cm Volume = 77 × 5 = 385 cm³
(ii) If height = 9 cm Volume = 77 × 9 = 693 cm³
(iii) If height = 13 cm Volume = 77 × 13 = 1001 cm³
To find the linear pattern representing the volume:
Let the height of the rectangular box be h cm.
Linear pattern for the volume = 77h
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.
[Concept: A linear expression can be formed by subtracting the quantity that decreases at a constant rate from the initial quantity. How to Solve: Find the number of pages read after the given number of days and subtract it from the total number of pages. Then replace the number of days with a variable to write the required linear pattern.]
Solution:
Total pages in the book = 500 Pages read every day = 20
The number of pages left after 15 days:
Pages read in 15 days = 20 × 15 = 300 pages
Pages left = 500 − 300 = 200 pages
Therefore,
200 pages will be left after 15 days.
To express this as a linear pattern:
Pages left
= Total pages – (pages read per day x Number of days)
= 500 – (20 x d)
= 500 – 20d
Let d represent the number of days.
Linear pattern for pages left = 500 − 20d
Exercise Set 2.4
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.
[Concept: A linear relationship is formed when a quantity increases by a fixed amount at regular intervals. Such a relationship can be represented by a linear expression. How to Solve: Find the height after the given number of months by adding the monthly growth to the initial height. Then prepare a table and use the initial height and monthly growth to write the linear expression.]
(i) Find the height after 7 months.
Solution:
Initial height of the plant = 1.75 feet
Growth every month = 0.5 feet
Height after 7 months
= 1.75 + (0.5 × 7)
= 1.75 + 3.5
= 5.25 feet
Therefore, the height of the plant after 7 months is 5.25 feet.
(ii) Make a table of values for t varying from 0 to 10 months.
Solution:
| Months, t | Height, h (feet) |
| 0 | 1.75 |
| 1 | 2.25 |
| 2 | 2.75 |
| 3 | 3.25 |
| 4 | 3.75 |
| 5 | 4.25 |
| 6 | 4.75 |
| 7 | 5.25 |
| 8 | 5.75 |
| 9 | 6.25 |
| 10 | 6.75 |
The height increases by 0.5 feet every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Solution:
Initial height of the plant = 1.75 feet
Growth every month = 0.5 feet
Height after t months
= Initial height + (Growth per month × t)
= 1.75 + 0.5t
Thus,
h = 1.75 + 0.5t
The expression represents linear growth because the height increases by the same amount (0.5 feet) every month.
2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
[Concept: A linear relationship is formed when a quantity decreases by a fixed amount at regular intervals. Such a relationship can be represented by a linear expression. How to Solve: Find the value after the given number of years by subtracting the yearly depreciation from the initial value. Then prepare a table and use the initial value and yearly depreciation to write the linear expression.]
(i) Find the value of the phone after 3 years.
Solution:
Initial value of the phone = ₹10,000
Depreciation every year = ₹800
Value after 3 years
= 10000 − (800 × 3)
= 10000 − 2400
= ₹7600
Therefore, the value of the phone after 3 years is ₹7600.
(ii) Make a table of values for t varying from 0 to 8 years.
Solution:
| Years, t | Value, v (₹) |
| 0 | 10000 |
| 1 | 9200 |
| 2 | 8400 |
| 3 | 7600 |
| 4 | 6800 |
| 5 | 6000 |
| 6 | 5200 |
| 7 | 4400 |
| 8 | 3600 |
The value decreases ₹800 every year.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Solution:
Initial value of the phone = ₹10,000
Depreciation every year = ₹800
Value after t years
= Initial value − (Depreciation per year × t)
= 10000 − 800t
Thus,
v = 10000 − 800t
The expression represents linear decay because the value decreases by the same amount (₹800) every year.
3. The initial population (शुरुआती आबादी) of a village is 750. Every year, 50 people move from a nearby city 9पास के शहर से 50 लोग आते हैं) to the village.
(i) Find the population of the (6 साल बाद गाँव की आबादी) village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years (t के 0 से 10 साल तक बदलने के लिए मानों की तालिका) and show how the population, P, increases every year (दिखाइए कि जनसंख्या P हर साल कैसे बढ़ती है).
(iii) Find an expression that relates P and t (P और t को जोड़ने वाला व्यंजक), and explain why it represents linear growth.
[Concept: A linear relationship is formed when a quantity increases by a fixed amount at regular intervals. Such a relationship can be represented by a linear expression. How to Solve: Find the population after the given number of years by adding the yearly increase to the initial population. Then prepare a table and use the initial population and yearly increase to write the linear expression.]
Solution: (i)
Initial population of the village = 750
Increase in population every year = 50
Population after 6 years
= 750 + (50 × 6)
= 750 + 300
= 1050
Therefore, the population of the village after 6 years is 1050.
Solution: (ii)
| Years, t | Population, P |
| 0 | 750 |
| 1 | 800 |
| 2 | 850 |
| 3 | 900 |
| 4 | 950 |
| 5 | 1000 |
| 6 | 1050 |
| 7 | 1100 |
| 8 | 1150 |
| 9 | 1200 |
| 10 | 1250 |
From the table, we observe that the population increases by 50 people every year.
Solution: (iii)
Initial population = 750
Increase in population every year = 50
Population after t years
= Initial population + (Increase per year × t)
= 750 + (50 × t)
= 750 + 50t
Therefore, the required expression is
P = 750 + 50t
The expression represents linear growth because the population increases by the same amount (50 people) every year.
4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
[Concept: A linear relationship is formed when a quantity decreases by a fixed amount at regular intervals. Such a relationship can be represented by a linear expression. How to Solve: Subtract the daily reduction from the initial balance to form a linear equation. Then use the equation to find the required values and prepare the table.]
Solution: (i)
Initial balance = ₹600
Balance reduced every day = ₹15
Remaining balance after x days
= Initial balance − (Reduction per day x )
= 600 − (15× )
= 600 − 15x
Therefore, the required equation is
b(x) = 600 − 15x
The equation represents linear decay because the balance decreases by ₹15 every day.
Solution: (ii)
The balance runs out when
b(x) = 0
Substitute b(x) = 600 − 15x:
600 − 15x = 0
15x = 600
x = 600/15
x = 40
Therefore, the balance will run out after 40 days.
Solution: (iii)
| Days, x | Remaining Balance, b(x) Rupees |
| 1 | 585 |
| 2 | 570 |
| 3 | 555 |
| 4 | 540 |
| 5 | 525 |
| 6 | 510 |
| 7 | 495 |
| 8 | 480 |
| 9 | 465 |
| 10 | 450 |
From the table, we observe that the balance decreases by ₹15 every day.
Exercise Set 2.5
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 ₹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.
[Concept: A linear equation of the form y = ax + b represents the relationship between two variables, where a is the rate of change (slope) and b is the fixed value (constant term). How to Solve: Substitute the given values of x and y into the equation y = ax + b to form two linear equations. Then solve the equations to find the values of a and b.]
Solution:
According to the question, the relation between the monthly bill (y) and the number of modules (x) is given by:
y = ax + b
From the given observations:
For x = 10, y = 400:
⇒ 400 = 10a + b — (Equation 1)
For x = 14, y = 500:
⇒ 500 = 14a + b — (Equation 2)
Subtracting Equation 1 from Equation 2:
(14a + b) − (10a + b) = 500 − 400
⇒ 4a = 100
⇒ a = 100 / 4
⇒ a = 25
Substituting the value of a = 25 in Equation 1:
400 = 10(25) + b
⇒ 400 = 250 + b
⇒ b = 400 − 250
⇒ b = 150
Therefore, the values are a = 25 and b = 150.
2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
[Concept: A linear equation of the form y = ax + b represents the relationship between two variables, where a is the rate of change (slope) and b is the fixed value (constant term). How to Solve: Form two linear equations by substituting the given values into y = ax + b. Then solve the equations to determine the values of a and b.]
Solution:
According to the question, the relation between the monthly bill (y) and the hours of badminton court use (x) is given by: y = ax + b
From the given observations:
1. For x = 10, y = 800: 800 = 10a + b — (Equation 1)
2. For x = 15, y = 1100: 1100 = 15a + b — (Equation 2)
Subtracting Equation 1 from Equation 2:
(15a + b) − (10a + b) = 1100 − 800
⇒ 5a = 300
⇒ a = 300 / 5
⇒ a = 60
Substituting the value of a = 60 in Equation 1:
800 = 10(60) + b
⇒ 800 = 600 + b
⇒ b = 800 − 600
⇒ b = 200
Therefore, the values are a = 60 and b = 200.
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.
[Concept: A linear equation can be used to represent the relationship between two related quantities. The constants in the equation can be determined using two given pairs of corresponding values. How to Solve: Substitute the given temperature values into the equation °C = a°F + b to form two linear equations. Then solve the equations to find the values of a and b.]
Solution:
According to the question, the relation between the temperature in Celsius (°C) and Fahrenheit (°F) is given by: °C = a(°F) + b
From the given observations:
Ice melts at 0°C and 32°F (For °F = 32, °C = 0):
0 = a(32) + b
⇒ 32a + b = 0 — (Equation 1)
Water boils at 100°C and 212°F (For °F = 212, °C = 100):
100 = a(212) + b
⇒ 212a + b = 100 — (Equation 2)
Subtracting Equation 1 from Equation 2:
(212a + b) − (32a + b) = 100 − 0
⇒ 180a = 100
⇒ a = 100 / 180
⇒ a = 5 / 9
Substituting the value of a = 5 / 9 in Equation 1:
32 x (5 / 9) + b = 0
⇒ 160 / 9 + b = 0
⇒ b = −160 / 9
Therefore, the values are a = 5 / 9 and b = −160 / 9.
Exercise Set 2.6
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
[Concept: In the linear equation y = ax + b, the coefficient a determines the slope (inclination) of the line, while the constant b determines the y-intercept, i.e., the point where the line cuts the y-axis. How to Solve: Draw the graph of each equation on the same coordinate plane. Compare the graphs carefully and observe how changing the values of a and b affects the slope and the position of the line.]
(i) y = 4x, y = 2x, y = x
Solution:
Plot the graphs of the given equations on the same coordinate plane.
Observation:
All three lines pass through the origin because b = 0.
As the value of a increases, the line becomes steeper.
(ii) y = – 6x, y = – 3x, y = -x
Solution:
Plot the graphs of the given equations on the same coordinate plane.
Observation:
All three lines pass through the origin because b = 0.
Since a is negative, all the lines slope downward from left to right.
A larger numerical value of a gives a steeper line.
(iii) y = 5x, y = –5x
Solution:
Plot the graphs of the given equations on the same coordinate plane.
Observation:
Both lines pass through the origin because b = 0.
The lines have the same steepness because the magnitude of a is the same.
One line rises while the other falls because their slopes have opposite signs.
(iv) y = 3x –1, y = 3x, y = 3x + 1
Solution:
Plot the graphs of the given equations on the same coordinate plane.
Observation:
All three lines have the same slope because a = 3.
The lines are parallel to each other.
Changing the value of b shifts the line upward or downward without changing its slope.
(v) y = -2x -3, y = -2x, y = 2x + 3
Solution:
Plot the graphs of the given equations on the same coordinate plane.
Observation:
The first two lines have the same slope because a = −2. Hence, they are parallel to each other.
The values of b are different (−3 and 0), so the lines intersect the y-axis at different points.
In the third equation, both a and b are different. Therefore, its slope and y-intercept are different from the first two lines, so it is neither parallel to them nor passes through the same y-intercept.
This shows that a determines the slope of the line, while b determines its y-intercept.
End-of-Chapter Exercises
1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is –7.
[Concept: The degree of a polynomial is determined by the highest power of the variable. The coefficient of a term is the numerical factor multiplied by that variable. How to Solve: Write a polynomial whose highest power of x is 3 and make sure that the coefficient of the x² term is –7]
Solution:
A polynomial of degree 3 must have the highest power of x equal to 3.
The coefficient of the x² term should be –7.
One such polynomial is
x³ – 7 x² + 5x – 2 …. (The coefficient of x and the constant term may be chosen freely.)
Therefore, one required polynomial is = x³ – 7 x² + 5x – 2
[Note: Any polynomial of degree 3 having the coefficient of x² equal to –7 is a correct answer.
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
[Concept: The value of a polynomial is obtained by substituting the given value of the variable into the polynomial and simplifying the expression. How to Solve: Replace the variable with the given value and perform the arithmetic operations carefully.]
(i) 5x² – 3x + 7 if x = 1
Solution:
Given polynomial:
5x² – 3x + 7
Substitute x = 1
= 5 (1)² – 3(1) + 7
= 5 – 3 + 7
= 9
Therefore, the value of the polynomial is 9.
(ii) 4t³ -t² +6 if t = a
Solution:
Given polynomial:
4t³ -t² +6
Substitute t = a
= 4a³ -a² +6
Since no numerical value of a is given, the expression cannot be simplified further.
Therefore, the value of the polynomial is 4a³ -a² +6.
3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.
[Concept: A word problem can be translated into a linear equation by representing the unknown quantity with a variable. How to Solve: Let the required number be a variable. Form the linear equation using the given conditions, then solve the equation to find the unknown number.]
Solution:
Let the required number be x.
According to the question,
5/2 + 2/3 = -7/12
Subtract 2/3 from both sides:
5/2 = -7/12 – 2/3
= -7/12 – 8/12
= -15/12
= -5/4
Multiply both sides by 2/5:
x = -5/4 X 2/5
x = -1/2
Therefore, the required number is -1/2.
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?
[Concept: Relationships between unknown quantities can be represented using variables and linear equations. How to Solve: Let the smaller number be a variable. Express the other number using the given relationship, form the equation according to the given condition, and solve it to find both numbers.]
Solution:
Let the smaller number be x.
Then the larger number is 5x.
After adding 21 to both numbers:
Smaller number becomes x+21.
Larger number becomes 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 x 7 = 35
Therefore, the required numbers are 7 and 35.
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.
[Concept: A linear pattern is formed when a quantity increases by a fixed amount at regular intervals. Such a relationship can be represented by a linear expression. How to Solve: Find the total amount after the given time by adding the monthly savings to the initial amount. Then write a linear expression using the initial amount and the monthly savings.]
Solution:
Initial amount = ₹800
Amount saved every month = ₹250
(i) Amount after 6 months
Amount after 6 months
= 800 + (250 × 6)
= 800 + 1500
= ₹2300
Therefore, the amount after 6 months is ₹2300.
(ii) Amount after 2 years
Since, 2 years=24 months
Amount after 24 months
= 800 + (250 × 24)
= 800 + 6000
= ₹6800
Therefore, the amount after 2 years is ₹6800.
Linear Pattern
Let n represent the number of months.
Amount after n months
= Initial amount + (Monthly saving × n)
= 800 + 250n
Therefore, the required linear pattern is 800 + 250n
| Verification: After adding 21, the numbers become 28 and 56, and 56 = 2 × 28. Hence, the answer is verified. |
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.
[Concept: A two-digit number can be represented using its tens digit and units digit. The given conditions can be converted into linear equations to find the unknown digits. How to Solve: Let the tens digit and the units digit be variables. Form the original and interchanged numbers, write the equations using the given conditions, and solve them to find the required digits.]
Solution:
Let the tens digit be x and the units digit be y.
Then,
Original number = 10x + y
Interchanged number = 10y + x
According to the question, the digits differ by 3.
x – y = 3…(i)
Also, the sum of the original number and the interchanged number is 143.
(10x + y) + (10y + x) = 143
11x + 11y = 143
Dividing both sides by 11,
x + y = 13…(ii)
Now, add equations (i) and (ii):
(x − y) + (x + y) = 3 + 13
2x = 16
x = 8
Substitute the value of x = 8 in equation (i):
8 – y = 3
y = 5
Hence, the tens digit is 8 and the units digit is 5.
Therefore, Original number = 85
Interchanged number = 58
Therefore, the required numbers are 85 and 58.
7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis. Are any of the lines parallel?
(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
[Concept: Every linear equation can be written in the form y = mx + c, where m is the slope and c is the y-intercept. How to Solve: Rewrite each equation in the form y = mx + c. Identify the values of m and c, draw the graph, locate the point where the line cuts the y-axis, and compare the slopes to determine whether any lines are parallel.]
(i) y = −3x + 4
Solution:
The given equation is already in the form y = mx + c.
Therefore,
Slope,
m = −3
y-intercept,
c = 4
The line cuts the y-axis at (0,4).
(ii) 2y = 4x + 7
Solution:
Write the equation in the form y = mx + c.
2y = 4x + 7 (Dividing both sides by 2)
y = 2x + 7/2
Therefore,
Slope,
m = 2
y-intercept,
c = 7/2
The line cuts the y-axis at (0, 7/2)
(iii) 5y = 6x − 10
Solution:
Write the equation in the form y = mx + c.
5y = 6x – 10 (Dividing both sides by 5)
y = 6/5x – 2
Therefore,
Slope,
m = 6/5
y-intercept,
c = −2
The line cuts the y-axis at (0, −2).
(iv) 3y = 6x − 11
Solution:
Write the equation in the form y = mx + c.
3y = 6x – 11 (Dividing both sides by 3,)
y = 2x – 11/3
Therefore,
Slope,
m = 2
y-intercept,
c = −11/3
The line cuts the y-axis at (0, -11/3)
Observation:
From the graph, we observe that the slopes of the four lines are:
Equation (i): −3
Equation (ii): 2
Equation (iii): 6/5
Equation (iv): 2
Since equations (ii) and (iv) have the same slope, their graphs are parallel.
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 is given by
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.
[Concept: A linear equation can be used to relate two variables. By substituting the given value of one variable into the equation, the value of the other variable can be found. How to Solve: Substitute the given value into the linear equation and simplify step by step. If the value of the other variable is given, substitute it into the equation and solve for the required variable.]
Solution:
The relation between Kelvin and Fahrenheit is: y = 9/5 (x -273) + 32
(i) When x = 313K
Substitute x = 313 into the given equation:
y = 9/5 (313 – 273) + 32
y = 9/5 (40) + 32
y = 72 + 32
y = 104
Therefore, the temperature of the liquid is 104∘F
(ii) When y = 158∘F
Substitute y = 158 into the given equation.
158 = 9/5 (x -273) + 32
Subtract 32 from both sides.
158 – 32 = 9/5 (x -273) + 32 -32
126 = 9/5 (x -273)
Multiply both sides by 5/9
126 x 5/9 = x – 273
70 = x – 273
Add 273 to both sides.
x = 70 + 273
x = 343
Therefore, the temperature of the liquid is 343 K.
[Note: When solving an equation, the objective is to isolate the required variable. This is done by removing, step by step, all the numbers and operations associated with the variable. The same operation is performed on both sides of the equation to keep it balanced and obtain the correct value of the variable.]
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.
[Concept: Work done is equal to the product of force and distance. When the force is constant, the relationship between work and distance is linear. How to Solve: Use the formula W = F × d. Substitute the given constant force to obtain the linear equation. Find the work done for the given distance and verify it using the graph.]
Solution:
The work done is the product of force and distance travelled in the direction of the force.
Therefore,
Work = Force × Distance
Let,
w = work done
d = distance travelled
The constant force is 3 units.
Therefore,
w = 3d
This is the required linear equation.
| Distance, d | 0 | 1 | 2 | 3 | 4 |
| Work, w = 3d | 0 | 3 | 6 | 9 | 12 |
Work done when the distance travelled is 2 units
Substitute d = 2 into the equation.
W = 3 × 2
W = 6
Therefore, the work done is 6 units.
| Verification: On the graph, the point (2,6) lies on the straight line representing the equation w = 3d. Hence, the graph also verifies that the work done is 6 units. |
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.
[Concept: A linear polynomial is of the form p(x) = ax + b, where a is the coefficient of x and b is the constant term. If two points on its graph are known, the polynomial can be determined by finding the values of a and b. How to Solve: Substitute the given points into the equation p(x) = ax + b to form two linear equations. Solve them to find the values of a and b. Then determine the x-intercept and y-intercept, and verify the results using the graph.]
Solution:
The given linear polynomial is: p(x) = ax + b
Since the graph passes through the point (1,5),
5 = a(1) + b
a + b = 5
a + b = 5 …. (i)
Since the graph also passes through the point (3, 11),
11 = a(3) + b
3a + b = 11
3a + b = 11 ….. (ii)
Subtracting equation (i) from equation (ii),
(3a + b) − (a + b) = 11 − 5
2a = 6
a = 3
Substituting a = 3 in equation (i),
3 + b = 5
b = 2
Therefore,
p(x) = 3x + 2
(ii) Coordinates where the graph cuts the axes
x-axis
On the x-axis, p(x) = 0
3x + 2 = 0
3x = -2
x = -2/3
Therefore, the graph cuts the x-axis at (-2/3, 0)
y-axis
On the y-axis, x = 0
p(0) = 3(0) + 2 = 2
Therefore, the graph cuts the y-axis at (0, 2)
(iii)
| Verification: From the graph, the line passes through the points (1, 5) and (3, 11), and cuts the x-axis at (− 2/3, 0) and y-axis at (0,2). Hence, the graph verifies the required polynomial and its intercepts. |
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).
[Concept: A linear polynomial is of the form ax + b. The given conditions are used to form linear equations involving the unknown coefficients. Solving these equations gives the required polynomials. How to Solve: Use each given condition to form equations involving the unknown coefficients. Solve the equations step by step and substitute the obtained values to determine the required polynomials.]
Solution:
The given polynomials are: p(x) = ax + b and q(x) = cx + d.
From condition (i):
Since
p(0) = 5
we have
a(0) + b = 5
b = 5.
From condition (ii):
The polynomial: p(x) − q(x) [cuts the x-axis at (3, 0)]
Therefore,
p(3) − q(3) = 0
Substituting x = 3,
(3a + b) − (3c + d) = 0
Using b = 5,
3a + 5 − 3c − d = 0
3a − 3c − d = −5 ………. (i)
From condition (iii):
The sum of the two polynomials is
p(x) + q(x) = 6x + 4
Substituting the expressions for p(x) and q(x),
(ax + b) + (cx + d) = 6x + 4
Since the equality holds for all real values of x, compare the coefficients of like terms.
Coefficient of x:
a + c = 6 ………. (ii)
Constant terms:
b + d = 4
Since b = 5,
5 + d = 4
d = −1
Substituting d = −1 in equation (i),
3a − 3c − (−1) = −5
3a − 3c + 1 = −5
3a − 3c = −6
a − c = −2 ………. (iii)
Adding equations (ii) and (iii),
(a + c) + (a − c) = 6 + (−2)
2a = 4
a = 2
Substituting a = 2 in equation (ii),
2 + c = 6
c = 4
Thus,
a = 2, b = 5, c = 4, d = −1
Therefore,
p(x) = ax + b and q(x) = cx + d.
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 matchsticks will be required at these stages?
(ii) Complete the following table
| Stage No. | 1 | 2 | 3 | 4 | 5 | … | n |
| No. 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.
[Concept: In a growing pattern, observe how the number of matchsticks changes from one stage to the next. Since each new hexagon shares one side with the previous hexagon, fewer matchsticks are required to add the next hexagon. By identifying the pattern, we can write a rule for any stage. How to Solve: Count the number of matchsticks in the first few stages. Observe how many matchsticks are added at each new stage. Complete the table, identify the pattern, and write a rule to find the number of matchsticks required for the nth stage.]
Solution:
(i) A hexagon is made using 6 matchsticks.
Each new hexagon shares one side with the previous hexagon, so only 5 new matchsticks are needed for every additional stage.
Therefore,
Stage 4 = 21 matchsticks
Stage 5 = 26 matchsticks
(ii)
(iii) The first stage requires 6 matchsticks.
Each new stage adds 5 matchsticks because one side is shared with the previous hexagon.
Thus, the sequence of matchsticks is
6, 11, 16, 21, 26, …
This is an arithmetic pattern with
First term = 6
Common difference = 5
Therefore,
Number of matchsticks in the nth stage
= 6 + (n − 1) × 5
= 6 + 5n − 5
= 5n + 1
Therefore, the required rule is
Number of matchsticks = 5n + 1
(iv) By using the rule,
Number of matchsticks = 5n + 1
Substitute n = 15.
Number of matchsticks
= 5 × 15 + 1
= 75 + 1
= 76
Therefore, 76 matchsticks will be required for the 15th stage.
(v) The rule for the pattern is
Number of matchsticks = 5n + 1
Suppose 200 matchsticks form a stage.
Then,
5n + 1 = 200
Subtracting 1 from both sides,
5n = 199
n = 199/5 = 39.8
Since the stage number must be a whole number, 39.8 is not possible.
Therefore, 200 matchsticks cannot form any stage of this pattern.
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.
[Concept: A linear polynomial is of the form ax + b. Two points determine a unique linear polynomial. Parallel lines have the same slope but different y-intercepts. How to Solve: First, determine p(x) using the two given points. Since q(x) is parallel to p(x), both have the same coefficient of x (same slope). Then use the given point on q(x) to find its constant term. Finally, find the x-intercepts by putting y = 0.]
Solution:
The given polynomial is
p(x) = ax + b
Since the graph of p(x) passes through the points (2, 3) and (6, 11),
For the point (2, 3),
3 = 2a + b ………. (i)
For the point (6, 11),
11 = 6a + b ………. (ii)
Subtracting equation (i) from equation (ii),
(6a + b) − (2a + b) = 11 − 3
4a = 8
a = 2
Substituting a = 2 in equation (i),
3 = 2(2) + b
3 = 4 + b
b = −1
Therefore,
p(x) = 2x − 1
Now,
q(x) = cx + d
Since the graph of q(x) is parallel to the graph of p(x),
c = 2
The graph of q(x) passes through the point (4, −1).
Substituting x = 4 and y = −1,
−1 = 2(4) + d
−1 = 8 + d
d = −9
Therefore,
q(x) = 2x − 9
Finding the x-intercepts
For p(x),
2x − 1 = 0
2x = 1
x = 1/2
Therefore, the graph of p(x) cuts the x-axis at
(1/2, 0)
For q(x),
2x − 9 = 0
2x = 9
x = 9/2
Therefore, the graph of q(x) cuts the x-axis at
(9/2, 0)
Therefore,
The required polynomials are
p(x) = 2x − 1
q(x) = 2x − 9
The graphs cut the x-axis at
(1/2, 0) and (9/2, 0), respectively.
14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
[Concept: A linear function of the form f(x) = ax + b has a slope, a y-intercept, and an x-intercept. By comparing the given function with the standard form and finding its intercepts, we can identify the common properties shared by all such linear functions. How to Solve: Compare the given function with the standard form f(x) = mx + c to identify the slope and y-intercept. Then find the x-intercept by putting f(x) = 0. Finally, write the common properties that are true for every function of the given form.]
Solution:
Slope: The slope is a. Since a > 0, all the lines have a positive slope.
Therefore, all the graphs rise from left to right.
y-intercept:
Putting x = 0,
f(0) = a
Hence, every graph cuts the y-axis at the point (0, a). Since a > 0, all the graphs cut the y-axis above the origin.
x-intercept:
To find the x-intercept, put
f(x) = 0
ax + a = 0
a(x + 1) = 0
Since a > 0, a ≠ 0.
Therefore,
x + 1 = 0
x = −1
Hence, every graph cuts the x-axis at the fixed point (−1, 0).
Therefore,
All linear functions of the form f(x) = ax + a, where a > 0, have these common properties:
• All have a positive slope.
• All rise from left to right.
• All cut the y-axis above the origin.
• All pass through the fixed point (−1, 0).
Frequently Asked Questions (FAQs):
1. What is a linear polynomial?
Answer: A linear polynomial is an algebraic expression where the highest degree of the variable is 1. Its general form is ax + b (where a is not 0), such as 3z + 7 or 2x + 3.
2. What is the degree of a constant polynomial?
Answer: The degree of a non-zero constant polynomial is 0. For example, the constant polynomial 8 can be written as 8x⁰, since x⁰ = 1 (for x ≠ 0). Therefore, 8x⁰ = 8, and the highest power of the variable is 0.
3. What is the difference between linear growth and linear decay?
Answer: Linear growth occurs when a quantity increases by a constant amount over equal intervals. Linear decay occurs when a quantity decreases by a constant amount over equal intervals.
4. What do a and b represent in the equation y = ax + b?
Answer: In the linear equation y = ax + b, a represents the slope of the line, while b represents the y-intercept, which is the point where the line cuts the y-axis.
5. How can you identify parallel lines from their equations?
Answer: Two lines are parallel if they have the same slope but different y-intercepts. Their equations are of the form y = ax + b, where a is the same for both lines, but b is different.
6. Where does the graph of y = ax pass through when b = 0?
Answer: When b = 0, the linear equation becomes y = ax. The graph of any equation in this form always passes through the origin, which has coordinates (0, 0).
7. What does a univariate polynomial mean?
Answer: A univariate polynomial is an algebraic expression that involves only one variable (such as x, y, or z) along with its powers. Examples include x² + 5x + 1.
8. Does a matchstick pattern form a linear relationship?
Answer: Yes, if the number of matchsticks increases by a constant difference at each stage, the pattern forms a linear relationship represented by expressions like 2n – 1 or 5n + 1.
9. What is the difference between quadratic and cubic polynomials?
Answer: A quadratic polynomial has a highest degree of 2, like x² + 5x + 1. A cubic polynomial has a highest degree of 3, such as 5y³ + y² – 8.
10. How many points are needed to draw the graph of a linear equation?
Answer: To plot and draw the graph of a linear equation (straight line) on a coordinate plane, you need to identify at least two points that satisfy the equation.
11. How do you find the value of a polynomial?
Answer: To find the value of a polynomial, substitute the given value of the variable into the polynomial and simplify the expression using the correct order of operations. Example: For p(x) = 5x − 3, when x = 2, p(2) = 5(2) − 3 = 10 − 3 = 7.
