Introduction — Part 1

Chapter 2: Introduction to Linear Polynomials

Page 1 of 6: Introduction — Part 1 — Full Concept Coverage

Welcome to the world of algebra! Before we dive into "polynomials," we need to build a strong foundation. Think of this lesson as learning the alphabet before you start writing words and sentences. We'll explore the basic building blocks of algebra: expressions, variables, and constants.

Imagine you're at a fair. The entry ticket costs a fixed ₹50. Inside, each ride costs ₹20. If you decide to take a few rides, how would you calculate your total spending? You'd take the fixed ticket price (₹50) and add the cost of the rides. If you take x rides, the ride cost is 20 × x. So, your total spending is 20x + 50.

This simple expression, 20x + 50, is an algebraic expression. It's a powerful mathematical sentence that combines numbers, letters, and operations to represent a real-world situation. In this chapter, we will master the art of understanding and building these expressions.

{{FORMULA: expr=ax + by + c | symbols=a,b:coefficients, x,y:variables, c:constant}}

Definitions & Key Concepts

Before we proceed, let's formally define the components we'll be working with. Understanding this vocabulary is crucial for success in algebra.

The Logic: Building an Algebraic Expression

How do these pieces fit together? Let's build an expression from the ground up to understand the logic.

1. Start with Numbers: In mathematics, we begin with numbers we know, like 7, -15, or ½. These are constants because their value is fixed.

2. Introduce the Unknown: We often need to work with quantities that are unknown or can change. We use a symbol, typically a letter like x, to represent this unknown. This is a variable.

3. Form a Term: When we multiply a constant and a variable, we form a term. For example, multiplying the constant 8 by the variable x gives us the term 8x. The constant part, 8, is now called the coefficient of x.

   {{VISUAL: diagram: An illustration breaking down the expression 7x - 4. An arrow from "7x" points to a label "Term". An arrow from the "7" in "7x" points to a label "Coefficient". An arrow from the "x" in "7x" points to a label "Variable". An arrow from "-4" points to labels "Term" and "Constant".}}

4. A Constant is also a Term: It's important to remember that a constant by itself, like -5, is also considered a term. You can think of it as -5x⁰, and since any non-zero number to the power of 0 is 1, it simplifies to -5.

5. Combine Terms: An algebraic expression is created when we connect one or more terms using addition or subtraction. For example, by combining the terms 8x, -2y, and 7, we can form the expression:

   8x - 2y + 7
   

This expression is a mathematical phrase that contains three distinct terms. This process of combining variables and constants allows us to translate real-world problems into a mathematical format we can solve.

{{KEY: type=concept | title=Terms are Building Blocks | text=Remember that terms are the individual parts of an expression separated by addition (+) or subtraction (−) signs. The sign to the left of a number is part of its identity as a coefficient or a constant.}}

Solved Examples

Let's apply these concepts to some problems, starting from easy and moving to more challenging ones.

Example 1: Identifying Components (Easy)

Given: The algebraic expression 9x² - 5y + 7.

To Find:

1. The terms in the expression.
2. The coefficient of x².
3. The coefficient of y.
4. The constant term.

Solution:

1. We look for parts of the expression separated by + or -. The terms are the individual blocks.

   9x², -5y, 7
   

2. The coefficient of x² is the number multiplied by x².

   9
   

3. The coefficient of y is the number multiplied by y. Remember to include the sign.

   -5
   

4. The constant term is the term without any variables.

   7
   

Final Answer: The terms are 9x², -5y, and 7. The coefficient of x² is 9, the coefficient of y is -5, and the constant term is 7.

Example 2: Forming an Expression from Words (Medium)

Given: Rahul's age is x years. His father is 5 years more than twice Rahul's age. His sister is 3 years younger than half of Rahul's age.

To Find: An algebraic expression for the sum of the ages of Rahul, his father, and his sister.

Solution:

1. First, let's write an expression for each person's age.

   • Rahul's age = x
   • Father's age = "5 more than twice Rahul's age" → 2x + 5
   • Sister's age = "3 younger than half of Rahul's age" → (x/2) - 3

2. Now, we need to find the sum of their ages. This means adding the three expressions together.

   Sum = (Rahul's age) + (Father's age) + (Sister's age)
   

3. Substitute the expressions we found in step 1.

   Sum = x + (2x + 5) + (x/2 - 3)
   

4. To simplify, we can group the like terms (terms with x) and the constant terms.

   Sum = (x + 2x + x/2) + (5 - 3)
   

5. Combine the terms to get the final expression.

   Sum = 3.5x + 2
   

   Or, using fractions: (7/2)x + 2

Final Answer: The algebraic expression for the sum of their ages is (7/2)x + 2.

Example 3: Analyzing a Complex Expression (Hard)

Given: The algebraic expression (x²y/3) - 4yz + 0.7z - √5.

To Find:

1. The number of terms.
2. The coefficient of the term containing yz.
3. The coefficient of the term containing only z.
4. The constant term.

Solution:

1. Let's identify the terms, which are separated by + and - signs.

   Terms: (x²y/3), -4yz, 0.7z, -√5
   

   There are four distinct terms.

2. The term containing yz is -4yz. The numerical part is its coefficient.

   Coefficient of yz = -4
   

3. The term containing only z is 0.7z. Its coefficient is the decimal number multiplied by it.

   Coefficient of z = 0.7
   

4. The constant term is the one without any variables. Here, √5 is just a number (an irrational one), so -√5 is the constant.

   Constant term = -√5
   

Final Answer: The expression has 4 terms. The coefficient of yz is -4, the coefficient of z is 0.7, and the constant term is -√5.

Example 4: Finding an Unknown Coefficient (Tricky)

Given: An algebraic expression ky² - 8y + (3k - 1). The sum of all its numerical coefficients and the constant term is equal to 10.

To Find: The value of the variable k.

Solution:

1. First, identify the coefficients and the constant term in the expression ky² - 8y + (3k - 1).

   • Coefficient of y² is k.
   • Coefficient of y is -8.
   • The constant term is the entire expression in the parenthesis, (3k - 1), as it has no y variable attached.

2. The problem states that the sum of these parts equals 10. Let's write this as an equation.

   (Coefficient of y²) + (Coefficient of y) + (Constant term) = 10
   

3. Substitute the values we identified.

   k + (-8) + (3k - 1) = 10
   

4. Now, solve this simple linear equation for k. Combine the k terms and the constant terms on the left side.

   (k + 3k) + (-8 - 1) = 10
   

5. Simplify the equation.

   4k - 9 = 10
   

6. Add 9 to both sides to isolate the term with k.

   4k = 19
   

7. Divide by 4 to find the value of k.

   k = 19/4
   

Final Answer: The value of k is 19/4.

Tips & Tricks

Use these shortcuts to quickly and accurately analyze algebraic expressions.

Common Mistakes to Avoid

Many students make these small errors. Being aware of them is the first step to avoiding them!

Brain-Teaser Questions

Test your understanding with these slightly trickier questions.

1. An expression is formed by multiplying a variable x by itself, then subtracting the product of 5 and another variable y, and finally adding the number π. How many terms does this expression have?

   💡 Answer: The expression is x² - 5y + π. It has three terms: x², -5y, and π. Remember, π is just a constant (approx. 3.14159), not a variable.

2. In the term -(p²q³)/7, what is the numerical coefficient?

   💡 Answer: The term can be rewritten as (-1/7) × p²q³. Therefore, the numerical coefficient is -1/7.

3. If a, b, and c are variables, is the expression (a+b)/c a single term or two terms? Justify your answer.

   💡 Answer: It is a single term. The division operation binds (a+b) and c together. While the numerator (a+b) contains two terms, the entire fraction acts as one unit in a larger expression. For example, in (a+b)/c + d, the two terms are (a+b)/c and d.

Mini Cheatsheet for Revision

Here is a super-compact summary of today's lesson. Screenshot this for last-minute revision!

Introduction — Part 2

Introduction to Linear Polynomials — Part 2

A World Built on Patterns

Have you ever noticed how taxi fares work? There's a fixed booking charge plus a price per kilometre. If you travel x km, the total fare follows a simple formula: base charge + rate × x. That's a linear polynomial at work — one of the most powerful mathematical tools hiding in plain sight around us.

Polynomials are special algebraic expressions where a variable appears with whole-number powers (like x, x², x³...). When we focus on expressions with just one variable, we call them univariate polynomials. The highest power of the variable gives us the degree of the polynomial.

{{VISUAL: diagram:visual comparison showing four cards labeled "degree 0" (just "5"), "degree 1" ("2x + 3"), "degree 2" ("x² - 4x + 7"), "degree 3" ("y³ + y - 9") with the highest power term highlighted in each}}

Classification by Degree

Polynomials are categorized based on their degree:

• Constant polynomials (degree 0): Just a number, like 8 or -5
• Linear polynomials (degree 1): Like 3x + 7 — the kind taxi fares follow
• Quadratic polynomials (degree 2): Such as x² + 5x + 6
• Cubic polynomials (degree 3): For example, 2y³ - y² + 4y - 1

{{KEY: type=concept | title=The Degree Rule | text=The degree is ALWAYS the highest power of the variable that appears with a non-zero coefficient. In 4x³ - 2x + 9, the degree is 3, not "3 + 1 + 0".}}

This chapter zooms in on linear polynomials — the simplest yet most practical type. You'll see them in billing systems, speed calculations, temperature conversions, and hundreds of everyday formulas.

{{VISUAL: diagram:real-world icons (taxi meter, thermometer, shopping cart with price tags) connected by arrows to their corresponding linear polynomial expressions}}

In the next section, we'll crack open linear polynomials and discover why they're called "linear" in the first place.

Word count: 298 ✓

Linear Polynomials — Part 1

Linear Polynomials — Part 1

Concept Introduction

Linear polynomials are the simplest yet most widely used polynomials in mathematics. They form the foundation for understanding relationships that change at a constant rate — a pattern we encounter everywhere in daily life.

Consider a simple scenario: you're planning a birthday party at a gaming arcade. The arcade charges a fixed entry fee of ₹200, plus ₹50 for every game you play. If you play m games, your total cost becomes 200 + 50m. Notice how this expression has only one variable (m) and its highest power is 1. This is a linear polynomial — an algebraic expression of degree 1.

Linear polynomials describe situations where the change is uniform and predictable. When you play one more game, your cost increases by exactly ₹50. When a square's side increases by 0.5 cm, its perimeter increases by exactly 2 cm. This constant rate of change is the hallmark of linear relationships.

In this section, we'll explore how linear polynomials emerge naturally from real-world patterns, learn to recognize their characteristic features, and master techniques to work with them confidently.

{{FORMULA: expr=ax + b | symbols=a:coefficient (rate of change), x:variable, b:constant term}}

Definitions & Formulas

{{KEY: type=concept | title=The Defining Feature | text=In a linear polynomial, the variable appears only to the first power. No x², no √x, no fractions with x in the denominator — just plain x multiplied by a coefficient and added to a constant.}}

Understanding the Logic: Why Linear Polynomials Work

Let's build the concept step-by-step through logical reasoning:

1. Start with the basic form

Every linear polynomial can be written as:

ax + b

where a and b are constants, and a ≠ 0 (because if a = 0, we'd just have the constant b, making it degree 0).

2. Interpret the coefficient

The coefficient a tells us how fast the value changes. For every unit increase in x, the polynomial's value changes by a units. This is the slope or rate of change.

When x increases by 1: value increases by a

3. Interpret the constant term

The constant b is the starting value — the value of the polynomial when x = 0. Substitute x = 0 into ax + b:

a(0) + b = b

4. Recognize the pattern

Create a table of values for 2x + 3:

Notice the constant difference of 2 (which is the coefficient of x). This uniform spacing is unique to linear patterns.

5. Compare with non-linear polynomials

For x² + 1:

The differences keep changing — this is a quadratic pattern, not linear.

6. The input-output perspective

Think of a linear polynomial as a machine: you input x, the machine multiplies it by a, adds b, and outputs the result. This process creates a function — for every input, there's exactly one output.

Solved Examples

Example 1: Finding the Value (Easy)

Given: Linear polynomial 5x - 3

To Find: Value when x = 4

Solution:

1. Write the polynomial and substitute the given value of x.

5x - 3

2. Replace x with 4.

5(4) - 3

3. Perform multiplication first.

20 - 3

4. Subtract to get the final value.

= 17

Final Answer: 17

Example 2: Identifying Pattern Terms (Medium)

Given: A pattern where stage n has 3n + 2 square tiles

To Find: (i) Number of tiles in stage 6, (ii) Which stage has 29 tiles?

Solution:

Part (i): Number of tiles in stage 6

1. Substitute n = 6 into the expression.

3n + 2 = 3(6) + 2

2. Calculate the value.

= 18 + 2 = 20

Answer (i): 20 tiles

Part (ii): Stage with 29 tiles

1. Set the expression equal to 29.

3n + 2 = 29

2. Subtract 2 from both sides.

3n = 27

3. Divide both sides by 3.

n = 9

Answer (ii): Stage 9

Example 3: Real-Life Application (Medium-Hard)

Given: A taxi charges ₹30 as base fare plus ₹12 per kilometer. Ravi paid ₹210.

To Find: How many kilometers did he travel?

Solution:

1. Let the distance traveled be x km. The total fare is represented by the linear polynomial.

Total fare = 12x + 30

2. Set this equal to the amount Ravi paid.

12x + 30 = 210

3. Subtract 30 from both sides to isolate the term with x.

12x = 180

4. Divide both sides by 12.

x = 15

Final Answer: 15 km

Example 4: Age Problem (Hard)

Given: Priya's mother is 3 times Priya's current age. In 5 years, their ages will add up to 70 years.

To Find: Their current ages

Solution:

1. Let Priya's current age be x years. Then mother's current age is 3x years.

Priya = x, Mother = 3x

2. In 5 years, Priya will be x + 5 and mother will be 3x + 5.

Priya in 5 years = x + 5
Mother in 5 years = 3x + 5

3. Their ages will add up to 70 in 5 years.

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

4. Simplify by combining like terms.

4x + 10 = 70

5. Subtract 10 from both sides.

4x = 60

6. Divide by 4.

x = 15

7. Find mother's current age.

Mother = 3x = 3(15) = 45

Final Answer: Priya: 15 years, Mother: 45 years

Tips & Tricks

{{KEY: type=exam_tip | title=Spotting Linear Polynomials | text=Quick test: If you can write it as 'something × x + something else' with no x², x³, √x, or x in denominator, it's linear. The 'something' (coefficient) tells you the rate of change.}}

Common Mistakes

Brain-Teaser Questions

Question 1: A number pattern follows the rule 5n - 7. The 8th term is 33. Verify this, then find which term equals 73.

💡 Answer: For n = 8: 5(8) - 7 = 40 - 7 = 33 ✓ Verified. For the term equaling 73: 5n - 7 = 73 → 5n = 80 → n = 16. The 16th term equals 73.

Question 2: Two linear polynomials 3x + k and 5x - 8 give the same value when x = 2. Find k.

💡 Answer: At x = 2: First polynomial = 3(2) + k = 6 + k. Second polynomial = 5(2) - 8 = 2. Setting equal: 6 + k = 2 → k = -4.

Question 3: A sequence of rectangles has length 2n + 1 and width n. Find the expression for the perimeter. For which value of n is the perimeter 36 units?

💡 Answer: Perimeter = 2(length + width) = 2(2n + 1 + n) = 2(3n + 1) = 6n + 2. For perimeter = 36: 6n + 2 = 36 → 6n = 34 → n = 34/6 = 17/3. Since n must be a whole number for a stage number, no integer stage has perimeter exactly 36. Stage 5 gives 32, stage 6 gives 38.

Mini Cheatsheet

End of Linear Polynomials — Part 1

Linear Polynomials — Part 2

Linear Polynomials — Part 2

Concept Introduction

In the previous section, we learned that linear polynomials are algebraic expressions of degree 1. Now we explore how these polynomials transform into linear equations when we equate them to constants, and how they can be visualized as input-output functions.

Consider a taxi service that charges ₹40 as a base fare plus ₹12 per kilometer. If you travel x kilometers, your total fare becomes 40 + 12x. This is a linear polynomial in x. Now, if you have exactly ₹160 to spend, the question "How far can I travel?" transforms the polynomial into a linear equation: 40 + 12x = 160. By solving this, we find x = 10 km. Notice how the polynomial acts as a function — for every input (distance), it produces an output (fare). This input-output relationship is fundamental to understanding how linear polynomials model real-world situations.

{{FORMULA: expr=ax + b | symbols=a:coefficient of x (rate of change), b:constant term (initial value)}}

{{KEY: type=concept | title=Linear Polynomial as a Function | text=A linear polynomial p(x) = ax + b can be thought of as a machine: input any value of x, and it outputs the value ax + b. This function property makes linear polynomials powerful tools for modeling relationships.}}

Definitions & Formulas

Derivation: From Polynomial to Equation to Solution

Let's understand the logical progression from a linear polynomial to finding its root.

Starting Point: We have a linear polynomial ax + b and need to find when it equals a constant c.

Step 1: Form the Equation

Equate the polynomial to the constant:

ax + b = c

Step 2: Isolate the Variable Term

Subtract the constant term b from both sides to isolate terms containing x:

ax = c - b

Step 3: Solve for the Variable

Divide both sides by the coefficient a (where a ≠ 0):

x = (c - b) / a

Step 4: Verify the Solution

Substitute the value back into the original equation to check:

a × [(c - b) / a] + b = c - b + b = c ✓

This four-step process transforms any linear polynomial equation into its solution. The key insight is that every linear equation has exactly one solution (as long as the coefficient of x is not zero).

Solved Examples

Example 1: Basic Equation Formation (Easy)

Given: The sum of a number and 15 is 42.

To Find: The number.

Solution:

1. Let the unknown number be x. The statement translates to:

x + 15 = 42

2. Subtract 15 from both sides to isolate x:

x = 42 - 15

3. Simplify:

x = 27

Final Answer: 27

Example 2: Input-Output Function Evaluation (Easy-Medium)

Given: Linear polynomial p(x) = 5x - 7.

To Find: Values of p(0), p(-2), and p(4).

Solution:

1. For p(0), substitute x = 0:

p(0) = 5(0) - 7 = -7

2. For p(-2), substitute x = -2:

p(-2) = 5(-2) - 7 = -10 - 7 = -17

3. For p(4), substitute x = 4:

p(4) = 5(4) - 7 = 20 - 7 = 13

Final Answer: p(0) = -7, p(-2) = -17, p(4) = 13

Example 3: Real-World Application (Medium)

Given: A mobile plan charges ₹200 as monthly rental plus ₹2 per minute of call. Rohan's bill for a month was ₹450.

To Find: How many minutes did Rohan talk?

Solution:

1. Let m be the number of minutes. The total bill forms a linear polynomial:

Total Bill = 200 + 2m

2. According to the problem, the bill equals ₹450:

200 + 2m = 450

3. Subtract 200 from both sides:

2m = 450 - 200 = 250

4. Divide both sides by 2:

m = 250 / 2 = 125

5. Verification: 200 + 2(125) = 200 + 250 = 450 ✓

Final Answer: 125 minutes

Example 4: Finding When Two Expressions Are Equal (Medium-Hard)

Given: Two linear polynomials p(x) = 3x + 8 and q(x) = 7x - 4.

To Find: The value of x for which p(x) = q(x).

Solution:

1. Set the two polynomials equal to each other:

3x + 8 = 7x - 4

2. Move all terms containing x to one side by subtracting 3x from both sides:

8 = 7x - 3x - 4

3. Simplify the right side:

8 = 4x - 4

4. Add 4 to both sides:

8 + 4 = 4x

12 = 4x

5. Divide by 4:

x = 12 / 4 = 3

6. Verification: p(3) = 3(3) + 8 = 17 and q(3) = 7(3) - 4 = 17 ✓

Final Answer: x = 3

Example 5: Word Problem with Ratios (Hard)

Given: The ratio of two numbers is 3:5. If their difference is 18, find both numbers.

To Find: The two numbers.

Solution:

1. Let the smaller number be 3x and the larger number be 5x (using ratio multipliers).

2. Their difference is given as 18:

5x - 3x = 18

3. Simplify:

2x = 18

4. Divide by 2:

x = 9

5. Find the actual numbers:

Smaller number = 3x = 3(9) = 27

Larger number = 5x = 5(9) = 45

6. Verification: Ratio = 27:45 = 3:5 ✓ and Difference = 45 - 27 = 18 ✓

Final Answer: 27 and 45

Example 6: Reverse Function Problem (Tricky)

Given: For the linear polynomial f(x) = 4x - 11, we know that f(a) = 25.

To Find: The value of a.

Solution:

1. Write the function equation:

f(a) = 4a - 11

2. We're told this equals 25:

4a - 11 = 25

3. Add 11 to both sides:

4a = 25 + 11 = 36

4. Divide by 4:

a = 36 / 4 = 9

5. Verification: f(9) = 4(9) - 11 = 36 - 11 = 25 ✓

Final Answer: a = 9

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Question 1

A linear polynomial p(x) satisfies p(2) = 7 and p(5) = 16. Find the polynomial p(x).

💡 Answer: The general form is p(x) = ax + b. Using p(2) = 7: 2a + b = 7. Using p(5) = 16: 5a + b = 16. Subtract the first from the second: 3a = 9, so a = 3. Substitute back: 2(3) + b = 7, so b = 1. Therefore p(x) = 3x + 1.

Question 2

If 3(x + 4) = 2(x + 10), find x. Then find the value of the expression 5x - 7.

💡 Answer: Expand: 3x + 12 = 2x + 20. Subtract 2x: x + 12 = 20. So x = 8. Now evaluate 5x - 7 = 5(8) - 7 = 40 - 7 = 33.

Question 3

A linear function f(x) has the property that increasing x by 1 increases f(x) by 4. If f(0) = -3, what is f(10)?

💡 Answer: Since the function increases by 4 for each unit increase in x, the coefficient of x is 4. The function is f(x) = 4x - 3 (since f(0) = -3). Therefore f(10) = 4(10) - 3 = 40 - 3 = 37.

Mini Cheatsheet

{{KEY: type=exam-tip | title=Master the Two Roles | text=Remember: A linear polynomial p(x) = ax + b plays TWO roles — (1) as a FUNCTION that produces outputs for inputs, and (2) as the LEFT side of an EQUATION when you set it equal to something. Practice both!}}

End of Page 4

Exploring linear patterns

Page 5: Exploring Linear Patterns

Introduction

Linear patterns surround us in everyday life, often hidden in plain sight. Consider a growing garden: a gardener plants a row of saplings with equal spacing. Each week, she adds two more saplings to extend the row. After Week 1, there are 3 saplings; after Week 2, there are 5; after Week 3, there are 7. This predictable growth forms a linear pattern — a sequence where each term increases (or decreases) by a constant amount.

Such patterns aren't limited to gardens. Bank balances growing with fixed monthly deposits, water tanks draining at steady rates, taxi fares calculated per kilometre — all follow linear patterns. In mathematics, we represent these using linear expressions, enabling us to predict future values, find unknown terms, and model real-world phenomena systematically.

{{FORMULA: expr=nth term = a + (n - 1)d | symbols=a:first term, n:position of term, d:common difference}}

Core Definitions & Variables

Understanding Linear Patterns: Step-by-Step Logic

Linear patterns follow a systematic structure. Let's break down how to identify and express them mathematically.

1. Identify the pattern type

Examine the sequence of numbers. Check if the difference between consecutive terms remains constant. For example, in 5, 8, 11, 14, ..., the difference is always 3. This confirms a linear pattern.

2. Determine the common difference (d)

Subtract any term from the term immediately following it:

d = (second term) - (first term)

If d > 0, the pattern shows growth. If d < 0, it shows decay.

3. Identify the first term (a)

The first term is simply the value at position n = 1. In practical problems, this might be an initial amount, starting height, or base fare.

4. Establish the general formula

The nth term of any linear pattern can be expressed as:

nth term = a + (n - 1) × d

Simplify this to the form an + b by expanding and combining like terms.

5. Verify with known terms

Substitute n = 1, 2, 3 into your formula and check if the results match the pattern. This validation ensures accuracy.

6. Use the formula for predictions

Once verified, use the formula to find any term's value (forward prediction) or determine which position holds a specific value (reverse calculation).

{{KEY: type=concept | title=Linear Pattern Recognition | text=A sequence is linear if and only if the difference between consecutive terms is constant. This difference becomes the coefficient of n in the general expression.}}

Solved Examples

Example 1: Basic Pattern Recognition (Easy)

Given: A pattern of square tiles grows as follows — Stage 1 has 1 tile, Stage 2 has 3 tiles, Stage 3 has 5 tiles, Stage 4 has 7 tiles.

To Find: The number of tiles in Stage 10 and the general formula for Stage n.

Solution:

1. Calculate the common difference between consecutive stages.

d = 3 - 1 = 2

2. Identify the first term.

a = 1

3. Apply the nth term formula.

nth term = a + (n - 1) × d

nth term = 1 + (n - 1) × 2

4. Simplify the expression.

nth term = 1 + 2n - 2 = 2n - 1

5. Substitute n = 10 to find Stage 10.

Stage 10 = 2(10) - 1 = 19 tiles

Final Answer: General formula: 2n - 1 | Stage 10: 19 tiles

Example 2: Pocket Money Problem (Medium)

Given: Bela starts with ₹100 as pocket money. She spends ₹5 every day.

To Find: (i) Amount left after 8 days, (ii) After how many days will she have ₹40 left?

Solution:

1. Determine the pattern of amount remaining. On Day 0, amount = ₹100. Each day, amount decreases by ₹5 (linear decay).

d = -5 (negative indicates decay)

2. Establish the general expression for amount left on Day n.

Amount left = 100 - 5n

3. Calculate amount after 8 days by substituting n = 8.

Amount = 100 - 5(8) = 100 - 40 = ₹60

4. Find when amount equals ₹40 by setting up an equation.

100 - 5n = 40

5. Solve for n.

5n = 100 - 40 = 60

n = 60 ÷ 5 = 12 days

Final Answer: (i) ₹60 after 8 days | (ii) 12 days to reach ₹40

Example 3: Auto-Rickshaw Fare Calculation (Medium-Hard)

Given: An auto-rickshaw charges ₹25 for the first 2 km, then ₹15 per km thereafter.

To Find: (i) Fare for 10 km, (ii) Distance travelled if fare is ₹130, (iii) General formula for n km (where n ≥ 2).

Solution:

1. Understand the fare structure. First 2 km cost ₹25 (fixed). Each additional km costs ₹15.

2. For n km (where n ≥ 2), additional km travelled = (n - 2).

Fare = 25 + 15 × (n - 2)

3. Simplify the expression.

Fare = 25 + 15n - 30 = 15n - 5

4. Calculate fare for 10 km.

Fare = 15(10) - 5 = 150 - 5 = ₹145

5. Find distance for fare ₹130.

15n - 5 = 130

15n = 135

n = 9 km

Final Answer: (i) ₹145 for 10 km | (ii) 9 km for ₹130 | (iii) Fare = 15n - 5 (n ≥ 2)

Example 4: Rectangular Area Pattern (Tricky)

Given: A rectangle has a fixed length of 13 cm. Its breadth forms a pattern: 12 cm, 10 cm, 8 cm, 6 cm, ...

To Find: (i) Area pattern formula, (ii) Area when breadth is 2 cm, (iii) Which term has breadth 0 cm?

Solution:

1. Identify the breadth pattern. Starting breadth = 12 cm, decreasing by 2 cm each time.

d = -2

2. Establish breadth formula for the nth term.

Breadth = 12 + (n - 1) × (-2) = 12 - 2n + 2 = 14 - 2n

3. Calculate area using Area = Length × Breadth.

Area = 13 × (14 - 2n) = 182 - 26n

4. Find area when breadth = 2 cm. First, find which n gives breadth = 2.

14 - 2n = 2

2n = 12, so n = 6

Area = 182 - 26(6) = 182 - 156 = 26 cm²

5. Find when breadth = 0.

14 - 2n = 0

n = 7

Final Answer: (i) Area = 182 - 26n cm² | (ii) 26 cm² when breadth = 2 cm | (iii) 7th term has breadth 0 cm

Tips & Tricks

{{KEY: type=formula | title=Universal Linear Pattern Formula | text=Any linear pattern can be written as nth term = an + b, where 'a' is the common difference and 'b' is adjusted based on the first term.}}

Common Mistakes

Brain-Teaser Questions

Question 1: A staircase pattern is built with matchsticks. Stage 1 uses 4 matchsticks, Stage 2 uses 7, Stage 3 uses 10. If you have exactly 100 matchsticks, can you complete a stage perfectly? If yes, which stage?

💡 Answer: The pattern is 4, 7, 10, ... with d = 3. Formula: 3n + 1. Set 3n + 1 = 100, giving 3n = 99, so n = 33. Yes, Stage 33 uses exactly 100 matchsticks.

Question 2: A water tank has 500 litres initially. A pump removes 25 litres every hour. Simultaneously, rain adds 10 litres every hour. After how many hours will the tank have 350 litres?

💡 Answer: Net decrease per hour = 25 - 10 = 15 litres. Amount after n hours: 500 - 15n. Set 500 - 15n = 350, giving 15n = 150, so n = 10 hours.

Question 3: In a growing pattern, the 5th term is 23 and the 8th term is 35. Find the first term and the 20th term.

💡 Answer: Common difference d = (35 - 23) ÷ (8 - 5) = 12 ÷ 3 = 4. Using a + 4d = 23, we get a + 16 = 23, so a = 7. Formula: 7 + 4(n - 1) = 4n + 3. 20th term = 4(20) + 3 = 83.

Mini Cheatsheet: Linear Patterns at a Glance

Reflection: Linear patterns form the backbone of algebraic thinking, bridging arithmetic sequences and real-world modelling. Mastering their recognition and representation equips you to tackle diverse problems — from predicting savings growth to optimizing travel costs. As you move forward, notice how many daily scenarios follow these predictable, linear rhythms.

Linear growth and linear decay & Summary

Page 6: Linear Growth, Linear Decay & Chapter Summary

Welcome to the final part of our journey into Linear Polynomials! We've seen how these simple expressions can represent costs, perimeters, and patterns. Now, we'll explore one of their most powerful real-world applications: modeling situations where things grow or shrink at a steady rate. Think of your savings growing each month or your phone battery draining while you play a game. These are examples of linear growth and linear decay.

By the end of this page, you will be able to model real-life scenarios of constant change using linear polynomials and have a complete summary of the entire chapter for quick revision.

{{FORMULA: expr=V = mt + c | symbols=V:Final Value, m:Rate of Change, t:Time/Steps, c:Initial Value}}

Understanding Linear Growth and Decay

In many situations, the change from one step to the next is constant. For example, a plant might grow 2 cm every week. Or a water tank might leak 5 litres every hour. This constant rate of change is the signature of a linear relationship.

• Linear Growth occurs when a quantity increases by the same amount in each unit of time.
• Linear Decay (or decline) occurs when a quantity decreases by the same amount in each unit of time.

Because the change is constant, we can perfectly describe these situations using linear polynomials, where the variable represents time or steps, and the polynomial's value represents the quantity we are measuring.

{{VISUAL: diagram: A simple coordinate plane with two lines starting from the same point on the y-axis. One line goes up to the right, labeled "Linear Growth (m > 0)". The other line goes down to the right, labeled "Linear Decay (m < 0)".}}

Definitions & General Form

The linear polynomial mn + c can be used to model these situations. Here’s what each part means:

Logic: Why the Sign of the Coefficient Matters

How does a simple linear polynomial capture both growth and decay? The secret lies in the sign of the rate of change (m).

1. Let's consider the general form for the value V after n steps:

V = mn + c

2. Here, c is the initial value. It's the amount you start with before any change happens (at n = 0).

3. The term mn represents the total change that has occurred after n steps.

4. Case 1: Growth. If the quantity is increasing, the rate of change m must be positive (m > 0). As n increases (1, 2, 3...), the total change mn also increases, adding to the initial value c.

5. Case 2: Decay. If the quantity is decreasing, the rate of change m must be negative (m < 0). As n increases, the total change mn becomes a larger negative number, subtracting from the initial value c.

{{KEY: type=concept | title=The Sign is the Story | text=The sign of the coefficient m tells you the entire story. A positive m means growth. A negative m means decay. The constant c is always your starting point.}}

Solved Examples

Let's apply this to some problems, starting from easy and moving to more challenging ones.

Example 1: Plant Growth (Easy)

Given: A sapling is 20 cm tall. It grows at a constant rate of 5 cm per month.

To Find: An expression for the plant's height after m months, and its height after 6 months.

Solution:

1. Identify the initial value (c). The sapling starts at 20 cm. So, c = 20.

2. Identify the rate of change (m). It grows by 5 cm per month. Since it's growth, the rate is positive. So, m = +5.

3. Write the linear polynomial. The height H after m months is given by Hm = mn + c.

H = 5m + 20

4. Now, find the height after 6 months by substituting m = 6 into the expression.

H = 5 × 6 + 20

H = 30 + 20 = 50

Final Answer: The expression is 5m + 20, and the plant's height after 6 months will be 50 cm.

Example 2: Leaky Bucket (Medium)

Given: A bucket contains 12 litres of water. It starts leaking at a constant rate of 0.5 litres per minute.

To Find: The amount of water left in the bucket after 15 minutes.

Solution:

1. Identify the initial value (c). The bucket starts with 12 litres. So, c = 12.

2. Identify the rate of change (m). The water is decreasing, so this is decay. The rate is -0.5 litres per minute. So, m = -0.5.

3. Write the linear polynomial for the volume V left after t minutes.

V = -0.5t + 12

Or, more intuitively:

V = 12 - 0.5t

4. Calculate the volume left after 15 minutes by substituting t = 15.

V = 12 - (0.5 × 15)

V = 12 - 7.5

V = 4.5

Final Answer: The amount of water left after 15 minutes is 4.5 litres.

Example 3: Competing Gym Memberships (Hard)

Given:

• Gym A charges a ₹1500 joining fee and ₹500 per month.
• Gym B charges a ₹1000 joining fee and ₹600 per month.

To Find: After how many months will the total cost of both memberships be the same?

Solution:

1. First, create a linear polynomial for the cost of Gym A. The initial value c₁ = 1500 and the rate m₁ = 500. Let n be the number of months.

Cost_A = 500n + 1500

2. Next, create a linear polynomial for the cost of Gym B. The initial value c₂ = 1000 and the rate m₂ = 600.

Cost_B = 600n + 1000

3. To find when the costs are the same, we set the two expressions equal to each other. This creates a linear equation.

500n + 1500 = 600n + 1000

4. Now, solve the equation for n. Move the n terms to one side and constants to the other.

1500 - 1000 = 600n - 500n

500 = 100n

5. Isolate n.

n = 500 ÷ 100 = 5

Final Answer: The total cost for both gym memberships will be the same after 5 months.

Example 4: Car Depreciation (Tricky)

Given: A new car is purchased for ₹8,00,000. After 4 years, its resale value is ₹6,00,000. Assume the value decreases linearly.

To Find: The linear expression for the car's value and its value after 10 years.

Solution:

1. Identify the initial value (c). The starting price is ₹8,00,000. So, c = 800000.

2. This is a tricky step. We are not given the rate of decay (m) directly. We need to calculate it.

   • Total drop in value in 4 years = Initial Value - Value after 4 years.
   • Total drop = 8,00,000 - 6,00,000 = ₹2,00,000.

3. Calculate the rate of decay per year.

   • Rate m = Total Drop ÷ Number of Years
   • m = 2,00,000 ÷ 4 = 50,000 per year.
   • Since this is decay, the rate is negative: m = -50000.

4. Now, write the linear polynomial for the value V after y years.

V = -50000y + 800000

or

V = 800000 - 50000y

5. Finally, find the value after 10 years by substituting y = 10.

V = 800000 - (50000 × 10)

V = 800000 - 500000

V = 300000

Final Answer: The linear expression is V = 800000 - 50000y, and the car's value after 10 years will be ₹3,00,000.

Tips & Tricks

Use these shortcuts to quickly model linear growth and decay problems.

Common Mistakes to Avoid

Many students understand the concept but make small errors in setting up the expression. Here’s what to watch out for.

Brain-Teaser Questions

Ready for a challenge? Test your understanding with these slightly harder problems.

1. Two factories start producing widgets. Factory A starts with a stock of 500 widgets and produces 75 widgets per hour. Factory B starts with a stock of 800 widgets and produces 50 widgets per hour. After how many hours will Factory A have more widgets in stock than Factory B?

   💡 Answer: Let h be the number of hours. Factory A's stock: A(h) = 75h + 500 Factory B's stock: B(h) = 50h + 800 We want to find when A(h) > B(h). Let's first find when they are equal: 75h + 500 = 50h + 800 → 25h = 300 → h = 12. At 12 hours, they are equal. Since Factory A produces faster, it will have more widgets after 12 hours. So, for any time h > 12 hours.

2. A submarine is at a depth of 50 meters below sea level. It descends (goes down) at a rate of 20 meters per minute for 10 minutes. Then, it ascends (goes up) at a rate of 15 meters per minute. What is its depth after a total of 15 minutes from the start?

   💡 Answer: Phase 1 (Descent for 10 min): Depth D = Initial Depth + (Rate × Time). We consider depth below sea level as positive. D₁ = 50 + (20 × 10) = 50 + 200 = 250 meters. Phase 2 (Ascent for 5 min): The ascent starts from the new depth of 250 meters. Ascent reduces the depth. D₂ = 250 - (15 × 5) = 250 - 75 = 175 meters. Its final depth is 175 meters below sea level.

3. The sum of the first n odd numbers (1, 3, 5, ...) is given by the expression n². The sum of the first n even numbers (2, 4, 6, ...) is given by n² + n. A student calculates the sum of the first k odd numbers and the sum of the first k even numbers. The difference between these two sums is 50. What is the value of k?

   💡 Answer: This seems quadratic, but the difference is linear! Sum of evens: E = k² + k Sum of odds: O = k² Difference = E - O = (k² + k) - k² = k The problem states this difference is 50. Therefore, k = 50.

Mini Cheatsheet: Chapter 2 Summary

This table summarizes all the key concepts from our chapter on Linear Polynomials. Screenshot this for your last-minute revision!