In this chapter, we study how to use letters like a, b, c to represent unknown numbers and create mathematical expressions with them. This topic forms the foundation for higher-level algebra and helps students understand patterns, relationships, and problem-solving in a much easier way.
The Notion of Letter-Numbers
Introduction to Letter-Numbers
Letter-numbers are basically letters that we use to represent unknown numbers or variables in mathematics. Think of them as placeholders for numbers we don’t know yet. These letters help us express mathematical relations and patterns in a much more convenient way.
When we write mathematical expressions that contain these letter-numbers, we call them algebraic expressions. Instead of writing long phrases like “Aftab’s age” every time, we can simply use a letter like ‘a’ to represent it. This makes our work much easier and cleaner.
For example:
- If Shabnam is 3 years older than Aftab, we can write: s = a + 3
- Here ‘s’ represents Shabnam’s age and ‘a’ represents Aftab’s age
Basic Concept
The main idea behind letter-numbers is to make mathematical relationships easier to write and understand. When we choose appropriate letters to represent unknown quantities, we can write the mathematical relationship using these letters. The best part is that these expressions can be used to find actual values by substituting real numbers for the letters.
Practical Applications
Letter-numbers have many real-life uses:
• Age relationships: Finding one person’s age when another person’s age is known • Pattern recognition: Expressing sequences and repeated structures
• Cost calculations: Finding total costs when quantities keep changing • Perimeter formulas: Expressing measurements of different geometric shapes
Questions and Answers
Q1: Write formulas for the perimeter of different shapes:
| Shape | Formula | Explanation |
|---|---|---|
| Triangle with all sides equal | P = 3s | s = side length |
| Regular pentagon | P = 5s | s = side length |
| Regular hexagon | P = 6s | s = side length |
Q2: Munirathna has a 20 m long pipe. He joins another pipe of length k meters. What is the expression for combined length?
Answer: Combined length = 20 + k meters
Q3: What is the total amount if Krithika has x notes of ₹100, y notes of ₹20, and z notes of ₹5?
Answer: Total amount = 100x + 20y + 5z
Q4: A flour mill takes 10 seconds to start and 8 seconds per kg to grind. Time to grind y kg?
Answer: Time = 10 + 8y seconds
Q5: Write algebraic expressions for these phrases:
• 5 more than a number: n + 5 • 4 less than a number: n – 4
• 2 less than 13 times a number: 13n – 2 • 13 less than 2 times a number: 2n – 13
Q6: Describe real situations for these expressions:
• 8x + 3y: Cost of 8 items at ₹x each and 3 items at ₹y each • 15j – 2k: 15 times j minus 2 times k (like points gained minus penalty points)
Revisiting Arithmetic Expressions
Rules for Expressions
When working with expressions, there are some imp rules we need to remember:
• Terms can be added in any order (this is called swapping property) • Numbers can be grouped conveniently for easier calculation • Brackets with negative signs need careful handling • Distributive property applies: multiple of a sum equals sum of multiples
Order of Operations
The order in which we solve expressions is very imp:
- Multiplication and division come before addition and subtraction
- Brackets are solved first
- Negative signs outside brackets affect all terms inside
Evaluation Process
Here’s how to evaluate expressions step by step:
• Convert all terms to numbers where possible • Apply swapping and grouping for easier calculation
• Follow proper order of operations • Remember that algebraic expressions take number values when letters are replaced by numbers
Questions and Answers
Q1: Evaluate these arithmetic expressions:
| Expression | Solution | Answer |
|---|---|---|
| 23 – 10 × 2 | 23 – 20 | 3 |
| 83 + 28 – 13 + 32 | (83 + 32) + (28 – 13) = 115 + 15 | 130 |
| 34 – 14 + 20 | 20 + 20 | 40 |
| 42 + 15 – (8 – 7) | 42 + 15 – 1 | 56 |
| 68 – (18 + 13) | 68 – 31 | 37 |
| 7 × 4 + 9 × 6 | 28 + 54 | 82 |
| 20 + 8 × (16 – 6) | 20 + 8 × 10 = 20 + 80 | 100 |
Omission of Multiplication Symbol
Standard Practice
In algebra, we have a special way of writing multiplication. Instead of writing 4 × n, we simply write 4n by removing the multiplication symbol. This is standard practice and makes expressions look much cleaner.
Some imp points to remember: • The number is always written first, followed by the letter • This makes expressions more compact and easier to read • The multiplication is still there, just not written
Examples of Shortened Forms
Here are some examples of how we shorten multiplication:
• 5 × m becomes 5m • 7 × k becomes 7k
• 3 × (a + b) becomes 3(a + b)
Evaluation with Shortened Forms
When we need to find the value: • 5m when m = 2 means 5 × 2 = 10 • 7k when k = 4 means 7 × 4 = 28 • Always remember that multiplication is still implied even though we don’t write the symbol
Questions and Answers
Q1: Find values when letter-numbers are replaced:
| Expression | Given Value | Calculation | Answer |
|---|---|---|---|
| 10 – a | a = -4 | 10 – (-4) = 10 + 4 | 14 |
| 3d | d = 6 | 3 × 6 | 18 |
| 3s – 2 | s = 7 | 3 × 7 – 2 = 21 – 2 | 19 |
| 2r + 1 | r = 8 | 2 × 8 + 1 = 16 + 1 | 17 |
| 2j | j = 5 | 2 × 5 | 10 |
| 3(m + 1) | m = -6 | 3(-6 + 1) = 3(-5) | -15 |
| 2f – 2g | f = 3, g = 1 | 2(3) – 2(1) = 6 – 2 | 4 |
| 2t + b | t = 4, b = 3 | 2(4) + 3 = 8 + 3 | 11 |
| h – (3 – n) | h = 5, n = 6 | 5 – (3 – 6) = 5 – (-3) = 5 + 3 | 8 |
Simplification of Algebraic Expressions
Like Terms and Unlike Terms
This is a very imp concept in algebra. We need to understand the difference between like terms and unlike terms.
Like terms: Terms that have the same letter-numbers • Examples: 5c, 3c, 10c (all have ‘c’) • Examples: 2x, 7x, -4x (all have ‘x’)
Unlike terms: Terms that have different letter-numbers
• Examples: 18c and 11d (different letters) • Examples: 5x and 3y (different letters)
The imp rule is: Only like terms can be combined and simplified
Combining Like Terms
When we combine like terms, we add their coefficients: • 5c + 3c + 10c = (5 + 3 + 10)c = 18c • 7x – 2x + 4x = (7 – 2 + 4)x = 9x
This uses the distributive property in reverse form.
Distributive Property in Algebra
The distributive property works like this: • a(b + c) = ab + ac • This works with letter-numbers just like regular numbers • It’s useful for both expanding and factoring expressions
Simplification Steps
Follow these steps to simplify expressions:
- Remove brackets using distributive property
- Group like terms together
- Combine coefficients of like terms
- Write in simplest form
Working with Brackets
When dealing with brackets: • (a + b) – (c + d) = a + b – c – d • Negative sign outside bracket changes signs of all terms inside • Terms can be rearranged in any order for easier grouping
Questions and Answers
Q1: Simplify these expressions:
| Expression | Simplified Form |
|---|---|
| p + p + p + p | 4p |
| p + p + p + q | 3p + q |
| p + q + p – q | 2p |
| p – q + p – q | 2p – 2q |
| p + q – p + q | 2q |
| p + q – (p + q) | 0 |
| p – q – p – q | -2q |
| 2d – d – d – d | -d |
| 2d – d – d – c | -c |
| 2d – d – (d – c) | c |
| 2d – (d – d) – c | 2d – c |
| 2d – d – c – c | d – 2c |
Q2: Correct the mistakes in these expressions:
| Expression | Given Answer | Correct Answer | Explanation |
|---|---|---|---|
| 3a + 2b = 5 | Wrong | 3a + 2b | Cannot add unlike terms |
| 3b – 2b – b = 0 | Correct | 0 | Like terms combined properly |
| 6(p + 2) = 6p + 8 | Wrong | 6p + 12 | Distributive property error |
| (4x + 3y) – (3x + 4y) = x + y | Wrong | x – y | Sign error when removing brackets |
| 5 – (2 – 6z) = 3 – 6z | Wrong | 3 + 6z | Negative sign affects all terms |
| 2 + (x + 3) = 2x – 6 | Wrong | x + 5 | Simple addition error |
| 2y + (3y – 6) = -y + 6 | Wrong | 5y – 6 | Like terms not combined |
| 7p – p + 5q – 2q = 7p + 3q | Wrong | 6p + 3q | Coefficient calculation error |
| 5(2w + 3x + 4w) = 10w + 15x + 20w | Wrong | 30w + 15x | Like terms not combined |
| 3j + 6k + 9h + 12 = 3(j + 2k + 3h + 4) | Correct | 3(j + 2k + 3h + 4) | Factoring done correctly |
| 4(2r + 3s + 5) – 20 = -8r – 12s | Wrong | 8r + 12s | Sign and calculation errors |
Pick Patterns and Reveal Relationships
Formula Detective
Number machines are like puzzles that perform the same operations on different inputs. To solve them: • Find the pattern by studying input-output relationships • Express the pattern as an algebraic formula • Verify that your formula works for all given examples
Algebraic Expressions for Patterns
Repeating patterns can be described using algebraic expressions very nicely: • Position of nth occurrence can be found using formulas • Remainder after division helps identify pattern elements • This makes predicting future values much easier
Calendar Patterns
Calendar grids show some really interesting patterns. When we take any 2×2 square from a calendar:
• Diagonal sums are always equal in any 2×2 square • If top-left number is ‘a’, then:
- Top-right: a + 1
- Bottom-left: a + 7
- Bottom-right: a + 8 • Both diagonal sums equal 2a + 8
Matchstick Patterns
Sequential patterns can be expressed algebraically. For example: • Pattern: Step 1 has 3 matchsticks, each new step adds 2 • Formula for step y: 2y + 1 or 3 + 2(y – 1) • Both forms are equivalent when simplified
Imp Benefits of Algebraic Modeling
Using algebra to model patterns has many benefits: • Verifies if patterns hold for all cases • Avoids need to check infinite examples • Provides general proofs for mathematical relationships • Makes predictions for large numbers without counting
Practical Problem Solving
Questions and Answers
Q1: Food cost problem – Jowar roti costs ₹30, Pulao costs ₹20. Which expression describes total earnings for x roti and y pulao?
Answer: (a) 30x + 20y
Q2: Flower shop problem – p customers bought only champak, q bought only marigold, r bought both. How many flags were given?
Answer: (a) p + q + r
Q3: Snail climbing problem – climbs u cm during day, slips d cm at night for 10 days:
• Distance from start: 10u – 10d = 10(u – d) • If d > u: Snail moves backward overall
Q4: Cycling practice – starts with 5 km daily, increases by z km each week. Distance after 3 weeks?
| Week | Daily Distance | Weekly Distance |
|---|---|---|
| Week 1 | 5 km | 7 × 5 = 35 km |
| Week 2 | 5 + z km | 7 × (5 + z) = 35 + 7z km |
| Week 3 | 5 + 2z km | 7 × (5 + 2z) = 35 + 14z km |
Total: 35 + 35 + 7z + 35 + 14z = 105 + 21z km
Q5: Train journey – travels time t between stations, stops 2 minutes at each of 3 stations:
• If t = 4: Total time = 4 × 4 + 2 × 3 = 16 + 6 = 22 minutes • General expression: 4t + 6 minutes
Q6: Simplify these expressions:
| Expression | Simplified Form |
|---|---|
| 3a + 9b – 6 + 8a – 4b – 7a + 16 | 4a + 5b + 10 |
| 3(3a – 3b) – 8a – 4b – 16 | a – 13b – 16 |
| 2(2x – 3) + 8x + 12 | 12x + 6 |
| 8x – (2x – 3) + 12 | 6x + 15 |
| 8h – (5 + 7h) + 9 | h + 4 |
| 23 + 4(6m – 3n) – 8n – 3m – 18 | 21m – 20n + 5 |
Q7: Add these expressions:
| First Expression | Second Expression | Sum |
|---|---|---|
| 4d – 7c + 9 | 8c – 11 + 9d | 13d + c – 2 |
| -6f + 19 – 8s | -23 + 13f + 12s | 7f + 4s – 4 |
| 8d – 14c + 9 | 16c – (11 + 9d) | -d + 2c – 2 |
| 6f – 20 + 8s | 23 – 13f – 12s | -7f – 4s + 3 |
| 13m – 12n | 12n – 13m | 0 |
| -26m + 24n | 26m – 24n | 0 |
Q8: Subtract these expressions:
| From | Subtract | Result |
|---|---|---|
| 6a + 9b – 18 | 9a – 6b + 14 | -3a + 15b – 32 |
| 7y – 10 + 3x | -15x + 13 – 9y | 18x + 16y – 23 |
| 11 – 10g + 3h | 17g + 9 – 7h | -27g + 10h + 2 |
| 6a – (9b + 18) | 9a – 6b + 14 | -3a – 3b – 32 |
| -3y + 8 – 3x | 10x + 2 + 10y | -13x – 13y + 6 |
| 7h – 8g + 20 | 8g + 4h – 10 | -16g + 3h + 30 |
Q9: Pattern problems:
• Rope cutting: If rope is folded r times and cut once, pieces = r + 2 • Matchstick squares: For w squares, matchsticks needed = 4w • Traffic signal: Red-Yellow-Green pattern repeats every 3 positions
| Position | Calculation | Color |
|---|---|---|
| 90 | 90 ÷ 3 = 30 remainder 0 | Green |
| 190 | 190 ÷ 3 = 63 remainder 1 | Red |
| 343 | 343 ÷ 3 = 114 remainder 1 | Red |
Q10: Square pattern:
| Step | Number of Squares | Calculation |
|---|---|---|
| Step 4 | 16 | 4 × 4 = 16 |
| Step 10 | 100 | 10 × 10 = 100 |
| Step 50 | 2500 | 50 × 50 = 2500 |
| Step n | n² | General formula |
Vertices formula: 4n² vertices
Q11: Number grid pattern:
| Column | Formula | Example |
|---|---|---|
| Column 1 | 4n – 3 | Row 1: 4(1) – 3 = 1 |
| Column 2 | 4n – 2 | Row 1: 4(1) – 2 = 2 |
| Column 3 | 4n – 1 | Row 1: 4(1) – 1 = 3 |
| Column 4 | 4n | Row 1: 4(1) = 4 |
Finding positions: • Number 124: Row 31, Column 4 • Number 147: Row 37, Column 3
• Number 201: Row 51, Column 1
Imp Concepts
Letter-numbers represent unknown quantities in mathematical expressions and make our calculations much easier. Algebraic expressions follow the same rules as arithmetic expressions, which means we can apply everything we learned about numbers to algebra too.
Like terms can be combined, but unlike terms cannot be combined. This is a very imp rule to remember. Patterns and relationships become much easier to express and verify when we use algebraic expressions.
Practical Uses
The uses of letter-numbers are everywhere around us:
• Modeling real-world situations with variable quantities
• Creating formulas for geometric measurements
• Expressing mathematical patterns and sequences
• Solving problems involving unknown values
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