Expression using letter number class 7

Unlock the Secrets of Algebra: The Ultimate Guide to Expressions

Welcome to your complete guide for Chapter 4! Algebra isn’t about complicated math; it’s a superpower that lets you describe rules, find patterns, and solve puzzles in a simple and elegant way. Let’s master it together.

1. The Notion of Letter-Numbers

🔑 Key Concepts

Variable (or Letter-Number): A letter (like a, x, or n) that acts as a placeholder for a number that can change or is unknown.
Algebraic Expression: A mathematical phrase that combines numbers, variables, and operations (+, -, ×, ÷). For example, a + 3 is an algebraic expression.
Formula: A special type of algebraic expression that describes a relationship between quantities, like the formula for a perimeter.

Example: Aftab and Shabnam’s Age (Pages 81-82)

Problem: Shabnam is 3 years older than Aftab. Express this relationship using algebra and solve for specific ages.

Step 1: Define the Relationship. The core rule is: Shabnam’s Age = Aftab’s Age + 3.

Step 2: Assign Variables. Let Aftab’s age be a and Shabnam’s age be s.

Step 3: Write the Expression. The relationship becomes the formula: s = a + 3.

Step 4: Solve for Given Values.

If Aftab (a) is 18, Shabnam (s) is 18 + 3 = 21.
If Aftab (a) is 23, Shabnam (s) is 23 + 3 = 26.
Conversely, if Shabnam (s) is 20, we can find Aftab’s age with the expression a = s - 3. So, Aftab (a) is 20 – 3 = 17.

2. Building Expressions from Situations

💡 Pro-Tip: Translating Words to Algebra

Look for keywords! “More than” or “sum” usually means addition (+). “Less than” or “difference” means subtraction (-). “Times” or “product” means multiplication (×). “Per” or “divided by” means division (÷).

Figure it Out (Page 84)

1. Perimeter Formulas:

(a) Equilateral Triangle (side `s`): Perimeter = `3s`
(b) Regular Pentagon (side `L`): Perimeter = `5L`
(c) Regular Hexagon (side `x`): Perimeter = `6x`

2. Munirathna’s Pipe: His new pipe’s length is 20 + k metres.

4. Venkatalakshmi’s Flour Mill: The total time is a fixed start-up time (10s) plus a variable time that depends on the weight (`y` kg). The variable part is `8 × y`. The correct expression is `10 + 8y` (Option d).

5. Expressions from Phrases:

(a) 5 more than a number `n`: `n + 5`
(b) 4 less than a number `x`: `x – 4`
(c) 2 less than 13 times a number `y`: `13y – 2`
(d) 13 less than 2 times a number `p`: `2p – 13`

3. Simplifying Expressions: The Power of Like Terms

🔑 Key Concepts

Term: Parts of an expression separated by addition (+) or subtraction (-) signs. In 5x + 3y - 7, the terms are `5x`, `3y`, and `-7`.
Like Terms: Terms that have the exact same variable part. 5x and -2x are like terms. 3a and 10a are like terms.
Unlike Terms: Terms with different variable parts. 5x and 5y are unlike terms. You cannot add or subtract them.
Simplifying: The process of combining all like terms in an expression to make it as short and simple as possible.

Mind the Mistake, Mend the Mistake (Page 87)

This exercise reveals common errors when substituting values. The key is to remember that a number next to a variable implies multiplication and to follow the order of operations.

Problem	Common Mistake	Correct Solution & Value
1. `10 – a` if `a=-4`	10 – 4 = 6. (Forgetting to subtract a negative is addition).	10 – (-4) = 10 + 4 = 14
2. `3d` if `d=6`	36. (Just placing the digits together).	3 × d = 3 × 6 = 18
3. `3s – 2` if `s=7`	3(7-2) = 15. (Wrong order of operations).	(3 × 7) – 2 = 21 – 2 = 19
6. `3(m+1)` if `m=-6`	3(-6)+1 = -17. (Incorrect distribution).	3(-6 + 1) = 3(-5) = -15
9. `h-(3-n)` if `h=5, n=6`	5 – (3-6) = 5 – 3 = 2. (Sign error).	5 – (3 – 6) = 5 – (-3) = 5 + 3 = 8

Simplification Exercises (Page 103)

(a) `3a + 9b – 6 + 8a – 4b – 7a + 16`

Combine ‘a’ terms: `3a + 8a – 7a = 4a`
Combine ‘b’ terms: `9b – 4b = 5b`
Combine constants: `-6 + 16 = 10`
Result: `4a + 5b + 10`

(d) `8x – (2x – 3) + 12`

Remove brackets (note the sign change): `8x – 2x + 3 + 12`
Combine ‘x’ terms: `8x – 2x = 6x`
Combine constants: `3 + 12 = 15`
Result: `6x + 15`

4. Using Algebra to Reveal Patterns

Patterns in a Calendar (Pages 97-99)

Problem: Prove that in any 2×2 square on a calendar, the sums of the numbers on the diagonals are always equal.

Step 1: Represent the square with algebra. Let the top-left number be any number `a`.

The number to its right is `a + 1`.
The number below it is `a + 7`.
The number diagonal to it is `a + 8`.

Step 2: Write expressions for the diagonal sums.

Diagonal 1 Sum: `a + (a + 8)`
Diagonal 2 Sum: `(a + 1) + (a + 7)`

Step 3: Simplify both expressions.

Diagonal 1: `a + a + 8 = 2a + 8`
Diagonal 2: `a + 1 + a + 7 = 2a + 8`

Conclusion: ✅ Since both expressions simplify to `2a + 8`, the diagonal sums are always equal, no matter which 2×2 square you choose! This shows the power of algebra to prove a pattern is always true.

Matchstick Patterns (Pages 100-101)

Problem: Find a formula for the number of matchsticks (`m`) needed to make a chain of `y` triangles.

Step 1: Analyze the pattern.

Step 1: 1 triangle, 3 matchsticks.
Step 2: 2 triangles, 5 matchsticks.
Step 3: 3 triangles, 7 matchsticks.

We start with 3 matchsticks, and add 2 more for every new triangle.

Step 2: Find the formula.

Method 1: The first triangle needs 3. The remaining `(y – 1)` triangles each need 2 matchsticks.
Formula: `m = 3 + 2(y – 1)`. Simplifying this gives `m = 3 + 2y – 2 = 2y + 1`.

Method 2: Notice the number of matchsticks is always one more than twice the step number. (Step 1: 2(1)+1=3; Step 2: 2(2)+1=5).
Formula: `m = 2y + 1`.

Step 3: Solve for given values.

For Step 33: `m = 2(33) + 1 = 66 + 1 = 67` matchsticks.
For Step 84: `m = 2(84) + 1 = 168 + 1 = 169` matchsticks.
For Step 108: `m = 2(108) + 1 = 216 + 1 = 217` matchsticks.