Finding the Unknown

CHAPTER 7: FINDING THE UNKNOWN

7.1 Find the Unknowns

What is an Unknown?

An **unknown** is a quantity whose value we need to find. We often represent unknowns using **letter-numbers** (like x, y, n, etc.) to help us solve problems systematically.

Finding Unknowns Using Weighing Scales

A **weighing scale** shows us that if both sides are balanced, they have equal weight. This principle helps us find unknowns.

**Key Principle:** When we remove equal weights from both plates of a balanced weighing scale, the scale remains balanced. This same idea applies to equations.

**Real-life Example 1 (Market):** A vegetable seller uses a balance scale. On one side: 3 identical bags of potatoes. On the other side: weights totaling 12 kg. Each bag weighs: 12 ÷ 3 = 4 kg. We found the unknown weight!

**Real-life Example 2 (Cooking):** In a recipe, if 2 cups of sugar balance with 400 grams on a scale, then 1 cup of sugar = 400 ÷ 2 = 200 grams.

Finding Unknowns in Patterns

Sometimes unknowns appear in **patterns**. For matchstick arrangements:

Position 1: 2(1) + 1 = 3 matchsticks

Position 2: 2(2) + 1 = 5 matchsticks

Position 3: 2(3) + 1 = 7 matchsticks

Position n: 2n + 1 matchsticks

If we need to find which position uses 99 matchsticks, we need: 2n + 1 = 99

What is an Equation?

An **equation** is a statement of equality between two algebraic expressions, separated by an **equal sign (=)**.

**Examples of equations:**

3x + 4 = 7

20 = y – 3

a/3 = 50

2z + 4 = 5z – 14

**Parts of an equation:**

**Left Hand Side (LHS):** The expression on the left of the =

**Right Hand Side (RHS):** The expression on the right of the =

**Example:** In 2n + 1 = 99

LHS = 2n + 1

RHS = 99

Solving an Equation

**Solving** means finding the value(s) of the letter-number for which LHS = RHS.

**Real-life Example (Sports):** A cricket player scores some runs in the first match and 45 runs in the second match. If total runs = 95, then: runs₁ + 45 = 95, so runs₁ = 50 runs.

---

7.2 Solving Equations Systematically

Method 1: Trial and Error

**Trial and Error** means substituting different values until we find the one that makes LHS = RHS.

**Example:** Solve 2n + 1 = 99

Try n = 5: LHS = 2(5) + 1 = 11 ✗ (too small)

Try n = 30: LHS = 2(30) + 1 = 61 ✗ (still too small)

Try n = 49: LHS = 2(49) + 1 = 99 ✓ (equals RHS!)

Solution: n = 49

**Problem with this method:** It can be very inefficient and time-consuming, especially with large numbers.

Method 2: Using Properties of Equations (Systematic Approach)

**Key Property of Equations:** If we perform the same operation (add, subtract, multiply, or divide) on both sides of an equation, the equality is preserved.

**Mathematical Principle:** Since LHS = RHS, any operation done to both sides keeps them equal.

**Three Important Rules:**

#### Rule 1: Additive Inverse (Removing Addition/Subtraction Terms)

When a term is added or subtracted on one side, its **additive inverse** appears on the other side when we remove it.

**Formula:** If a + b = c, then a = c – b

**Example:** Solve 2y + 7 = 21

Subtract 7 from both sides: 2y + 7 – 7 = 21 – 7

Simplify: 2y = 14

Therefore: y = 14 ÷ 2 = 7

**Check:** 2(7) + 7 = 14 + 7 = 21 ✓

#### Rule 2: Multiplicative Inverse (Removing Multiplication Factors)

When one side is a product and we remove a factor, we divide the other side by that factor.

**Formula:** If ax = b, then x = b ÷ a

**Example:** Solve 5x = 25

Divide both sides by 5: 5x ÷ 5 = 25 ÷ 5

Simplify: x = 5

**Check:** 5(5) = 25 ✓

#### Rule 3: Multiplicative Inverse (Removing Division)

When one side is a quotient and we remove the divisor, we multiply the other side by that divisor.

**Formula:** If x/a = b, then x = b × a

**Example:** Solve u/15 = 6

Multiply both sides by 15: (u/15) × 15 = 6 × 15

Simplify: u = 90

**Check:** 90/15 = 6 ✓

Step-by-Step Procedure for Solving Equations

**Steps:**

1. Identify what we need to find (the unknown)

2. Use inverse operations to isolate the unknown on one side

3. Simplify step by step

4. Find the value of the unknown

5. **Always check** your solution by substituting back

Worked Examples

**Example 1:** Solve 3x – 10 = 35

```

3x – 10 = 35

Add 10 to both sides: 3x – 10 + 10 = 35 + 10

3x = 45

Divide by 3: x = 45 ÷ 3

x = 15

Check: 3(15) – 10 = 45 – 10 = 35 ✓

```

**Example 2:** Solve 5s = 3s

```

5s = 3s

Subtract 3s from both sides: 5s – 3s = 3s – 3s

2s = 0

s = 0

Check: 5(0) = 0 and 3(0) = 0, so 0 = 0 ✓

```

**Example 3:** Solve 3u – 7 = 2u + 3

```

3u – 7 = 2u + 3

Subtract 2u from both sides: 3u – 2u – 7 = 2u – 2u + 3

u – 7 = 3

Add 7 to both sides: u – 7 + 7 = 3 + 7

u = 10

Check: 3(10) – 7 = 30 – 7 = 23, and 2(10) + 3 = 20 + 3 = 23 ✓

```

**Example 4:** Solve 4(m + 6) – 8 = 2m – 4

```

4(m + 6) – 8 = 2m – 4

Expand: 4m + 24 – 8 = 2m – 4

4m + 16 = 2m – 4

Subtract 2m: 4m – 2m + 16 = 2m – 2m – 4

2m + 16 = – 4

Subtract 16: 2m + 16 – 16 = –4 – 16

2m = –20

Divide by 2: m = –10

Check: 4(–10 + 6) – 8 = 4(–4) – 8 = –16 – 8 = –24

and 2(–10) – 4 = –20 – 4 = –24 ✓

```

**Example 5:** Solve u/15 = 6

```

u/15 = 6

Multiply both sides by 15: (u/15) × 15 = 6 × 15

u = 90

Check: 90/15 = 6 ✓

```

Equations with Unknowns on Both Sides

**Strategy:** Move all unknown terms to one side and all constants to the other side.

**Example:** Solve 6y + 7 = 4y + 21

```

Step 1: Subtract 4y from both sides (to get unknowns on one side)

6y – 4y + 7 = 4y – 4y + 21

2y + 7 = 21

Step 2: Subtract 7 from both sides (to isolate the unknown term)

2y + 7 – 7 = 21 – 7

2y = 14

Step 3: Divide by 2

y = 14 ÷ 2 = 7

Check: 6(7) + 7 = 42 + 7 = 49, and 4(7) + 21 = 28 + 21 = 49 ✓

```

---

Solving Real-World Problems Using Equations

Example 1: Pattern Problems

**Problem:** Ranjana creates a sequence of square tile arrangements. In Step k, the number of tiles needed is 3k + 1. Can she make an arrangement using exactly 100 tiles? If yes, which step?

**Solution:**

```

Set up equation: 3k + 1 = 100

Subtract 1: 3k = 100 – 1

3k = 99

Divide by 3: k = 99 ÷ 3

k = 33

Check: 3(33) + 1 = 99 + 1 = 100 ✓

Answer: Yes, it will be in Step 33.

```

Example 2: Shopping/Party Planning Problem

**Problem:** Madhubanti wants to buy snacks for a party. Each plate costs ₹25. Delivery charge is ₹50 (fixed). She has ₹500 total. There are 5 family members. How many friends can she invite?

**Method 1 (Arithmetic):**

```

Total cost = (Number of plates × 25) + 50

Out of ₹500, subtract delivery charge: 500 – 50 = ₹450

Number of plates = 450 ÷ 25 = 18

Number of friends = 18 – 5 family members = 13 friends

```

**Method 2 (Using Equation - Let p = total people):**

```

25p + 50 = 500

Subtract 50: 25p = 450

Divide by 25: p = 18

Friends = 18 – 5 = 13

```

**Method 3 (Using Equation - Let f = number of friends):**

```

25(f + 5) + 50 = 500

25f + 125 + 50 = 500

25f + 175 = 500

Subtract 175: 25f = 325

Divide by 25: f = 13 friends

```

Example 3: Savings Problem

**Problem:** Jahnavi starts with ₹4000 and saves ₹650/month. Sunita starts with ₹5050 and saves ₹500/month. After how many months will they have equal savings?

**Solution:**

```

Let m = number of months

Jahnavi's savings after m months = 4000 + 650m

Sunita's savings after m months = 5050 + 500m

Set them equal:

4000 + 650m = 5050 + 500m

Subtract 500m from both sides:

4000 + 650m – 500m = 5050

4000 + 150m = 5050

Subtract 4000 from both sides:

150m = 5050 – 4000

150m = 1050

Divide by 150:

m = 1050 ÷ 150 = 7 months

Check:

Jahnavi: 4000 + 650(7) = 4000 + 4550 = 8550 ✓

Sunita: 5050 + 500(7) = 5050 + 3500 = 8550 ✓

```

Example 4: Multi-Step Problem

**Problem:** Solve 28(x + 4) + 300 = 1000

**Method 1 (Standard approach):**

```

28(x + 4) + 300 = 1000

Subtract 300: 28(x + 4) = 700

Divide by 28: x + 4 = 700 ÷ 28 = 25

Subtract 4: x = 21

```

**Method 2 (Divide by common factor first):**

```

28(x + 4) + 300 = 1000

All terms divisible by 4, so divide all by 4:

7(x + 4) + 75 = 250

7(x + 4) = 175

x + 4 = 25

x = 21

```

**Method 3 (Expand first):**

```

28(x + 4) + 300 = 1000

28x + 112 + 300 = 1000

28x + 412 = 1000

28x = 588

x = 588 ÷ 28 = 21

```

Example 5: Magic Trick Problem

**Problem:** Riyaz's magic trick:

1. Think of a number (x)

2. Subtract 3 → (x – 3)

3. Multiply by 4 → 4(x – 3) = 4x – 12

4. Add 8 → 4x – 12 + 8 = 4x – 4

Akash's final answer: 24. What was his starting number?

**Solution:**

```

4x – 4 = 24

4(x – 1) = 24

Divide by 4: x – 1 = 6

Add 1: x = 7

Check:

Start: 7

Subtract 3: 7 – 3 = 4

Multiply by 4: 4 × 4 = 16

Add 8: 16 + 8 = 24 ✓

```

**Rule discovered:** Starting number = (Final answer + 4) ÷ 4

Example 6: Problem with Two Unknowns

**Problem:** Ramesh and Suresh have 60 marbles between them. Ramesh has 30 more marbles than Suresh. How many does each have?

**Strategy:** Express one unknown in terms of the other.

**Solution:**

```

Let y = marbles with Suresh

Then Ramesh has = y + 30

Total: y + (y + 30) = 60

2y + 30 = 60

2y = 30

y = 15

Suresh has 15 marbles

Ramesh has 15 + 30 = 45 marbles

Check: 15 + 45 = 60 ✓

```

---

Generating Equations

What is Generating an Equation?

Instead of solving a given equation, we can **create equations** that have a specific solution.

**Example:** Write equations whose solution is y = 5

Equation 1: y + 1 = 6

Equation 2: 3y = 15

Equation 3: 2y – 5 = 5

Equation 4: y/5 = 1

Equation Chains

We can create chains of equations by performing the same operation on both sides:

**Chain 1 (Starting from y + 1 = 6):**

```

y + 1 = 6

Multiply by –1: –y – 1 = –6

Add y: –1 = –6 + y

Add 6: 5 = y

```

**Chain 2 (Starting from 3y = 15):**

```

3y = 15

Add 6: 3y + 6 = 21

Divide by 3: y + 2 = 7

Subtract 2: y = 5

```

**Key Observation:** When going from top to bottom, if we use an operation, the reverse operation takes us from bottom to top.

---

Important Concepts to Remember

1. Balance Principle

Just like a weighing scale:

If we remove equal weights from both sides, it stays balanced

If we add equal weights to both sides, it stays balanced

**Same applies to equations:** Do the same thing to both sides, equality is preserved

2. Inverse Operations

**Addition ↔ Subtraction**

**Multiplication ↔ Division**

Use inverse operations to **isolate** the unknown

3. Common Mistakes to Avoid

❌ Forgetting to perform the same operation on BOTH sides

❌ Making arithmetic errors when simplifying

❌ Not checking the final answer

❌ Losing track of negative signs

❌ Forgetting to simplify before solving

4. Always Check Your Answer

Substitute the solution back into the original equation:

If LHS = RHS, your solution is correct ✓

If LHS ≠ RHS, your solution is wrong ✗

5. Types of Equations

**Linear equations:** Highest power of unknown is 1 (e.g., 2x + 5 = 11)

**Equations with unknowns on both sides:** (e.g., 3x + 2 = x + 8)

**Equations with brackets:** Expand first (e.g., 2(x – 3) = 10)

**Equations with fractions:** Multiply to remove fractions (e.g., x/5 = 3)

---

Real-Life Applications

In Markets

Finding price of individual items when total cost is known

Calculating change

Determining discounts

In Sports

Finding average scores

Calculating total points

Determining goals needed to reach a target

In Daily Life

Party planning (snacks, decorations)

Sharing equally (marbles, candies)

Time and distance problems

Comparing savings or investments

---

Summary Table of Solving Techniques

| Operation on One Side | Remove By | Example |

|---|---|---|

| + (addition) | Subtract | 2x + 5 = 11 → 2x = 6 |

| − (subtraction) | Add | 3x − 2 = 10 → 3x = 12 |

| × (multiplication) | Divide | 5x = 20 → x = 4 |

| ÷ (division) | Multiply | x/3 = 4 → x = 12 |

| Brackets | Expand or divide | 2(x + 3) = 10 → 2x + 6 = 10 |

---

Practice Tips

1. **Read carefully:** Identify what the unknown represents

2. **Frame the equation:** Translate words into mathematical symbols

3. **Solve systematically:** Use inverse operations step by step

4. **Simplify:** Combine like terms before solving

5. **Check always:** Substitute back to verify

6. **Practice different types:** Mix up equation types in your practice

7. **Show all steps:** Don't skip steps, even if they seem obvious

---

Key Formulas and Rules

**Equation:** LHS = RHS (where LHS and RHS must be equal)

**Solving Rule 1:** If a + b = c, then a = c – b

**Solving Rule 2:** If a − b = c, then a = c + b

**Solving Rule 3:** If ax = b, then x = b/a (a ≠ 0)

**Solving Rule 4:** If x/a = b, then x = ab (a ≠ 0)

**Linear Pattern:** If position = n, number of items = an + b (general form)

**Checking:** Always substitute the value back: Does LHS = RHS?