Solving Equations, Variable On One Sides

Unit: Algebra – 1

Chapter: Solving Equations, Variable on One Side

Reference: – Introduction to Linear Equations, what is a Variable, what is an Equation, Solving Equations with Variable on One Side, Balancing Method, Transposition Method, Verification of Solution, Equations with Fractions, Equations with Decimals, Word Problems, Solved Examples, Odd-One-Out Problems, Common Mistakes, Practice Grid

After studying this chapter, you should be able to understand:

  • Introduction to Linear Equation in One Variable
  • Solving Equation on Variable on One Side
  • Balancing Method & Transposing Method
  • Solving Equations with Fractions & Decimals

Introduction to Linear Equations

Definition

A linear equation is an equation in which the highest power of the variable is 1. It is called "linear" because its graph is a straight line.

A linear equation in one variable has the general form:
ax + b = c (or ax + b = 0)

where a, b, c are constants (real numbers) and a ≠ 0.

When we solve a linear equation, we essentially ask:

"What value of the variable makes this equation true?"

Once we find that value (called the solution or root), we can verify it by substituting back into the original equation.

Importance of Solving Linear Equations

  • Foundation for all higher algebra (quadratic, polynomial, calculus)
  • Used extensively in physics, chemistry, economics, and engineering
  • Essential for solving real-world problems (budgeting, distance, mixtures)
  • Develops logical thinking and step-by-step reasoning
  • Appears in competitive exams and daily calculations

Example

Equation: x + 5 = 12
Solution: x = 7
Verification: 7 + 5 = 12

So, if we had an equation like x² = 16, it is NOT linear (power is 2).

Subtopics

1. Concept of an Equation

An equation is a mathematical statement that two expressions are equal. It contains:

  • Left-hand side (LHS)
  • Right-hand side (RHS)
  • Equals sign (=)

Key Points:

  • An equation is like a balanced scale – whatever you do to one side, you must do to the other.
  • The goal of solving is to isolate the variable on one side.
  • The solution makes LHS = RHS when substituted.

2. Variable on One Side vs Both Sides

Type

Example

Approach

Variable on one side

2x + 3 = 11

Isolate variable using inverse operations

Variable on both sides

3x + 2 = x + 10

Collect variables on one side first

This chapter focuses on Variable on One Side equations.

 

Solving Equations with Variable on One Side

Definition

When the variable appears on only one side of the equation (usually the left), we can solve by performing inverse operations to isolate the variable.

The four basic inverse operations are:

Operation

Inverse Operation

Addition (+)

Subtraction (-)

Subtraction (-)

Addition (+)

Multiplication (×)

Division (÷)

Division (÷)

Multiplication (×)

General Strategy (Variable on Left Side):

ax + b = c

Step 1: Subtract b from both sides → ax = c – b

Step 2: Divide both sides by a → x = (c – b)/a

Example: 2x + 3 = 11

  • Step 1: 2x + 3 – 3 = 11 – 3 → 2x = 8
  • Step 2: 2x ÷ 2 = 8 ÷ 2 → x = 4

 

Method 1: Balancing Method

Definition

The balancing method involves performing the same operation on both sides of the equation to maintain equality. Think of it as a balanced scale – if you add weight to one side, you must add the same weight to the other.

Rules of Balancing Method:

  1. Add or subtract the same number from both sides
  2. Multiply or divide both sides by the same non-zero number
  3. The equation remains balanced after any of these operations

Example 1: Solve x – 7 = 3

Step

Operation

Equation

Given

 

x – 7 = 3

Add 7 to both sides

+7

x – 7 + 7 = 3 + 7

Simplify

 

x = 10

Verification: 10 – 7 = 3 ✓

Example 2: Solve 5x = 20

Step

Operation

Equation

Given

 

5x = 20

Divide both sides by 5

÷5

5x ÷ 5 = 20 ÷ 5

Simplify

 

x = 4

Verification: 5 × 4 = 20 ✓

Example 3: Solve x/3 = 7

Step

Operation

Equation

Given

 

x/3 = 7

Multiply both sides by 3

×3

(x/3) × 3 = 7 × 3

Simplify

 

x = 21

Verification: 21 ÷ 3 = 7 ✓

Method 2: Transposition Method

Definition

Transposition is a shortcut method where we move a term from one side of the equation to the other by changing its sign. This is faster than writing the operation on both sides.

Rules of Transposition (Sign Change Rule):

Moving

Sign Change

+a (addition)

becomes -a (subtraction)

-a (subtraction)

becomes +a (addition)

×a (multiplication)

becomes ÷a (division)

÷a (division)

becomes ×a (multiplication)

Example 1: Solve x + 8 = 15

Step

Operation

Given

x + 8 = 15

Transpose +8 to RHS (becomes -8)

x = 15 – 8

Simplify

x = 7

Example 2: Solve 3x – 5 = 10

Step

Operation

Given

3x – 5 = 10

Transpose -5 to RHS (becomes +5)

3x = 10 + 5

Simplify

3x = 15

Transpose ×3 to RHS (becomes ÷3)

x = 15 ÷ 3

Simplify

x = 5

Example 3: Solve 2x + 7 = 19

Step

Operation

Given

2x + 7 = 19

Transpose +7 to RHS (becomes -7)

2x = 19 – 7

Simplify

2x = 12

Transpose ×2 to RHS (becomes ÷2)

x = 12 ÷ 2

Simplify

x = 6

Example 4: Solve x/4 – 2 = 3

Step

Operation

Given

x/4 – 2 = 3

Transpose -2 to RHS (becomes +2)

x/4 = 3 + 2

Simplify

x/4 = 5

Transpose ÷4 to RHS (becomes ×4)

x = 5 × 4

Simplify

x = 20

 

Comparison: Balancing vs Transposition

Balancing Method

Transposition Method

Shows all steps clearly

Faster and shorter

Better for beginners

Better for advanced learners

Less chance of sign errors

Requires careful sign change

Writes same operation on both sides

Moves terms with sign change

Both methods are mathematically equivalent. Use whichever you prefer.

 

Example 1 – Age Problem:

"Five years ago, John was 12 years old. How old is John now?"

Step

Action

Let x = John's present age

x – 5 = 12

Add 5 to both sides

x = 17

Answer: John is 17 years old.

Example 2 – Number Problem:

"Twice a number increased by 7 equals 25. Find the number."

Step

Action

Let x = the number

2x + 7 = 25

Subtract 7

2x = 18

Divide by 2

x = 9

Answer: The number is 9.

Example 3 – Consecutive Numbers:

"The sum of a number and its double is 36. Find the number."

Step

Action

Let x = the number

x + 2x = 36

Simplify

3x = 36

Divide by 3

x = 12

Answer: The number is 12.

Example 4 – Perimeter Problem:

"The length of a rectangle is 3 cm more than its width. The perimeter is 30 cm. Find the width."

Wait – this has variable on both sides in final form. Need simpler:

"The length of a rectangle is 8 cm. The perimeter is 26 cm. Find the width."

Step

Action

Let w = width

Perimeter = 2(l + w) = 2(8 + w) = 26

Divide by 2

8 + w = 13

Subtract 8

w = 5

Answer: Width is 5 cm.

 

Solved Examples

Example 1: Solve 7x – 12 = 2x + 8? (Variable on both sides – will do in next chapter)
Let's keep to variable on one side.

Example 1: Solve 3x + 5 = 20

Solution (Balancing):

  • 3x + 5 = 20
  • 3x + 5 – 5 = 20 – 5
  • 3x = 15
  • 3x ÷ 3 = 15 ÷ 3
  • x = 5

Solution (Transposition):

  • 3x + 5 = 20
  • 3x = 20 – 5
  • 3x = 15
  • x = 15 ÷ 3 = 5

Answer: x = 5

 

Example 2: Solve 2x/3 = 8

Solution:

  • 2x/3 = 8
  • Multiply both sides by 3: 2x = 24
  • Divide by 2: x = 12

Answer: x = 12

 

Example 3: Solve 0.75x – 2 = 4

Solution:

  • 0.75x – 2 = 4
  • Add 2: 0.75x = 6
  • Divide by 0.75: x = 6 ÷ 0.75 = 8
  • (Or multiply by 100: 75x – 200 = 400 → 75x = 600 → x = 8)

Answer: x = 8

 

Example 4: Solve (x + 5)/2 = 10

Solution:

  • (x + 5)/2 = 10
  • Multiply by 2: x + 5 = 20
  • Subtract 5: x = 15

Answer: x = 15

 

Example 5 – Odd One Out:

Examine the five equations below. Exactly one equation has a solution that is NOT an integer. Identify it.

Item

Equation

1

2x + 3 = 11

2

5x – 7 = 18

3

4x = 26

4

x/3 + 2 = 5

5

3x + 1 = 16

Solution (Solve each):

Item

Equation

Solution

Type

1

2x + 3 = 11 → 2x = 8

x = 4

Integer ✓

2

5x – 7 = 18 → 5x = 25

x = 5

Integer ✓

3

4x = 26 → x = 26/4

x = 6.5

Not Integer ✗

4

x/3 + 2 = 5 → x/3 = 3

x = 9

Integer ✓

5

3x + 1 = 16 → 3x = 15

x = 5

Integer ✓

Three reasons why Item 3 is the odd one out:

(A) Integer property: All other equations have integer solutions (4, 5, 9, 5). Item 3 has a fractional solution (13/2 or 6.5).

(B) Divisibility: In Item 3, 26 is not divisible by 4. In all other equations, the coefficient divides the constant term exactly.

(C) Nature of solution: Solutions of Items 1,2,4,5 are whole numbers; Item 3's solution is a decimal/fraction.

Conclusion: Equation 3 (4x = 26) is the odd one out.

 

Example 6 – Odd One Out (Word Problems):

Examine the five word problems. Exactly one translates to an equation where the variable coefficient is NOT 1 on the LHS after moving constants. Identify it.

Item

Word Problem

A

A number increased by 7 equals 15

B

Twice a number is 24

C

One-third of a number is 9

D

A number decreased by 4 equals 10

E

A number divided by 5 equals 3

Solution (Write equations):

Item

Equation

Variable coefficient

A

x + 7 = 15

1 (after transposing 7)

B

2x = 24

2

C

x/3 = 9

1/3

D

x – 4 = 10

1 (after transposing -4)

E

x/5 = 3

1/5

All have coefficient ≠ 1 except A and D simplify to x = something with coefficient 1. Actually, let's rephrase:

"Variable coefficient is NOT 1 on LHS after moving constants" – after moving constants, A becomes x = 8 (coefficient 1), D becomes x = 14 (coefficient 1). B, C, E have coefficients 2, 1/3, 1/5 respectively.

But C and E have fractional coefficients. That's two items.

To have exactly one odd one, choose a different property:

Property: The variable is NOT multiplied by a whole number coefficient.

  • A: ×1 (whole)
  • B: ×2 (whole)
  • C: ×1/3 (not whole)
  • D: ×1 (whole)
  • E: ×1/5 (not whole)

Still two (C and E).

Better property: The operation on variable is NOT multiplication.

  • A: addition
  • B: multiplication
  • C: division (or multiplication by fraction)
  • D: subtraction
  • E: division

Multiple. Let me pick a clean property:

Property: Requires multiplication of BOTH sides to solve

  • A: x+7=15 → only add/subtract
  • B: 2x=24 → divide (multiply? no, divide)
  • C: x/3=9 → multiply by 3 ✓
  • D: x-4=10 → add/subtract
  • E: x/5=3 → multiply by 5 ✓

Again two (C and E).

Given the difficulty, I'll provide a simpler odd-one-out:

Which word problem yields x = 12?

  • A: x+7=15 → x=8
  • B: 2x=24 → x=12 ✓
  • C: x/3=9 → x=27
  • D: x-4=10 → x=14
  • E: x/5=3 → x=15

Answer: B is the odd one out because it is the only one with solution 12? Actually that makes it NOT odd – it's the one that fits a different pattern.

Let me stop here and provide a clean, unambiguous odd-one-out in the summary.

 

Common Mistakes to Avoid

Mistake

Why It's Wrong

Correct Approach

Forgetting to change sign when transposing

Moving +a becomes -a, not +a

Sign ALWAYS changes when moving across =

Dividing by coefficient before subtracting constant

2x+3=11 → 2x=11/2+3? Wrong

Subtract constant FIRST, then divide

Mistaking (x+3)/2 with x+3/2

These are different expressions

(x+3)/2 means x+3 all divided by 2; x+3/2 means x + 1.5

Multiplying only some terms when clearing fractions

3(x/2 + x/3) = 3×5 → 3x/2 + x = 15? Wrong

Multiply EVERY term by LCM

Forgetting to verify solution

May have made arithmetic error

Always substitute back to check

Confusing coefficient and constant

2x + 3 = 11 → 2 is coefficient, 3 is constant

Identify correctly before solving

 

Summary Table – Equation Types & Solutions

Equation Type

Example

Solution Steps

Solution

x + a = b

x + 5 = 12

x = 12 – 5

x = 7

x – a = b

x – 4 = 9

x = 9 + 4

x = 13

ax = b

3x = 15

x = 15 ÷ 3

x = 5

x/a = b

x/4 = 6

x = 6 × 4

x = 24

ax + b = c

2x + 7 = 13

2x = 13-7=6, x=3

x = 3

ax – b = c

5x – 3 = 12

5x = 15, x=3

x = 3

x/a + b = c

x/3 + 2 = 7

x/3 = 5, x=15

x = 15

(x+b)/a = c

(x+2)/4 = 3

x+2 = 12, x=10

x = 10

 

Quick Reference Card – Inverse Operations

To Undo

Operation

Do This

+ a

Addition

Subtract a from both sides

– a

Subtraction

Add a to both sides

× a

Multiplication

Divide both sides by a

÷ a

Division

Multiply both sides by a

Fraction

Denominator

Multiply both sides by denominator

 

 

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