Solving Equations, Variable On Both Side

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Unit: Algebra – 1

Chapter: Solving Equations, Variable on Both Side

Reference: – Introduction to Variable on Both Sides, Collecting Like Terms, Moving Variables to One Side, Balancing Method, Transposition Method, Equations with Parentheses, Equations with Fractions and Decimals, Special Cases (No Solution, Infinite Solutions), Verification, 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 Both Side
• Collecting Variable Terms on Both side
• Equation with Parentheses, Fractions & Decimals

Introduction to Variable on Both Sides

Definition

In many linear equations, the variable appears on both sides of the equal’s sign. To solve such equations, we must first collect all variable terms on one side and all constant terms on the other.

General Form: ax + b = cx + d (where a, b, c, d are constants, a ≠ c or possibly a = c)

When we solve equations with variable on both sides, we essentially ask:

“How do I get all the x’s together on one side so I can isolate the variable?”

The answer: Add or subtract variable terms from both sides to move them.

Importance

• Builds on basic equation-solving skills
• Essential for solving real-world problems where variables appear in multiple places
• Foundation for systems of equations and advanced algebra

Example

Equation: 3x + 2 = x + 10
Step 1: Subtract x from both sides → 2x + 2 = 10
Step 2: Subtract 2 from both sides → 2x = 8
Step 3: Divide by 2 → x = 4

So, if we had 3x + 2 = 3x + 5, that simplifies to 2 = 5 (no solution).

Subtopics

1. Why Move Variables to One Side?

Having variables on both sides makes it impossible to isolate the variable directly. By moving all variable terms to one side, we convert the equation into the familiar variable on one side form.

Golden Rule: Whatever you do to one side, do to the other.

2. Which Side to Move Variables To?

You can move variables to either side, but it’s conventional to end with a positive coefficient.

Method 1: Balancing Method

Definition

Perform the same operation on both sides to eliminate variables from one side.

Example 1: Solve 5x – 3 = 2x + 9

Verification: LHS = 5(4)-3=20-3=17; RHS = 2(4)+9=8+9=17 ✓

Example 2: Solve 7x + 2 = 4x + 14

Verification: LHS=7(4)+2=28+2=30; RHS=4(4)+14=16+14=30 ✓

Method 2: Transposition Method

Definition

Move variable terms from one side to the other by changing their sign (just like constants).

Example 1: Solve 6x – 5 = 2x + 11

Example 2: Solve 9x + 3 = 5x + 19

Standard Step-by-Step Strategy

Step 1: Use the distributive property if parentheses exist.
Step 2: Collect variable terms on one side (usually left).
Step 3: Collect constant terms on the other side.
Step 4: Combine like terms.
Step 5: Divide by the coefficient of the variable.
Step 6: Verify the solution.

Equations with Parentheses

Definition

When parentheses appear, simplify both sides first by distributing before moving variables.

Example 1: Solve 3(x + 2) = 2x + 10

Verification: LHS=3(4+2)=3×6=18; RHS=2(4)+10=8+10=18 ✓

Example 2: Solve 2(3x – 1) = 4(x + 2)

Verification: LHS=2(15-1)=2×14=28; RHS=4(5+2)=4×7=28 ✓

Equations with Fractions

Definitions

Clear fractions by multiplying both sides by the LCM of all denominators.

Example 1: Solve x/2 + 3 = x/3 + 5

Verification: LHS=12/2+3=6+3=9; RHS=12/3+5=4+5=9 ✓

Example 2: Solve (2x – 1)/3 = (x + 2)/2

Verification: LHS=(16-1)/3=15/3=5; RHS=(8+2)/2=10/2=5

Special Cases

Case 1: No Solution

When the variable cancels out and the remaining statement is false.

Example: 2x + 3 = 2x + 7

Conclusion: No solution (∅). These equations are called inconsistent.

Case 2: Infinite Solutions (Identity)

When the variable cancels out and the remaining statement is true for all values.

Example: 3x + 4 = 3x + 4

Conclusion: Infinite solutions (all real numbers). These equations are called identities.

Summary Table – Special Cases

Solved Examples

Example 1: Solve 8x – 5 = 3x + 15

Solution:

• 8x – 5 = 3x + 15
• 8x – 3x – 5 = 15
• 5x – 5 = 15
• 5x = 20
• x = 4

Answer: x = 4

Example 2: Solve 4(x – 3) = 2(x + 5)

Solution:

• 4x – 12 = 2x + 10
• 4x – 2x = 10 + 12
• 2x = 22
• x = 11

Answer: x = 11

Example 3: Solve (x + 3)/4 = (x – 1)/2

Solution:

• Cross multiply: 2(x + 3) = 4(x – 1)
• 2x + 6 = 4x – 4
• 2x – 4x = -4 – 6
• -2x = -10
• x = 5

Answer: x = 5

Example 4: Solve 5x – 7 = 5x + 3

Solution:

• Subtract 5x from both sides: -7 = 3
• False statement

Answer: No solution

Example 5 – Odd One Out:

Examine the five equations below. Exactly one has a different number of solutions. Identify it.

Solution:

Items 2 and 3 are special cases. If “exactly one has a different number of solutions” – Item 2 has infinite, Item 3 has zero, Items 1,4,5 have one. That’s two special cases.

Perhaps the intended is exactly one has INFINITE solutions – then Item 2.

Three reasons why Item 2 is the odd one out:

(A) Solution count: Item 2 has infinitely many solutions; all others have either one solution or no solution.

(B) Identity property: Item 2 is an identity (true for all x); others are conditional equations.

(C) Simplification result: Item 2 simplifies to 0=0 (always true); Item 3 simplifies to 7=9 (always false); Items 1,4,5 simplify to x = constant.

Conclusion: Item 2 is the odd one out.

Common Mistakes to Avoid

Practice Grid

Quick Reference Card

Golden Rule: Always verify by substituting your answer back into the original equation.