Master Linear Equations in One Variable: Class 8 Maths Tutorial

Have you ever played a game of riddles where someone says, "I am thinking of a number. If I double it and add 5, I get 15. What is my number?"

Guess what? You were already solving Linear Equations in your head!

In Class 8 Mathematics, this chapter is one of your strongest building blocks. Once you master linear equations, algebra will become as simple and fun as solving a puzzle. Let us break down this concept step-by-step together!

1. What is a Linear Equation in One Variable?

To understand this big title, let us break it into three small parts:

1. Variable: A symbol (usually an alphabet like $x, y, z, a, b$) that represents an unknown value. Its value can change!
2. Linear: The highest power (exponent) of the variable in the expression is 1. For example, $x^1$ (written just as $x$). If you see $x^2$ or $y^3$, it is not linear!
3. Equation: A mathematical statement showing that two expressions are equal, using an equality sign ($=$).

Putting it all together: An equation which has only one variable, and the highest power of that variable is 1, is called a Linear Equation in One Variable.

Structure of an Equation:

Look at the equation:

$2x + 3 = 7$

• $x$ is the Variable.
• $2$ is the Coefficient of $x$.
• $3$ and $7$ are Constants.
• Everything to the left of the $=$ sign is the LHS (Left Hand Side): $2x + 3$.
• Everything to the right of the $=$ sign is the RHS (Right Hand Side): $7$.

2. The Golden Rule: The Weighing Balance Analogy

Think of an equation as a traditional two-pan weighing balance (like the ones used by fruit sellers).

• For the balance to stay horizontal, the weight on the Left Pan (LHS) must equal the weight on the Right Pan (RHS).
• The Rule: Whatever operation you perform on one side, you must perform the exact same operation on the other side to keep it balanced!

       [ LHS ] === ( = ) === [ RHS ]

• Add $5$ to LHS $\rightarrow$ Add $5$ to RHS.
• Subtract $2$ from LHS $\rightarrow$ Subtract $2$ from RHS.
• Multiply LHS by $3$ $\rightarrow$ Multiply RHS by $3$.
• Divide LHS by $4$ $\rightarrow$ Divide RHS by $4$.

3. How to Solve Linear Equations

Solving an equation means finding the value of the variable that makes $\text{LHS} = \text{RHS}$. This value is called the solution or root of the equation.

There are two main methods to solve equations:

Method A: Balancing Method (Doing the same on both sides)

Let us solve: $x - 5 = 7$

• Goal: Keep $x$ alone on the LHS.
• To remove $-5$ from LHS, we add $5$ to both sides.

$x - 5 + 5 = 7 + 5$ $x = 12$

Method B: Transposition Method (The Express Highway Route!)

Transposition means shifting a term from one side of the equality sign ($=$) to the other side. When a term crosses the highway ($=$ sign), its operation flips:

• $+$ becomes $-$
• $-$ becomes $+$
• $\times$ becomes $\div$
• $\div$ becomes $\times$

Example: Solve $3x + 4 = 19$

1. Step 1: Transpose $+4$ from LHS to RHS (it becomes $-4$). $3x = 19 - 4$ $3x = 15$

2. Step 2: Transpose $3$ (which is multiplied with $x$) to RHS (it goes into division). $x = \frac{15}{3}$ $x = 5$

Check your answer: Substitute $x = 5$ in $\text{LHS}$: $\text{LHS} = 3(5) + 4 = 15 + 4 = 19 = \text{RHS}$. Since $\text{LHS} = \text{RHS}$, our answer is 100% correct!

4. Solving Equations with Variables on Both Sides

Sometimes, variables appear on both LHS and RHS!

Strategy: Group all terms containing the variable on one side (usually LHS) and all constant numbers on the other side (RHS).

Example: Solve $5x - 3 = 2x + 9$

1. Step 1: Transpose $2x$ from RHS to LHS (it becomes $-2x$). $5x - 2x - 3 = 9$ $3x - 3 = 9$

2. Step 2: Transpose $-3$ from LHS to RHS (it becomes $+3$). $3x = 9 + 3$ $3x = 12$

3. Step 3: Divide by $3$. $x = \frac{12}{3} = 4$

5. Applications of Linear Equations (Word Problems)

Word problems convert English sentences into mathematical equations. Don't worry, here is our Secret 4-Step Recipe to master word problems!

1. Read & Identify: Read the problem carefully and identify what is unknown.
2. Assign Variable: Let the unknown quantity be '$x$'.
3. Form the Equation: Translate the words into mathematical statements using the conditions given.
4. Solve & Verify: Solve for '$x$' and check if it makes sense in the context of the question.

Common Student Mistakes to Avoid

1. Sign Errors in Algebraic Calculations: Mistakes in distributing negative signs across brackets or when transferring terms across the equals sign.
2. Formula Misapplication: Memorizing formulas without checking required units or conditions (e.g. using diameter instead of radius).
3. Skipping Intermediate Steps: Jumping directly to final numerical answers without showing step-by-step mathematical working, leading to partial credit loss in board exams.
4. Incorrect Unit Conversions: Forgetting to convert parameters into uniform SI units (e.g., cm to meters or minutes to seconds) before computing.

Practice Questions with Detailed Solutions

Let us test our learning with three important NCERT-style practice questions!

Question 1: Simple Linear Equation

Solve the equation for $x$: $\frac{2x + 3}{5} = \frac{x - 1}{2}$

Solution:

• Step 1: Cross-multiply the denominators across the '=' sign to remove fractions. $2 \cdot (2x + 3) = 5 \cdot (x - 1)$

• Step 2: Expand the brackets on both sides using the distributive property. $4x + 6 = 5x - 5$

• Step 3: Transpose variable terms to LHS and constant terms to RHS. $4x - 5x = -5 - 6$ $-1x = -11$

• Step 4: Divide both sides by $-1$. $x = 11$

Answer: $x = 11$

Question 2: Word Problem on Ages

The present age of Sahil's mother is three times the present age of Sahil. After 5 years, the sum of their ages will be 66 years. Find their present ages.

Solution:

• Step 1: Assign variables for present ages. Let Sahil's present age $= x \text{ years}$ Therefore, Sahil's mother's present age $= 3x \text{ years}$

• Step 2: Express ages after 5 years. Sahil's age after 5 years $= (x + 5) \text{ years}$ Mother's age after 5 years $= (3x + 5) \text{ years}$

• Step 3: Form the equation using the given condition. Sum of their ages after 5 years $= 66$ $(x + 5) + (3x + 5) = 66$

• Step 4: Solve the equation. $4x + 10 = 66$ $4x = 66 - 10$ $4x = 56$ $x = \frac{56}{4} = 14$

• Step 5: State the final answer.

  • Sahil's present age $= x = \mathbf{14 \text{ years}}$
  • Sahil's mother's present age $= 3x = 3 \times 14 = \mathbf{42 \text{ years}}$

Question 3: Word Problem on Consecutive Numbers

The sum of three consecutive multiples of $11$ is $363$. Find these multiples.

Solution:

• Step 1: Understand consecutive multiples. Multiples of 11 come at intervals of 11 (e.g., 11, 22, 33). Let the first multiple of 11 be $= x$ Then the second consecutive multiple $= x + 11$ And the third consecutive multiple $= x + 22$

• Step 2: Form the equation. $\text{Sum} = x + (x + 11) + (x + 22) = 363$

• Step 3: Solve the equation. $3x + 33 = 363$ $3x = 363 - 33$ $3x = 330$ $x = \frac{330}{3} = 110$

• Step 4: Find the three multiples.

  • First multiple $= x = \mathbf{110}$
  • Second multiple $= x + 11 = 110 + 11 = \mathbf{121}$
  • Third multiple $= x + 22 = 110 + 22 = \mathbf{132}$

(Check: $110 + 121 + 132 = 363$. Correct!)

Summary Key Points

1. A linear equation in one variable has one unknown alphabet, and its power is always 1.
2. Transposition Rules:
   • $+ \rightarrow -$
   • $- \rightarrow +$
   • $\times \rightarrow \div$
   • $\div \rightarrow \times$
3. Always verify your answer by substituting the value of $x$ back into the original equation!

Keep practicing, stay curious, and remember—maths is not about memorization; it's a superpower to solve daily real-world problems! Happy learning!

Exam Preparation & Frequently Asked Questions (FAQ)

Q1. How should I revise Linear Equations in One Variable for the Class 8 Mathematics examination?

Focus on mastering core textbook definitions, practicing 3-4 numerical problems daily with pen and paper, and reviewing previous year CBSE/NCERT board exam questions.

Q2. What are the key concepts that carry maximum marks in this chapter?

Pay special attention to core definitions, step-by-step derivations, solved textbook examples, and practical real-world applications outlined in your NCERT curriculum.

Q3. How can I avoid losing marks in long answer questions?

Always structure your answers with clear subheadings, write step-by-step working for numerical problems, state given values clearly, and highlight your final answers with correct SI units.