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Published 2026-08-28•Chapter: Linear Equations in One Variable

Linear Equations in One Variable - Solving linear equations and their applications in word problems

Hello my bright young mathematicians! Welcome to today’s lesson. 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!

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1. What is a Linear Equation in One Variable?

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

  • Variable: A symbol (usually an alphabet like $x, y, z, a, b$) that represents an unknown value. Its value can change!
  • 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!
  • 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.

    • $$x - 5 + 5 = 7 + 5$$

    $$x = 12$$

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    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 $3x + 4 = 19$

  • Step 1: Transpose $+4$ from LHS to RHS (it becomes $-4$).

    • $$3x = 19 - 4$$

    $$3x = 15$$

  • 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!*

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    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 $5x - 3 = 2x + 9$

  • Step 1: Transpose $2x$ from RHS to LHS (it becomes $-2x$).

    • $$5x - 2x - 3 = 9$$

    $$3x - 3 = 9$$

  • Step 2: Transpose $-3$ from LHS to RHS (it becomes $+3$).

    • $$3x = 9 + 3$$

    $$3x = 12$$

  • Step 3: Divide by $3$.

    • $$x = \frac{12}{3} = 4$$

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    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!

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

    • ---

    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: $x = 11$

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    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!)*

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    Summary Key Points

  • A linear equation in one variable has one unknown alphabet, and its power is always 1.
  • Transposition Rules:
  • $+ \rightarrow -$
  • $- \rightarrow +$
  • $\times \rightarrow \div$
  • $\div \rightarrow \times$
  • 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!