CBSE Class 8 Maths: Linear Equation Ex 2.1 (One Variable)

Welcome, Class 8 students! Are you ready to unravel the mystery of finding a hidden number? In this chapter, we dive into Linear Equations in One Variable, specifically focusing on Exercise 2.1 from your NCERT textbook. This might sound complex, but don't worry, it's all about finding a value that makes a statement true! Linear equations are fundamental building blocks of algebra and appear everywhere – from calculating how many chocolates you can buy to solving complex scientific problems. By understanding how to solve these equations, you'll develop crucial problem-solving skills that will serve you well in higher classes and real life. Get ready to master the art of finding the unknown and confidently tackle linear equation ex 2 1 class 8 ncert problems!

What are Linear Equations in One Variable?

A Linear Equation in One Variable is a mathematical statement that shows two expressions are equal, where one of the expressions contains a variable raised to the power of one (that's what 'linear' means) and there's only one type of unknown quantity (that's 'one variable'). Think of it like a balanced weighing scale!

For example, consider the equation x + 5 = 12. Here:

  • x is the variable – it's the unknown number we need to find.
  • 5 and 12 are constants – their values don't change.
  • The + sign and = sign show the relationship.
  • The part to the left of the = sign, x + 5, is called the Left Hand Side (LHS).
  • The part to the right of the = sign, 12, is called the Right Hand Side (RHS).

The goal of solving a linear equation is to find the unique value of the variable that makes the equation true. For x + 5 = 12, if x were 7, then 7 + 5 = 12, which is true. So, x = 7 is the solution. This section lays the groundwork for solving all the problems in linear equation ex 2 1 class 8 ncert.

Step-by-Step Method to Solve Linear Equations

  1. Step 1: Identify the Variable and Constants — Look at the equation and clearly identify the unknown variable (e.g., 'x', 'y', 'z') and all the constant numbers. Note which constants are on the same side as the variable and which are on the other side.
  2. Step 2: Isolate the Variable Term using Transposition — Your aim is to get all terms containing the variable on one side (usually the LHS) and all constant terms on the other side (usually the RHS). To move a term from one side of the equation to the other, you perform the opposite operation and change its sign. This process is called transposition. If a term is added, subtract it from both sides (or move it and change '+' to '-'). If a term is subtracted, add it to both sides (or move it and change '-' to '+').
  3. Step 3: Isolate the Variable Itself — Once you have the variable term (e.g., 3x or x/2) on one side and a single constant on the other, you need to find the value of the variable itself. If the variable is multiplied by a number, divide both sides by that number. If the variable is divided by a number, multiply both sides by that number.
  4. Step 4: Verify Your Solution (Optional but Recommended) — Substitute the value you found for the variable back into the original equation. If the LHS equals the RHS, your solution is correct!

Worked Examples from Ex 2.1

  • Example 1: Solve x - 2 = 7 Step 1: Identify variable (x), constants (-2, 7). Step 2: Transpose -2 to the RHS. It becomes +2. x = 7 + 2 Step 3: Perform the addition. x = 9 Step 4: Verify: Substitute x = 9 into x - 2 = 7 -> 9 - 2 = 7 -> 7 = 7 (True). Final Answer: x = 9
  • Example 2: Solve y + 3 = 10 Step 1: Identify variable (y), constants (3, 10). Step 2: Transpose +3 to the RHS. It becomes -3. y = 10 - 3 Step 3: Perform the subtraction. y = 7 Step 4: Verify: Substitute y = 7 into y + 3 = 10 -> 7 + 3 = 10 -> 10 = 10 (True). Final Answer: y = 7
  • Example 3: Solve 6 = z + 2 Step 1: Identify variable (z), constants (6, 2). Variable is on RHS. Step 2: Transpose +2 to the LHS. It becomes -2. 6 - 2 = z Step 3: Perform the subtraction. 4 = z or z = 4 Step 4: Verify: Substitute z = 4 into 6 = z + 2 -> 6 = 4 + 2 -> 6 = 6 (True). Final Answer: z = 4
  • Example 4: Solve 3/7 + x = 17/7 Step 1: Identify variable (x), constants (3/7, 17/7). Step 2: Transpose 3/7 to the RHS. It becomes -3/7. x = 17/7 - 3/7 Step 3: Subtract the fractions (they have the same denominator). x = (17 - 3) / 7 x = 14 / 7 x = 2 Step 4: Verify: Substitute x = 2 into 3/7 + x = 17/7 -> 3/7 + 2 = 17/7 -> 3/7 + 14/7 = 17/7 -> 17/7 = 17/7 (True). Final Answer: x = 2

Exam Tip: Avoiding Common Mistakes

When solving linear equations, especially those from linear equation ex 2 1 class 8 ncert, students often make a few common errors. Be careful with these:

  1. Sign Errors during Transposition: Always remember that when you move a term from one side of the equation to the other, its sign must change. x + 3 = 5 becomes x = 5 - 3, not x = 5 + 3. This is the most frequent mistake!
  2. Incorrect Operations: If a number is multiplying the variable (e.g., 2x), you must divide by that number. If it's dividing (e.g., x/2), you must multiply. Don't add or subtract!
  3. Forgetting Both Sides: Whatever operation you perform (adding, subtracting, multiplying, dividing), you must do it on both the Left Hand Side (LHS) and the Right Hand Side (RHS) to maintain the balance of the equation. Thinking of the equation as a balanced scale helps avoid this.
  4. Not Simplifying Fractions/Expressions: After solving, always simplify your answer to its lowest terms, especially if it's a fraction.

Practice Questions with Solutions

  • Q: Solve x + 3 = 10 A: Step 1: Identify x, +3, 10. Step 2: Transpose +3 to RHS. It becomes -3. x = 10 - 3 Step 3: Calculate the value. x = 7 Final answer: x = 7
  • Q: Solve m - 5 = -12 A: Step 1: Identify m, -5, -12. Step 2: Transpose -5 to RHS. It becomes +5. m = -12 + 5 Step 3: Calculate the value. m = -7 Final answer: m = -7
  • Q: Solve 3x = 18 A: Step 1: Identify 3x, 18. (3 is multiplying x) Step 2: Divide both sides by 3. x = 18 / 3 Step 3: Calculate the value. x = 6 Final answer: x = 6
  • Q: Solve t / 5 = 10 A: Step 1: Identify t/5, 10. (t is divided by 5) Step 2: Multiply both sides by 5. t = 10 * 5 Step 3: Calculate the value. t = 50 Final answer: t = 50
  • Q: Solve 2y + 9 = 4 A: Step 1: Identify 2y, +9, 4. Step 2: Transpose +9 to RHS. It becomes -9. 2y = 4 - 9 2y = -5 Step 3: Divide both sides by 2. y = -5 / 2 Final answer: y = -5/2

Frequently Asked Questions

What is a linear equation in one variable?

A linear equation in one variable is an algebraic equation where the highest power of the variable is one, and there is only one type of unknown variable. It represents a balanced statement of equality between two expressions.

Why is it important to change the sign when transposing a term?

Changing the sign is crucial because transposition is essentially performing the opposite operation on both sides of the equation. If a term is added on one side, you subtract it from both sides; when it moves, it appears as a subtraction on the other side. This maintains the equality of the equation.

How do I check if my solution to an equation is correct?

To check your solution, substitute the value you found for the variable back into the original equation. If the Left Hand Side (LHS) of the equation equals the Right Hand Side (RHS) after substitution, then your solution is correct.

Can linear equations have fractional or decimal answers?

Yes, absolutely! Linear equations in one variable can have solutions that are whole numbers, integers (positive or negative), fractions, or decimals. The type of answer depends on the specific numbers and operations involved in the equation.