Linear Equations in One Variable: Exercise 2.5 Class 8 NCERT Solutions

Welcome to your step-by-step guide for CBSE Class 8 Maths Chapter 2, Exercise 2.5! In this section, we move beyond basic algebraic statements and tackle more complex linear equations that contain fractional coefficients and multiple variables across denominators. Mastering the linear equation ex 2 5 class 8 ncert exercises is a fundamental milestone. It bridges the gap between simple integer calculations and advanced algebraic formulations you will encounter in higher classes. Here, you will learn how to reduce complex fractional expressions to simpler linear forms using the Least Common Multiple (LCM) method and cross-multiplication. With patient practice, clear mental models, and step-by-step guidance from YoLearn AI, you will learn to dismantle complex expressions easily and secure perfect marks in your CBSE assessments.

Understanding Reducible Linear Equations

Equations in Exercise 2.5 are often presented with variable expressions as numerators and constant integers as denominators. Solving them directly can be confusing because dealing with fractions at every step increases the likelihood of calculation errors. To solve these easily, our primary goal is to eliminate the denominators to create a simpler, single-line linear equation with integers. We achieve this by finding the Least Common Multiple (LCM) of all the denominators present on both sides of the equation. Multiplying every single term by this LCM clears the fractions instantly, leaving us with a familiar linear equation structure. Once cleared, we use basic distributive properties and transposition to isolate the variable and find its value.

Step-by-Step Method to Clear Fractions

  1. Identify all Denominators — Look at all the terms on both the Left-Hand Side (LHS) and Right-Hand Side (RHS) of the equation and list out their denominators.
  2. Find the Least Common Multiple (LCM) — Calculate the LCM of all the listed denominators. This single number is divisible by all the individual denominators.
  3. Multiply Every Term by the LCM — Multiply each term on both sides of the equation by the LCM. This clears out all fractional denominators, converting them into integers.
  4. Expand and Group Like Terms — Use the distributive law to open any brackets. Be highly cautious with negative signs outside parentheses. Bring all variable terms to one side and constant terms to the other.
  5. Solve and Verify — Simplify both sides to solve for the variable. Substitute your final answer back into the original equation to verify that LHS equals RHS.

Common Exam Trap: The Distributive Negative Sign

The most common mistake students make in Exercise 2.5 is failing to distribute a negative sign to all terms inside a numerator bracket. For example, when simplifying an expression like -(3t - 2)/4, if you multiply by an LCM (say 12), the negative sign must apply to the entire numerator: -3(3t - 2) = -9t + 6. Many students write -9t - 6 by mistake, which completely ruins the final calculation. Always place multi-term numerators inside protective brackets before clearing denominators!

Practice Questions with Solutions

  • Q: Solve the linear equation: (x/2) - (1/5) = (x/3) + (1/4) A: Step 1: Identify all denominators in the equation. They are 2, 5, 3, and 4. Step 2: Find the LCM of 2, 5, 3, and 4. The LCM is 60. Step 3: Multiply every term on both sides by 60: 60 (x/2) - 60 (1/5) = 60 (x/3) + 60 (1/4) Step 4: Simplify each term: 30x - 12 = 20x + 15 Step 5: Transpose variable terms to LHS and constant terms to RHS: 30x - 20x = 15 + 12 10x = 27 Step 6: Solve for x: x = 27/10 Final answer: x = 27/10 (or 2.7)
  • Q: Solve the linear equation: (x - 5)/3 = (x - 3)/5 A: Step 1: Since both sides are single fractions, we can cross-multiply directly to clear denominators. Step 2: Cross-multiply 5 with (x - 5) and 3 with (x - 3): 5 (x - 5) = 3 (x - 3) Step 3: Expand the brackets using the distributive property: 5x - 25 = 3x - 9 Step 4: Collect like terms by transposing 3x to LHS and -25 to RHS: 5x - 3x = -9 + 25 2x = 16 Step 5: Divide by 2 to isolate x: x = 16 / 2 x = 8 Final answer: x = 8
  • Q: Solve the equation: m - (m - 1)/2 = 1 - (m - 2)/3 A: Step 1: Note that the term 'm' can be written as m/1 and '1' as 1/1. The denominators are 1, 2, 1, and 3. The LCM of 2 and 3 is 6. Step 2: Multiply every single term by 6: 6 m - 6 [(m - 1)/2] = 6 1 - 6 [(m - 2)/3] Step 3: Simplify the fractions: 6m - 3(m - 1) = 6 - 2(m - 2) Step 4: Carefully expand the brackets, keeping sign changes in mind: 6m - 3m + 3 = 6 - 2m + 4 Step 5: Simplify both sides: 3m + 3 = 10 - 2m Step 6: Transpose -2m to LHS and 3 to RHS: 3m + 2m = 10 - 3 5m = 7 m = 7/5 Final answer: m = 7/5
  • Q: Solve the equation: 15(y - 4) - 2(y - 9) + 5(y + 6) = 0 A: Step 1: This equation doesn't have fractions, but requires careful expansion of brackets. Step 2: Apply the distributive property to expand each set of brackets: 15y - 60 - 2y + 18 + 5y + 30 = 0 (Notice how -2 * -9 became +18) Step 3: Group the variable terms together and the constant terms together: (15y - 2y + 5y) + (-60 + 18 + 30) = 0 Step 4: Simplify the grouped terms: 18y - 12 = 0 Step 5: Transpose -12 to the RHS: 18y = 12 Step 6: Solve for y: y = 12 / 18 Reduce the fraction to its lowest terms by dividing numerator and denominator by 6: y = 2/3 Final answer: y = 2/3

Frequently Asked Questions

What is the best way to avoid calculation errors in Exercise 2.5?

The most effective method is to find the LCM of all denominators first and multiply the entire equation by it. Also, always write multi-term numerators within parentheses to ensure negative signs are distributed correctly.

When should I use cross-multiplication instead of finding the LCM?

Cross-multiplication is perfect when you have a single fraction on the LHS and a single fraction on the RHS, like (a/b) = (c/d). If there are multiple separate terms on either side, find the LCM to clear fractions first.

How can I verify if my calculated value of the variable is correct?

Substitute your final numerical answer back into the original equation in place of the variable. Calculate the LHS and RHS separately; if both sides yield the exact same value, your solution is correct.