NCERT Solutions for Class 8 Maths: Linear Equation Ex 2.3

Welcome! So far, you've learned to solve linear equations where the variable (like 'x' or 'y') appears on just one side. Now, we're taking the next step! In this section, focusing on NCERT Exercise 2.3, we'll tackle equations where the variable is present on both sides of the equals sign. Think of an equation as a balanced scale. If you have unknown weights (the variables) on both pans, your goal is to rearrange them so all the unknown weights are on one side and all the known weights are on the other. This helps you find the value of the unknown weight. Mastering this technique is essential for algebra, as it allows you to solve a much wider range of real-world problems. By the end of this lesson, you will be able to confidently solve and verify any linear equation with variables on both sides.

Key Concept: Handling Variables on Both Sides

The main goal when solving an equation like 3x = 2x + 18 is to isolate the variable x. The new challenge here is that x appears on both the Left Hand Side (LHS) and the Right Hand Side (RHS). The strategy is to gather all the variable terms on one side and all the constant terms (the plain numbers) on the other. We do this by 'transposing'. Transposing is like moving a term across the '=' bridge. When a term crosses this bridge, it must change its operation sign. A positive term becomes negative, and a negative term becomes positive. For example, in 3x = 2x + 18, we want to move 2x from the RHS to the LHS. Since it is +2x on the right, it becomes -2x on the left. So, the equation becomes 3x - 2x = 18. Now it's easy to solve: x = 18. This method keeps the equation balanced and helps us find the solution systematically.

A Step-by-Step Guide to Solving These Equations

  1. Step 1: Simplify Each Side — If there are any brackets in the equation, use the distributive property to remove them first. For example, change 3(x-1) to 3x - 3.
  2. Step 2: Transpose Variable Terms to LHS — Move all terms containing the variable (like 5x, -2y) to the Left Hand Side (LHS). Remember to change their sign when you move them.
  3. Step 3: Transpose Constant Terms to RHS — Move all the constant numbers (like +4, -10) to the Right Hand Side (RHS). Again, change their sign as you move them.
  4. Step 4: Simplify and Solve — Combine the like terms on both sides of the equation. You will be left with something like ax = b. Finally, divide both sides by 'a' to find x.
  5. Step 5: Check Your Solution — Substitute the value you found for the variable back into the original equation. Calculate the value of the LHS and RHS separately. If LHS = RHS, your answer is correct!

Worked Examples: Solve and Check

  • Question 1: Solve 5t - 3 = 3t - 5 and check the result. Solution: Step 1: The given equation is 5t - 3 = 3t - 5. Step 2: Transpose the variable term 3t from RHS to LHS (it becomes -3t). 5t - 3t - 3 = -5 Step 3: Transpose the constant term -3 from LHS to RHS (it becomes +3). 5t - 3t = -5 + 3 Step 4: Simplify both sides. 2t = -2 Step 5: Divide by 2 to solve for t. t = -2 / 2 t = -1 Check: LHS = 5t - 3 = 5(-1) - 3 = -5 - 3 = -8 RHS = 3t - 5 = 3(-1) - 5 = -3 - 5 = -8 Since LHS = RHS, our solution is correct.
  • Question 2: Solve 8x + 4 = 3(x - 1) + 7 and check the result. Solution: Step 1: First, simplify the RHS by opening the bracket. 8x + 4 = 3x - 31 + 7 8x + 4 = 3x - 3 + 7 8x + 4 = 3x + 4 Step 2: Transpose the variable term 3x from RHS to LHS. 8x - 3x + 4 = 4 Step 3: Transpose the constant term +4 from LHS to RHS. 8x - 3x = 4 - 4 Step 4: Simplify both sides. 5x = 0 Step 5: Divide by 5. x = 0 / 5 x = 0 Check: LHS = 8x + 4 = 8(0) + 4 = 0 + 4 = 4 RHS = 3(x - 1) + 7 = 3(0 - 1) + 7 = 3(-1) + 7 = -3 + 7 = 4 Since LHS = RHS, our solution is correct.

Exam Tip: Checking Your Answer is a Superpower!

In an exam, it's easy to make small mistakes with positive and negative signs. The 'Check' step is your secret weapon to catch these errors! Always substitute your final answer back into the original equation. If the Left Hand Side equals the Right Hand Side, you can be 100% confident that you have scored full marks for that question. If they are not equal, it's a signal that you made a mistake somewhere. Go back and check your transposing and calculations carefully. Don't skip this step; it can be the difference between a good score and a great score!

Practice Questions with Solutions

  • Q: Solve for z: 4z + 3 = 6 + 2z A: Step 1: Write down the equation: 4z + 3 = 6 + 2z. Step 2: Transpose 2z to the LHS (becomes -2z) and 3 to the RHS (becomes -3). 4z - 2z = 6 - 3 Step 3: Simplify both sides. 2z = 3 Step 4: Divide by 2 to find z. z = 3/2 Final answer: z = 3/2
  • Q: Solve for x: 2x - 1 = 14 - x A: Step 1: Write down the equation: 2x - 1 = 14 - x. Step 2: Transpose -x to the LHS (becomes +x) and -1 to the RHS (becomes +1). 2x + x = 14 + 1 Step 3: Combine like terms. 3x = 15 Step 4: Divide by 3 to find x. x = 15 / 3 Final answer: x = 5
  • Q: Find the solution of 9p = 4p + 25 A: Step 1: The equation is 9p = 4p + 25. Step 2: Transpose the variable term 4p to the LHS. 9p - 4p = 25 Step 3: Simplify the LHS. 5p = 25 Step 4: Divide both sides by 5. p = 25 / 5 Final answer: p = 5
  • Q: Solve 2(y + 1) = 10 - y A: Step 1: First, expand the bracket on the LHS: 2y + 2 = 10 - y. Step 2: Transpose -y from RHS to LHS (becomes +y) and +2 from LHS to RHS (becomes -2). 2y + y = 10 - 2 Step 3: Simplify both sides. 3y = 8 Step 4: Divide by 3 to find y. y = 8/3 Final answer: y = 8/3

Frequently Asked Questions

What does it mean to 'transpose' a term in an equation?

Transposing means moving a term from one side of the equals sign to the other. When you transpose, you must change the sign of the term: addition becomes subtraction, subtraction becomes addition, and so on.

Why is it so important to check the solution?

Checking your solution by substituting the value back into the original equation confirms your answer is correct. If the Left Hand Side (LHS) equals the Right Hand Side (RHS), you can be certain your solution is right. It's the best way to catch mistakes during exams.

What should I do if the variable cancels out and I get a statement like 5 = 5?

If the variables cancel out and you are left with a true statement (like 5 = 5 or 0 = 0), the equation is an identity. This means that any real number is a valid solution for the variable.

What if the variable cancels out and I get a false statement like 3 = 7?

If the variables cancel out and you are left with a false statement (like 3 = 7), it means the equation has no solution. There is no value of the variable that can make the statement true.