Linear Equations in Two Variables: Reducing to Linear Form (NCERT Ex 3.6)

Welcome to a crucial topic in your CBSE Class 10 Maths journey: solving pairs of linear equations that don't initially appear linear! In NCERT Exercise 3.6, you'll learn a powerful technique to transform complex-looking equations into the familiar linear form you already know how to solve. Many real-world problems can be modeled by equations that, at first glance, seem non-linear. This exercise equips you with the skills to simplify these problems, making them solvable using methods like substitution, elimination, or cross-multiplication. By the end of this chapter, you will master the art of identifying suitable substitutions, solving the transformed linear systems, and finding the values of the original variables, preparing you for more advanced mathematical concepts and excelling in your board exams.

Understanding Equations Reducible to Linear Form

Often, you will encounter pairs of equations in two variables that are not linear in their original state. A linear equation is typically of the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero. The key characteristic is that the variables x and y have a degree of 1. However, some equations might involve terms like 1/x, 1/y, 1/(x+y), 1/(x-y), or even xy. These are not linear equations. The good news is that many such equations can be reduced to a pair of linear equations by making appropriate substitutions. The general strategy involves identifying common expressions involving the variables in both equations and replacing them with new variables (e.g., u and v). This transformation converts the non-linear system into a standard pair of linear equations in u and v, which you can then solve using any of the algebraic methods you've learned previously (substitution, elimination, or cross-multiplication). Once you find the values of u and v, you simply substitute them back into your original substitution definitions to find the values of x and y. This technique is not just an academic exercise; it's a fundamental problem-solving approach in various scientific and engineering fields where complex relationships need simplification.

Step-by-Step Method for Solving Reducible Equations

  1. Identify Suitable Substitutions — Look for terms or expressions involving variables that are repeated or can be simplified. For example, if you see 1/x or 1/y in both equations, consider substituting u = 1/x and v = 1/y. If you see 1/(x+y) and 1/(x-y), substitute u = 1/(x+y) and v = 1/(x-y).
  2. Formulate the New Linear Equations — Replace the identified expressions with your new variables (e.g., u and v) in both original equations. This will transform the non-linear system into a pair of linear equations in terms of u and v.
  3. Solve the Linear System — Solve the new pair of linear equations for u and v using any of the algebraic methods you are comfortable with: the substitution method, the elimination method, or the cross-multiplication method. Choose the method that seems most efficient for the given equations.
  4. Substitute Back to Find Original Variables — Once you have the values of u and v, substitute these back into your original substitution definitions (e.g., 1/x = u, 1/y = v). Solve these simpler equations to find the values of x and y.
  5. Verify Your Solution — Always check your final values of x and y by substituting them into the original equations. Both equations must be satisfied. This step helps catch any calculation errors made during the process.

Solved Examples: Applying the Reduction Method

  • Example 1: Solve the following pair of equations: 1. 2/x + 3/y = 13 2. 5/x - 4/y = -2 Solution: Step 1: Let 1/x = u and 1/y = v. Step 2: Substitute these into the given equations: 1. 2u + 3v = 13 (Equation 3) 2. 5u - 4v = -2 (Equation 4) Step 3: Solve the new linear system using elimination. Multiply Equation 3 by 4 and Equation 4 by 3: (2u + 3v = 13) 4 => 8u + 12v = 52 (5u - 4v = -2) 3 => 15u - 12v = -6 Add the two new equations: (8u + 12v) + (15u - 12v) = 52 + (-6) 23u = 46 u = 46 / 23 = 2 Step 4: Substitute u = 2 into Equation 3: 2(2) + 3v = 13 4 + 3v = 13 3v = 9 v = 3 Step 5: Substitute back to find x and y: 1/x = u => 1/x = 2 => x = 1/2 1/y = v => 1/y = 3 => y = 1/3 Step 6: Verification: For (1/2, 1/3) in 2/x + 3/y = 13: 2/(1/2) + 3/(1/3) = 22 + 33 = 4 + 9 = 13 (Correct) For (1/2, 1/3) in 5/x - 4/y = -2: 5/(1/2) - 4/(1/3) = 52 - 43 = 10 - 12 = -2 (Correct) Final answer: x = 1/2, y = 1/3
  • Example 2: Solve for x and y: 1. 6/(x+y) + 3/(x-y) = 3 2. 1/(x+y) + 1/(x-y) = 5/9 Solution: Step 1: Let 1/(x+y) = u and 1/(x-y) = v. Step 2: Substitute these into the given equations: 1. 6u + 3v = 3 (Divide by 3: 2u + v = 1) (Equation 3) 2. u + v = 5/9 (Equation 4) Step 3: Solve the new linear system using elimination. Subtract Equation 4 from Equation 3: (2u + v) - (u + v) = 1 - 5/9 u = 4/9 Step 4: Substitute u = 4/9 into Equation 4: 4/9 + v = 5/9 v = 5/9 - 4/9 = 1/9 Step 5: Substitute back to find x and y: 1/(x+y) = u => 1/(x+y) = 4/9 => x+y = 9/4 (Equation 5) 1/(x-y) = v => 1/(x-y) = 1/9 => x-y = 9 (Equation 6) Step 6: Solve the new linear system (Equations 5 and 6) for x and y. Add Equation 5 and Equation 6: (x+y) + (x-y) = 9/4 + 9 2x = 9/4 + 36/4 = 45/4 x = 45/8 Step 7: Substitute x = 45/8 into Equation 5: 45/8 + y = 9/4 y = 9/4 - 45/8 = 18/8 - 45/8 = -27/8 Final answer: x = 45/8, y = -27/8

Exam Tips and Common Mistakes to Avoid

When solving problems involving equations reducible to linear form, students often make a few recurring errors. A common mistake is forgetting the back-substitution step. After finding the values for your new variables (u and v), it's easy to stop there, but remember your goal is to find x and y! Always explicitly substitute u and v back into your initial definitions (1/x=u, etc.) to find the original variables. Another frequent error involves algebraic mistakes when solving the transformed linear system for u and v, particularly with fractions or negative signs. Be meticulous with your calculations. Lastly, always verify your final answer by plugging the calculated x and y values back into the original non-linear equations. If both equations are satisfied, your solution is correct. This verification step is a powerful way to self-correct during exams and secure full marks.

Practice Questions with Solutions

  • Q: Solve for x and y: 1. 10/(x+y) + 2/(x-y) = 4 2. 15/(x+y) - 5/(x-y) = -2 A: Step 1: Let 1/(x+y) = u and 1/(x-y) = v. Step 2: Rewrite the equations as: 1. 10u + 2v = 4 (Divide by 2: 5u + v = 2) (Equation 3) 2. 15u - 5v = -2 (Equation 4) Step 3: Solve the new linear system. Multiply Equation 3 by 5: (5u + v = 2) * 5 => 25u + 5v = 10 (Equation 5) Add Equation 4 and Equation 5: (15u - 5v) + (25u + 5v) = -2 + 10 40u = 8 => u = 8/40 = 1/5 Step 4: Substitute u = 1/5 into Equation 3: 5(1/5) + v = 2 => 1 + v = 2 => v = 1 Step 5: Substitute back to find x and y: 1/(x+y) = u => 1/(x+y) = 1/5 => x+y = 5 (Equation 6) 1/(x-y) = v => 1/(x-y) = 1 => x-y = 1 (Equation 7) Step 6: Add Equation 6 and Equation 7: (x+y) + (x-y) = 5 + 1 => 2x = 6 => x = 3 Step 7: Substitute x = 3 into Equation 6: 3 + y = 5 => y = 2 Final answer: x = 3, y = 2
  • Q: Solve for x and y: 1. 7x - 2y = 5xy 2. 8x + 7y = 15xy A: Step 1: Divide both equations by xy (assuming x ≠ 0, y ≠ 0): 1. 7/y - 2/x = 5 (Equation 1a) 2. 8/y + 7/x = 15 (Equation 2a) Step 2: Let 1/x = u and 1/y = v. Step 3: Rewrite the equations as: 1. 7v - 2u = 5 (Equation 3) 2. 8v + 7u = 15 (Equation 4) Step 4: Solve the new linear system. Multiply Equation 3 by 7 and Equation 4 by 2: (7v - 2u = 5) 7 => 49v - 14u = 35 (8v + 7u = 15) 2 => 16v + 14u = 30 Add the two new equations: (49v - 14u) + (16v + 14u) = 35 + 30 65v = 65 => v = 1 Step 5: Substitute v = 1 into Equation 3: 7(1) - 2u = 5 => 7 - 2u = 5 => 2u = 2 => u = 1 Step 6: Substitute back to find x and y: 1/x = u => 1/x = 1 => x = 1 1/y = v => 1/y = 1 => y = 1 Final answer: x = 1, y = 1
  • Q: Solve for x and y: 1. 1/(2x) + 1/(3y) = 2 2. 1/(3x) + 1/(2y) = 13/6 A: Step 1: Rewrite the equations for clarity: 1. (1/2)(1/x) + (1/3)(1/y) = 2 2. (1/3)(1/x) + (1/2)(1/y) = 13/6 Step 2: Let 1/x = u and 1/y = v. Step 3: Substitute these into the given equations: 1. u/2 + v/3 = 2 (Multiply by 6: 3u + 2v = 12) (Equation 3) 2. u/3 + v/2 = 13/6 (Multiply by 6: 2u + 3v = 13) (Equation 4) Step 4: Solve the new linear system. Multiply Equation 3 by 3 and Equation 4 by 2: (3u + 2v = 12) 3 => 9u + 6v = 36 (2u + 3v = 13) 2 => 4u + 6v = 26 Subtract the second new equation from the first: (9u + 6v) - (4u + 6v) = 36 - 26 5u = 10 => u = 2 Step 5: Substitute u = 2 into Equation 3: 3(2) + 2v = 12 => 6 + 2v = 12 => 2v = 6 => v = 3 Step 6: Substitute back to find x and y: 1/x = u => 1/x = 2 => x = 1/2 1/y = v => 1/y = 3 => y = 1/3 Final answer: x = 1/2, y = 1/3
  • Q: Solve for x and y: 1. 1/(x-y) + 1/(x+y) = 7/6 2. 1/(x-y) - 1/(x+y) = 1/6 A: Step 1: Let 1/(x-y) = u and 1/(x+y) = v. Step 2: Rewrite the equations as: 1. u + v = 7/6 (Equation 3) 2. u - v = 1/6 (Equation 4) Step 3: Solve the new linear system. Add Equation 3 and Equation 4: (u + v) + (u - v) = 7/6 + 1/6 2u = 8/6 => 2u = 4/3 => u = 4/6 = 2/3 Step 4: Substitute u = 2/3 into Equation 3: 2/3 + v = 7/6 => v = 7/6 - 2/3 => v = 7/6 - 4/6 => v = 3/6 = 1/2 Step 5: Substitute back to find x and y: 1/(x-y) = u => 1/(x-y) = 2/3 => x-y = 3/2 (Equation 5) 1/(x+y) = v => 1/(x+y) = 1/2 => x+y = 2 (Equation 6) Step 6: Add Equation 5 and Equation 6: (x-y) + (x+y) = 3/2 + 2 2x = 3/2 + 4/2 => 2x = 7/2 => x = 7/4 Step 7: Substitute x = 7/4 into Equation 6: 7/4 + y = 2 => y = 2 - 7/4 => y = 8/4 - 7/4 => y = 1/4 Final answer: x = 7/4, y = 1/4
  • Q: Solve for x and y: 1. 2/(√x) + 3/(√y) = 2 2. 4/(√x) - 9/(√y) = -1 A: Step 1: Let 1/√x = u and 1/√y = v. Step 2: Rewrite the equations as: 1. 2u + 3v = 2 (Equation 3) 2. 4u - 9v = -1 (Equation 4) Step 3: Solve the new linear system. Multiply Equation 3 by 3: (2u + 3v = 2) * 3 => 6u + 9v = 6 (Equation 5) Add Equation 4 and Equation 5: (4u - 9v) + (6u + 9v) = -1 + 6 10u = 5 => u = 1/2 Step 4: Substitute u = 1/2 into Equation 3: 2(1/2) + 3v = 2 => 1 + 3v = 2 => 3v = 1 => v = 1/3 Step 5: Substitute back to find x and y: 1/√x = u => 1/√x = 1/2 => √x = 2 => x = 2^2 = 4 1/√y = v => 1/√y = 1/3 => √y = 3 => y = 3^2 = 9 Final answer: x = 4, y = 9

Frequently Asked Questions

What does it mean for an equation to be 'reducible to a pair of linear equations'?

It means the equation, while not linear in its original form (e.g., involving `1/x` or `1/y`), can be transformed into a standard linear equation by making suitable substitutions for certain expressions. This allows us to use familiar methods to solve it.

Why do we use new variables like 'u' and 'v'?

Using new variables (like `u = 1/x` or `v = 1/(x+y)`) simplifies the complex equations into a straightforward linear system in terms of `u` and `v`. This makes it much easier to solve using techniques like elimination or substitution, which are designed for linear equations.

What happens if one of the original variables (x or y) is zero?

If an equation has terms like `1/x` or `1/y`, it implies that `x` cannot be zero and `y` cannot be zero, respectively. Similarly, `1/(x+y)` implies `x+y ≠ 0`. Always be mindful of the domain of the original equations.

Can I always use the elimination method for the new linear equations?

You can use any algebraic method (substitution, elimination, or cross-multiplication) that you find most convenient and efficient for solving the new linear system in `u` and `v`. The choice often depends on the coefficients and the structure of the transformed equations.