NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6

With the NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6, students can enhance their skills for solving a pair of non-linear equations in two variables. The exercise deals with the equations that don’t seem to be linear at first look but are transformable into one by performing some mathematical operations. Understanding these mathematical operations is essential for solving such non-linear equations and real-life problems. Here, you can find all the key concepts and an overview of the exercise, along with NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6 PDF.

1.0Download NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6 : Free PDF 

2.0Introduction to Equations Reducible to a Pair of Linear Equations: 

Before getting resourceful insights into the method of reducing non-linear equations, it is important to understand the meaning of these equations. Most of the time, the equations provided in a problem are also not linear in their original form. Such equations may have terms such as x², y²  {\x^2 , y^2} or even variable products (such as xy), which are not directly solvable via standard methods for linear equations. However, if proper mathematical operations like substitutions are performed, such non-linear equations can be made linear. Once these equations are reduced to linear form, they can be solved via standard methods such as substitution or elimination.

3.0Key Concepts of Exercise 3.6: Overview 

Exercise 3.6 goes beyond the traditional linear equations; that is, it primarily focuses on handling non-linear equations not solved with standard methods. Let’s see some key concepts and insightful tips used in solving the exercise: 

4.0Equations Reducible to Linear Form: Overview

The reducible equations may seem very complex; however, with the right approach, this can be made easy. Here is how students should approach solving these questions: 

• Identifying the Non-Linear Equations: The first step for solving these equations is to identify the non-linear part or parts of the equation. These parts can be in any form, such that they can be quadratic, cubic, product of variables, or even a fraction of variables. This step is crucial to know the need for substitution for further operations. 
• Making Suitable Substitutions: Once you have identified the non-linear parts of the equations, substitute them with a proper linear form. For example, if you are provided with an equation 1xor 1y, substitute 1x=u, and 1y=v, \frac{1}{x}\textit{or}\frac{1}{y}, substitute\frac{1}{x} = u, \textit{and} \frac{1}{y} = vor as per the requirement of the question. 
• Solving the Reduced Linear System: Now that the non-linear equations are reduced and transformed into linear ones, solve them using any one of these methods: 
• • Substitution Method
  • Elimination Method
  • Cross multiplication method 
  • Graphical representation of a pair of equations. 

All of these methods mentioned above can be found in other sections of the same chapter 3. 

• Back Substituting to Find the Final Solution: After getting the values of assumed variables, back substitute the value of these variables in the original equations. This step is done to find the solutions to the non-linear equations. 

Verifying the Solutions:

It is always necessary to check the solutions against the initial equations to confirm there are no errors in the solutions. Substitution can sometimes bring in solutions that don't satisfy the initial equation, so checking is an essential step. This step can essentially help in situations where you don’t know the answer to the question but want to verify its validity, like in the exam hall. 

5.0Exercise 3.6: Overview

• The exercise is filled with different complex questions of reducible equations, which need to be solved using the method mentioned above. 
• The exercise also includes the solutions of real-life scenarios of reducible equations, like in a situation of a boat and its distance and speed. 

To enhance your problem-solving skills and do well in the exams, try practicing NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6, to get sure results and improvement.

6.0NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6 : Detailed Solutions

1. Solve the following pairs of equations by reducing them to a pair of linear equations :

(i) 1/(2x) + 1/(3y) = 2, 1/(3x) + 1/(2y) = 13/6

(ii) 2/√x + 3/√y = 2, 4/√x - 9/√y = -1

(iii) 4/x + 3y = 14, 3/x - 4y = 23

(iv) 5/(x-1) + 1/(y-2) = 2, 6/(x-1) - 3/(y-2) = 1

(v) (7x-2y)/xy = 5, (8x+7y)/xy = 15

(vi) 6x + 3y = 6xy, 2x + 4y = 5xy

(vii) 10/(x+y) + 2/(x-y) = 4, 15/(x+y) - 5/(x-y) = -2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4, 1/(2(3x+y)) - 1/(2(3x-y)) = -1/8

Solution:

(i) 1/(2x) + 1/(3y) = 2, 1/(3x) + 1/(2y) = 13/6

Substituting 1/x = u and 1/y = v

We get, (1/2)u + (1/3)v = 2, (1/3)u + (1/2)v = 13/6

Multiplying by 6 on both sides, we get

=> 3u + 2v = 12 ...(1)

2u + 3v = 13 ...(2)

Multiplying (1) by 3 and (2) by 2, then subtracting the latter from the first, we get

3(3u + 2v) - 2(2u + 3v) = 3 × 12 - 2 × 13

=> 9u + 6v - 4u - 6v = 36 - 26

=> 5u = 10

=> u = 2

Then substituting u = 2 in (1), we get

6 + 2v = 12

=> v = 3

Now, u = 2 and v = 3

=> 1/x = 2 and 1/y = 3 => x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2 ...(1)

4/√x - 9/√y = -1 ...(2)

Take 1/√x = a, 1/√y = b, we get

2a + 3b = 2 ...(3)

4a - 9b = -1 ...(4)

Multiplying (3) by 3, we get

6a + 9b = 6 ...(5)

Adding (4) and (5), we get

10a = 5

=> a = 1/2

Substituting a = 1/2 in (3), we get

3b = 1

=> b = 1/3

Now, 1/√x = a = 1/2 and 1/√y = b = 1/3

=> √x = 2, √y = 3

Squaring, we get

x = 4, y = 9

(iii) 4/x + 3y = 14

3/x - 4y = 23

Take, 1/x = a

4a + 3y = 14 ...(3)

3a - 4y = 23 ...(4)

Multiplying (3) by 4 and (4) by 3

16a + 12y = 56

9a - 12y = 69

Adding both, we get

25a = 125

=> a = 5

Substituting a in (3), we get

20 + 3y = 14

=> 3y = -6

=> y = -2

As, 1/x = a = 5

=> x = 1/5

Hence, x = 1/5 and y = -2

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) - 3/(y-2) = 1

Take, 1/(x-1) = a and 1/(y-2) = b

5a + b = 2 ...(3)

6a - 3b = 1 ...(4)

Multiplying (3) by 3, we get

15a + 3b = 6

6a - 3b = 1

Adding both, we get

21a = 7

a = 1/3

So, by solving, b = 1/3

As, a = 1/3 = 1/(x-1)

=> x - 1 = 3 => x = 4

and b = 1/3 = 1/(y-2)

=> y - 2 = 3

=> y = 5

(v) (7x-2y)/xy = 5, (8x+7y)/xy = 15

By solving, we get

7/y - 2/x = 5

8/y + 7/x = 15

Taking 1/y = u, 1/x = v

7u - 2v = 5 ...(3)

8u + 7v = 15 ...(4)

Multiplying (3) by 7 and (4) by 2, we get

49u - 14v = 35

16u + 14v = 30

Adding, we get

65u = 65

u = 1

By solving, we get v = 1

As, u = 1 = 1/y => y = 1

And, v = 1 = 1/x => x = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

By solving, we get

6/y + 3/x = 6

2/y + 4/x = 5

Take, 1/y = u, 1/x = v, we get

6u + 3v = 6 ...(3)

2u + 4v = 5 ...(4)

Multiply (4) by 3, we get

6u + 12v = 15 ...(5)

Subtract (3) from (5), we get

9v = 9

v = 1

By solving we get u = 1/2

As, 1/x = v = 1 => x = 1

And 1/y = u = 1/2

=> y = 2

(vii) 10/(x+y) + 2/(x-y) = 4, 15/(x+y) - 5/(x-y) = -2

Take 1/(x+y) = u and 1/(x-y) = v

10u + 2v = 4 ...(1)

15u - 5v = -2 ...(2)

Multiply (1) by 5 and (2) by 2, we get

50u + 10v = 20

30u - 10v = -4

Adding, we get

80u = 16

u = 1/5

By solving, we get v = 1

As, 1/(x+y) = u = 1/5

=> x + y = 5 ...(3)

And 1/(x-y) = v = 1

=> x - y = 1 ...(4)

Adding (3) and (4), we get

2x = 6

=> x = 3 and y = 2

Hence, x = 3, y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4 ...(1)

1/(2(3x+y)) - 1/(2(3x-y)) = -1/8 ...(2)

Let, 1/(3x+y) = u, 1/(3x-y) = v

u + v = 3/4 ...(3)

u/2 - v/2 = -1/8 ...(4)

From (4), we get

u - v = -1/4 ...(5)

Solving (3) and (5), we get:

2u = 1/2

u = 1/4, v = 1/2

So, 1/(3x + y) = 1/4  =>  3x + y = 4

1/(3x - y) = 1/2  =>  3x - y = 2

Solving these equations, we get:

x = 1, y = 1

2. Formulate the following problems as a pair of linear equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

(i) Speed of Ritu in still water = x km/hr

Speed of current = y km/hr

Then speed downstream = (x+y) km/hr

speed upstream = (x-y) km/hr

20/(x+y) = 2 and 4/(x-y) = 2

=> x+y = 10  ...(1)

x-y = 2   ...(2)

From (1) and (2)

x+y = 10

x-y = 2

2x = 12 => x = 6

From (1)

6+y = 10

y = 4

So the speed of Ritu in still water = 6 km/hr and speed of current = 4 km/hr.

(ii) Let 1 woman finish the work in x days and let 1 man finish the work in y days.

Work of 1 woman in 1 day = 1/x

Work of 1 man in 1 day = 1/y

Work of 2 women and 5 men in one day = 2/x + 5/y

The number of days required for complete work = xy/(5x+2y)

We are given that xy/(5x+2y) = 4

Similarly, in second case xy/(6x+3y) = 3

xy = 4(5x+2y)

xy = 3(6x+3y)

By solving, we get

xy = 20x+8y

xy = 18x+9y

=> 20/y + 8/x = 1 ...(5)

18/y + 9/x = 1 ...(6)

Putting 1/y = u, 1/x = v

20u + 8v = 1

18u + 9v = 1

Multiply (5) by 9 and (6) by 10

180u + 72v = 9

180u + 90v = 10

On subtracting, we get

18v = 1

v = 1/18

and u = 1/36

As u = 1/36 = 1/y => y = 36 days

And v = 1/18 = 1/x => x = 18 days

(iii) Let the speed of train be x km/hr and the speed of bus be y km/hr

So, 60/x + 240/y = 4 ...(1)

100/x + 200/y = 25/6 ...(2)

Let 1/x = u, 1/y = v

60u + 240v = 4 ...(3)

100u + 200v = 25/6 ...(4)

Solving (3) and (4), we get

u = 1/60, v = 1/80

So, 1/x = 1/60 and 1/y = 1/80

∴ x = 60 km/hr and y = 80 km/hr

7.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.6

1. Helps students plot linear equations on graphs and understand points of intersection.
2. Explains the types of solutions (unique, no solution, infinite) clearly through visual and algebraic methods.
3. Prepares students for board exam questions with similar patterns and step-by-step methods.
4. Enhances analytical thinking and accuracy in solving equation-based problems.
5. Builds a base for advanced algebra in higher classes and competitive exams like JEE, NTSE, etc.