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NCERT Solutions for Exercise 3.6 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

NCERT Solutions for Exercise 3.6 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Edited By Ramraj Saini | Updated on Nov 13, 2023 03:23 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6

NCERT Solutions for class 10 maths ex 3.6 Pair of Linear Equations in two variables is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 3.6 class 10 consists of two questions, each with subparts. The first is straightforward, requiring only direct substitutions of the mathematical formula, whereas the second involves word problems. In this exercise, the majority of the variables are in fractional form, meaning that they are nonlinear, since the variables are now infraction, we can’t take the value of the denominator as zero because it will be against the definition of the fraction.

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  1. NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6
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  4. More About Ncert Solutions For Class 10 Maths Exercise 3.6
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  6. NCERT Solutions of Class 10 Subject Wise
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NCERT Solutions for Exercise 3.6 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables
NCERT Solutions for Exercise 3.6 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

These class 10 maths ex 3.6 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.6

Q1(i) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}

Answer:

Given Equations,

\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}

Let,

\frac{1}{x}=p\:and\:\frac{1}{y}=q

Now, our equation becomes

\frac{p}{2}+\frac{q}{3}=2

\Rightarrow 3p+2q=12........(1)

And

\frac{p}{3}+\frac{q}{2}=\frac{13}{6}

\Rightarrow 2p+3q=13..........(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{p}{(2)(-13)-(3)(-12)}=\frac{q}{(-12)(2)-(-13)(3)}=\frac{1}{(3)(3)-(2)(2)}

\frac{p}{-26+36}=\frac{q}{-24+39}=\frac{1}{9-4}

\frac{p}{10}=\frac{q}{15}=\frac{1}{5}

p=2,\:and\:q=3

And Hence,

x=\frac{1}{2}\:and\:y=\frac{1}{3}.

Q1(ii) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1

Answer:

Given Equations,

\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1

Let,

\frac{1}{\sqrt{x}}=p\:and\:\frac{1}{\sqrt{y}}=q

Now, our equation becomes

2p+3q=2........(1)

And

4p-9q=-1..........(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{p}{(3)(1)-(-9)(-2)}=\frac{q}{(-2)(4)-(1)(2)}=\frac{1}{(2)(-9)-(4)(3)}

\frac{p}{3-18}=\frac{q}{-8-2}=\frac{1}{-18-12}

\frac{p}{-15}=\frac{q}{-10}=\frac{1}{-30}

p=\frac{1}{2},\:and\:q=\frac{1}{3}

So,

p=\frac{1}{2}=\frac{1}{\sqrt{x}}\Rightarrow x=4

q=\frac{1}{3}=\frac{1}{\sqrt{y}}\Rightarrow y=9 .

And hence

x=4\:and\:y=9.

Q1(iii) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23

Answer:

Given Equations,

\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23

Let,

\frac{1}{x}=p\:and\:y=q

Now, our equation becomes

\Rightarrow 4p+3q=14........(1)

And

\Rightarrow 3p-4q=23..........(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{p}{(3)(-23)-(-4)(-14)}=\frac{q}{(-14)(3)-(-23)(4)}=\frac{1}{(4)(-4)-(3)(3)}

\frac{p}{-69-56}=\frac{q}{-42+92}=\frac{1}{-16-9}

\frac{p}{-125}=\frac{q}{50}=-\frac{1}{25}

p=5,\:and\:q=-2

And Hence,

x=\frac{1}{5}\:and\:y=-2.

Q1(iv) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1

Answer:

Given Equations,

\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1

Let,

\frac{1}{x-1}=p\:and\:\frac{1}{y-2}=q

Now, our equation becomes

5p+q=2........(1)

And

6p-3q=1..........(2)

Multiplying (1) by 3 we get

15p+3q=6..........(3)

Now, adding (2) and (3) we get

21p=7

\Rightarrow p=\frac{1}{3}

Putting this in (2)

6\left ( \frac{1}{3} \right )-3q=1

\Rightarrow 3q=1

\Rightarrow q=\frac{1}{3}

Now,

p=\frac{1}{3}=\frac{1}{x-1}\Rightarrow x-1=3\Rightarrow x=4

q=\frac{1}{3}=\frac{1}{y-2}\Rightarrow y-2=3\Rightarrow x=5

Hence,

x=4,\:and\:y=5.

Q1(v) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\\frac{7x - 2y}{xy} = 5\\ \frac{8x + 7y}{xy} = 15

Answer:

Given Equations,

\\\frac{7x - 2y}{xy} = 5\\\\\Rightarrow\frac{7}{y} -\frac{2}{x}=5\\ \frac{8x + 7y}{xy} = 15\\\Rightarrow \frac{8}{y}+\frac{7}{x}=15

Let,

\frac{1}{x}=p\:and\:\frac{1}{y}=q

Now, our equation becomes

7q-2p=5........(1)

And

8q+7p=15..........(2)

By Cross Multiplication method,

\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{q}{(-2)(-15)-(7)(-5)}=\frac{p}{(-5)(8)-(-15)(7)}=\frac{1}{(7)(7)-(8)(-2)}

\frac{q}{30+35}=\frac{p}{-40+105}=\frac{1}{49 +16}

\frac{q}{65}=\frac{p}{65}=\frac{1}{65}

p=1,\:and\:q=1

And Hence,

x=1\:and\:y=1.

Q1(vi) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\6x + 3y = 6xy\\ 2x + 4y = 5 xy

Answer:

Given Equations,

\\6x + 3y = 6xy\\\Rightarrow \frac{6x}{xy}+\frac{3y}{xy}=6\\\\\Rightarrow \frac{6}{y}+\frac{3}{x}=6\\and\\\ 2x + 4y = 5 xy\\\Rightarrow \frac{2x}{xy}+\frac{4y}{xy}=5\\\Rightarrow \frac{2}{y}+\frac{4}{x}=5

Let,

\frac{1}{x}=p\:and\:\frac{1}{y}=q

Now, our equation becomes

6q+3p=6........(1)

And

2q+4p=5..........(2)

By Cross Multiplication method,

\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{q}{(3)(-5)-(-6)(4)}=\frac{p}{(6)(2)-(6)(-5)}=\frac{1}{(6)(4)-(3)(2)}

\frac{q}{-15+24}=\frac{p}{-12+30}=\frac{1}{24 -6}

\frac{q}{9}=\frac{p}{18}=\frac{1}{18}

q=\frac{1}{2}\:and\:p=1

And Hence,

x=1\:and\:y=2.

Q1(vii) Solve the following pairs of equations by reducing them to a pair of linear equations:
\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2


Answer:

Given Equations,

\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2

Let,

\frac{1}{x+y}=p\:and\:\frac{1}{x-y}=q

Now, our equation becomes

10p+2q=4........(1)

And

15p-5q=-2..........(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{p}{(2)(2)-(-5)(-4)}=\frac{q}{(-4)(15)-(2)(10)}=\frac{1}{(10)(-5)-(15)(2)}

\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}

\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}

p=\frac{1}{5},\:and\:q=1

Now,

p=\frac{1}{5}=\frac{1}{x+y}

\Rightarrow x+y=5........(3)

And,

q=1=\frac{1}{x-y}

\Rightarrow x-y=1...........(4)

Adding (3) and (4) we get,

\Rightarrow 2x=6

\Rightarrow x=3

Putting this value in (3) we get,

3+y=5

\Rightarrow y=2

And Hence,

x=3\:and\:y=2.

Q1(viii) Solve the following pairs of equations by reducing them to a pair of linear equations:

\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}


Answer:

Given Equations,

\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}

Let,

\frac{1}{3x+y}=p\:and\:\frac{1}{3x-y}=q

Now, our equation becomes

p+q=\frac{3}{4}.........(1)

And

\\\frac{p}{2}-\frac{q}{2}=\frac{-1}{8}\\\\p-q=\frac{-1}{4}..........(2)

Now, Adding (1) and (2), we get

2p=\frac{3}{4}-\frac{1}{4}

\Rightarrow 2p=\frac{2}{4}

\Rightarrow p=\frac{1}{4}

Putting this value in (1)

\frac{1}{4}+q=\frac{3}{4}

\Rightarrow q=\frac{3}{4}-\frac{1}{4}

\Rightarrow q=\frac{2}{4}

\Rightarrow q=\frac{1}{2}

Now,

p=\frac{1}{4}=\frac{1}{3x+y}

\Rightarrow 3x+y=4...........(3)

And

q=\frac{1}{2}=\frac{1}{3x-y}

\Rightarrow 3x-y=2............(4)

Now, Adding (3) and (4), we get

6x=4+2

\Rightarrow 6x=6

\Rightarrow x=1

Putting this value in (3),

3(1)+y=4

\Rightarrow y=4-3

\Rightarrow y=1

Hence,

x=1,\:and\:y=1

Q2(i) Formulate the following problems as a pair of equations and hence find their solutions: 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.

Answer:

Let the speed of Ritu in still water be x and speed of current be y,

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction (downstream)= x +y

the relative speed when they are going in the opposite direction (upstream)= x - y

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

x+y=\frac{20}{2}

\Rightarrow x+y=10.........(1)

And,

x-y=\frac{4}{2}

\Rightarrow x-y=2...........(2)

Now, Adding (1) and (2), we get

2x=10+2

\Rightarrow 2x=12

\Rightarrow x=6

Putting this in (2)

6-y=2

\Rightarrow y=6-2

\Rightarrow y=4

Hence,

x=6\:and\:y=4.

Hence Speed of Ritu in still water is 6 km/hour and the speed of the current is 4 km/hour

Q2(ii) Formulate the following problems as a pair of equations and hence find their solutions: 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.

Answer:

Let the number of days taken by woman and man be x and y respectively,

The proportion of Work done by a woman in a single day

=\frac{1}{x }

The proportion of Work done by a man in a single day

=\frac{1}{y }

Now, According to the question,

4\left ( \frac{2}{x}+\frac{5}{y} \right )=1

\Rightarrow \left ( \frac{2}{x}+\frac{5}{y} \right )=\frac{1}{4}

Also,

3\left ( \frac{3}{x}+\frac{6}{y} \right )=1

\Rightarrow \left ( \frac{3}{x}+\frac{6}{y} \right )=\frac{1}{3}

Let,

\frac{1}{x}=p\:and\:\frac{1}{y}=q

Now, our equation becomes

2p+5q=\frac{1}{4}

8p+20q=1........(1)

And

3p+6p=\frac{1}{3}

\Rightarrow 9p+18p=1.............(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{p}{(20)(-1)-(18)(-1)}=\frac{q}{(-1)(9)-(8)(-1)}=\frac{1}{(8)(18)-(20)(9)}

\frac{p}{-20+18}=\frac{q}{-9+8}=\frac{1}{146 -60}

\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}

p=\frac{1}{18},\:and\:q=\frac{1}{36}

So,

x=18\:and\:y=36.

Q2(iii) Formulate the following problems as a pair of equations and hence find their solutions: 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.

Answer:

Let the speed of the train and bus be u and v respectively

Now According to the question,

\frac{60}{u}+\frac{240}{v}=4

And

\frac{100}{u}+\frac{200}{v}=4+\frac{1}{6}

\Rightarrow \frac{100}{u}+\frac{200}{v}=\frac{25}{6}

Let,

\frac{1}{u}=p\:and\:\frac{1}{v}=q

Now, our equation becomes

60p+140q=4

\Rightarrow 15p+60q=1.........(1)

And

100p+200q=\frac{25}{6}

\Rightarrow 4p+8q=\frac{1}{6}

\Rightarrow 24p+48q=1..........(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{q}{(60)(-1)-(48)(-1)}=\frac{p}{(-1)(24)-(-1)(15)}=\frac{1}{(15)(48)-(60)(24)}

\frac{p}{-60+48}=\frac{q}{-24+15}=\frac{1}{720-1440}

\frac{p}{-12}=\frac{q}{-9}=\frac{1}{-720}

p=\frac{12}{720}=\frac{1}{60},\:and\:q=\frac{9}{720}=\frac{1}{80}

And Hence,

x=60\:and\:y=80

Hence the speed of the train and bus are 60 km/hour and 80 km/hour respectively.

More About Ncert Solutions For Class 10 Maths Exercise 3.6

NCERT solutions for Class 10 Maths exercise 3.6- consists of two extremely important questions. In order to answer the first question, you must reduce the pair of equations to a pair of linear equations. Students must express problems as a pair of equations in order to determine their answers to the second question. Furthermore, exercise 3.6 is a more advanced chapter exercise. Above all, this task is an excellent technique to assess a student's ability to solve linear equations in two variables. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 3.6

  • NCERT book Exercise 3.6 Class 10 Maths, is based on the basic idea of solving a pair of equations with fractional variables.
  • NCERT solutions for Class 10 Maths chapter 3 exercise 3.6 helps to solve complex non-linear equations by simplifying it to simple linear equations.
  • If you go through the NCERT solution for Class 10 Maths chapter 3 exercise 3.6, we can see that we will have to use the previous solving methods like substitution and elimination. Thus, this part helps us to clear our previous concepts too.

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NCERT Solutions of Class 10 Subject Wise

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Frequently Asked Question (FAQs)

1. What are the methods used for solving linear equation in two variables?

We use substitution method, graphical and elimination method etc.

2. Which one of the methods is easier compared to the other between Graphical and Elimination Method?

Because it is time-efficient and does not take a long time to complete. Hence, the elimination method is preferred.

3. What is the basic concept behind NCERT solutions for Class 10 Maths 3 exercise 3.4?

In this exercise, the majority of the variables are in fractional form, meaning that they are nonlinear, since the variables are now infractions, we can’t take the value of the denominator as zero because it will be against the definition of the fraction.

4. How do we solve the majority of the variables which are in fractional form?

We substitute the fraction variable as another new variable; thus, we get two new linear equations that we can solve easily.

5. Is there a role for substitution in the elimination method?

Yes, we can get the value of the second equation by substituting the value of the first variable into any of the equations.

6. Do we need any kind of substitution in order to get the final answer?

Yes, we need substitution in order to get the final answer.

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558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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