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

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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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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.

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.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (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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sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024

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hello Zaid,

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024

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Central Board of Secondary Education Class 10th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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

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

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.6

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