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.

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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## Assess NCERT Solutions for Class 10 Maths chapter 3 exercise 3.6

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

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

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,

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

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

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

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

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

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,

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

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

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,

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

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

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,

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

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

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

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,

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

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

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,

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

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

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,

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

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.

Also see-

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## Subject Wise NCERT Exemplar Solutions

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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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

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.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

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In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

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