NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

# NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

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

## NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.4

NCERT Solutions for class 10 maths ex 3.4 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 3.4 class 10 deals with the graphical and algebraic representations of linear equations in two variables and their different type of solutions such as a unique solution, no solutions or infinitely many solutions, using the five methods namely the Graphical method, substitution method, cross Multiplication method, elimination method and determinant method.

These class 10 maths ex 3.4 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 3.4

Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excrcise: 3.4

$x + y =5 \ \textup{and} \ 2x - 3y = 4$

Elimination Method:

Given, equations

$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$

Now, multiplying (1) by 3 we, get

$\\3x +3 y =15............(3)$

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

$\\2x-3y+3x +3 y =4+15$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value in (1) we, get

$\frac{19}{5}+y=5$

$\Rightarrow y=5-\frac{19}{5}$

$\Rightarrow y=\frac{6}{5}$

Hence,

$x=\frac{19}{5}\:and\:y=\frac{6}{5}$

Substitution method :

Given, equations

$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$

Now, from (1) we have,

$y=5-x.......(3)$

substituting this value in (2)

$2x-3(5-x)=4$

$\Rightarrow 2x-15+3x=4$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value of x in (3)

$\Rightarrow y=5-x=5-\frac{19}{5}=\frac{6}{5}$

Hence,

$x=\frac{19}{5}\:and\:y=\frac{6}{5}$

$3x + 4 y = 10 \ \textup{and} \ 2x - 2y = 2$

Elimination Method:

Given, equations

$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$

Now, multiplying (2) by 2 we, get

$\\4x -4 y =4............(3)$

Now, Adding (1) and (3), we get

$\\3x+4y+4x -4 y =10+4$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Putting this value in (2) we, get

$2(2)-2y=2$

$\Rightarrow 2y=2$

$\Rightarrow y=1$

Hence,

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

Substitution method :

Given, equations

$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$

Now, from (2) we have,

$y=\frac{2x-2}{2}=x-1.......(3)$

substituting this value in (1)

$3x+4(x-1)=10$

$\Rightarrow 3x+4x-4=10$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=x-1=2-1=1$

Hence,

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

Elimination Method:

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, multiplying (1) by 3 we, get

$\\9x -15 y -12=0............(3)$

Now, Subtracting (3) from (2), we get

$9x-2y-7-9x+15y+12=0$

$\Rightarrow 13y+5=0$

$\Rightarrow y=\frac{-5}{13}$

Putting this value in (1) we, get

$3x-5(\frac{-5}{13})-4=0$

$\Rightarrow 3x=4-\frac{25}{13}$

$\Rightarrow 3x=\frac{27}{13}$

$\Rightarrow x=\frac{9}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Substitution method :

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, from (2) we have,

$y=\frac{9x-7}{2}.......(3)$

substituting this value in (1)

$3x-5\left(\frac{9x-7}{2} \right )-4=0$

$\Rightarrow 6x-45x+35-8=0$

$\Rightarrow -39x+27=0$

$\Rightarrow x=\frac{27}{39}=\frac{9}{13}$

Substituting this value of x in (3)

$\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Elimination Method:

Given, equations

$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$

Now, multiplying (2) by 2 we, get

$\\2x - \frac{2y}{3} =6............(3)$

Now, Adding (1) and (3), we get

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

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Putting this value in (2) we, get

$2-\frac{y}{3}=3$

$\Rightarrow \frac{y}{3}=-1$

$\Rightarrow y=-3$

Hence,

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

Substitution method :

Given, equations

$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$

Now, from (2) we have,

$y=3(x-3)......(3)$

substituting this value in (1)

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

$\Rightarrow \frac{x}{2}+2x-6=-1$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=3(x-3)=3(2-1)=-3$

Hence,

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

Let the numerator of the fraction be x and denominator is y,

Now, According to the question,

$\frac{x+1}{y-1}=1$

$\Rightarrow x+1=y-1$

$\Rightarrow x-y=-2.........(1)$

Also,

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow 2x=y+1$

$\Rightarrow 2x-y=1..........(2)$

Now, Subtracting (1) from (2) we get

$x=3$

Putting this value in (1)

$3-y=-2$

$\Rightarrow y=5$

Hence

$x=3\:and\:y=5$

And the fraction is

$\frac{3}{5}$

Let the age of Nuri be x and age of Sonu be y.

Now, According to the question

$x-5=3(y-5)$

$\Rightarrow x-5=3y-15$

$\Rightarrow x-3y=-10.........(1)$

Also,

$x+10=2 (y+10)$

$\Rightarrow x+10=2y+20$

$\Rightarrow x-2y=10........(2)$

Now, Subtracting (1) from (2), we get

$y=20$

putting this value in (2)

$x-2(20)=10$

$\Rightarrow x=50$

Hence the age of Nuri is 50 and the age of Nuri is 20.

Let the unit digit of the number be x and 10's digit be y.

Now, According to the question,

$x+y=9.......(1)$ 24323

Also

$9(10y+x)=2(10x+y)$

$\Rightarrow 90y+9x=20x+2y$

$\Rightarrow 88y-11x=0$

$\Rightarrow 8y-x=0.........(2)$

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

$\Rightarrow 9y=9$

$\Rightarrow y=1$

now putting this value in (1)

$x+1=9$

$\Rightarrow x=8$

Hence the number is 18.

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, According to the question,

$x+y=25..........(1)$

And

$50x+100y=2000$

$\Rightarrow x+2y=40.............(2)$

Now, Subtracting(1) from (2), we get

$y=15$

Putting this value in (1).

$x+15=25$

$\Rightarrow x=10$

Hence Meena received 10 50 Rs notes and 15 100 Rs notes.

Let fixed charge be x and per day charge is y.

Now, According to the question,

$x+4y=27...........(1)$

And

$x+2y=21...........(2)$

Now, Subtracting (2) from (1). we get,

$4y-2y=27-21$

$\Rightarrow 2y=6$

$\Rightarrow y=3$

Putting this in (1)

$x+4(3)=27$

$\Rightarrow x=27-12=15$

Hence the fixed charge is 15 Rs and the per day charge is 3 Rs.

## More About NCERT Solutions for Class 10 Maths Exercise 3.4

In the elimination method from NCERT solutions for Class 10 Maths exercise 3.4 we pick a variable which we need to eliminate then we multiply both the equations with non-zero constants such that the coefficient of the variable we picked becomes numerically equal but with opposite signs (one negative the other positive) so that they get cancelled when we add the equation.

Thus, we are left with the other variable which we can solve easily and get the value of the remaining variable. Once we get the value of the other variable, we substitute it to any of the equations to get the value of the first variable.

In this example, we will eliminate ‘y’ from the equation

x + 2y = 4 ..........(1)

2x - y = 3 ............(2)

We multiply equation (I) with 1 and equation (II) with 2 to make the coefficient of ‘y’ numerically equal.

x + 2y = 4 ..........(3)

4x - 2y = 6 ............(4)

Now add these two equations we get.

5x = 10

x = 10/2 = 2

Now substituting the value of ‘x’ on equation (I) we get.

(2) + 2y = 4

2y = 2

y = 1

So, we have ‘x’ as 2 and ‘y’ as 1.

The elimination method can only be applied if the ratios of the coefficient are different. It doesn’t solve the equations if the ratios are the same.

In the above example, the ratio of equation (I) is 1:2 and the ratio of Equation (II) is 2: -1 which is different. 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 3.4

• Class 10 Maths chapter 3 exercise 3.4 introduces us to a more convenient, easier and time-efficient method for solving a pair of linear equations by elimination method.
• The elimination from NCERT solutions for Class 10 Maths exercise 3.4 method is quite conceptual and this method makes dealing with higher-level questions easier.
• Apart from solving through the algebraic method and graphical method the elimination method of Class 10 Maths chapter 3 exercise 3.4 is quite an easier and time-efficient method to solve the pair of linear equations.
• In some exams, the question asks only the value of one variable so by using the elimination method of Class 10th Maths chapter 3 exercise 3.4 we can eliminate the second variable and end up with the value of the first (required) variable only.

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. Choose the convenient and time efficient method to solve linear a pair of linear equations?
1. Graphical Method

2. Elimination method

Elimination method as it is time-efficient and not lengthy.

2. What is the basic concept of elimination method according to NCERT solutions for Class 10 Maths 3 exercise 3.4?

We multiply both the equations with some non-zero constant to eliminate one of the variables by adding both of the new equations.

3. Why do we multiply both the equations with some non-zero variable?

In order to make the coefficient of one of the variables numerically equal but with opposite signs so that they get cancelled when we add them together.

4. Does substitution play any role in elimination method?

Yes, once we find the value of the first variable, we substitute its value in any of the equation to get the value of the second equation.

5. Why graphical method is not always useful?

It doesn’t provide us the desired point if the values are irrational.

6. Can we apply elimination method in the equation?

It doesn’t provide us the desired point if the values are irrational.

7. Can we apply elimination method in the equation?

It doesn’t provide us the desired point if the values are irrational.

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### Questions related to CBSE Class 10th

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