NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLE EX . 3.4 QUESTION NO. 1 ALL SOLUTIONS

दोस्तों अब हम Class 10 Maths Exercise 3.4 answers NCERT Solutions Question number 1 all के सारे हल देखेंगे।

EX 3.4 CLASS 10 QUESTION 1 PART 4

CLASS 10 MATHS CHAPTER 3 EXERCISE 3.4 QUESTION 2 IN HINDI

प्रश्न - ( 1 ) निम्न समीकरणों के युग्म को विलोपन विधि तथा प्रतिस्थापन विधि से हल कीजिये। कोनसी विधि अधिक उपयुक्त हे।

( I ) x + y = 5 और 2x - 3y = 4

विलोपन विधि :

x + y = 5 -----> ( 1 )

2x - 3y = 4 ------> ( 2 )

समीकरण ( 1 ) में 3 से गुणा करने पर ,

x का यह मान समीकरण (1 ) में रखने पर ,

x + y = 5 -----> ( 1 )

19 / 5 + y = 5

y = 5 - 19 / 5

y = (25 - 19 ) / 5

y = 6 / 5

प्रतिस्थापन विधि : समीकरण (1 ) से , x + y = 5 -----> ( 1 )

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

x का यह मान समीकरण (2 ) में रखने पर ,

2x - 3y = 4 ------> ( 2 )

2 (5 - y ) - 3y = 4

10 - 2y -3y = 4

10 - 5y = 4

- 5y = 4 -10

-5y = -6

y = 6 / 5

y का यह मान समीकरण (3) में रखने पर ,

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

x = 5 - 6 / 5

x = (25 - 6 ) / 5

x = 19 / 5

विलोपन विधि : (ii) 3x + 4y = 10 -----> (1)

2 x - 2y = 2 ------> (2)

समीकरण ( 2 ) को 2 से गुणा करने पर , ( ताकि y का गुणांक बराबर हो जाये। )

4 x - 4y = 4 -------> (3)

समीकरण (1) और (3) को जोड़ने पर ,

7 x = 14

x = 14/ 7

x = 2

x का मान समीकरण (1 ) में रखने पर ,

3x + 4y = 10

3 * 2 + 4y = 10

6 + 4y = 10

4y = 10 - 6

4y = 4

y = 4 / 4 = 1

प्रतिस्थापन विधि : 3x + 4y = 10 -----> (1)

2 x - 2y = 2 ------> (2)

समीकरण (2) से , 2x = 2 + 2y

x =( 2 + 2y ) / 2

x = 2 ( 1 + y ) / 2

x = 1 + y -------> (3)

समीकरण (3) से x का मान समीकरण (1) में रखने पर ,

3x + 4y = 10

3 (1 + y ) + 4y = 10

3 + 3y + 4y = 10

3 + 7y = 10

7y = 10 - 3

7y = 7

y = 7 / 7

y = 1

y का मान समीकरण (3) में रखने पर ,

x = 1 + y -------> (3)

x = 1 + 1

x = 2

(iii ) विलोपन विधि :

3x - 5y -4 = 0 => 3x - 5y = 4 ------> (1 )

9x = 2y + 7 => 9x - 2y = 7 -------> (2)

समीकरण (1 ) में 3 से गुणा करने पर ,

                                      y = - 5 / 13

              समीकरण (1 ) में y का मान रखने पर ,

                                                                3x - 5y = 4

                                                              3 x - 5 * (-5 / 13 ) = 4

                                                              39 x +25 = 52

                                                              39 x = 52 -25

                                                              39 x = 27

                                                                  x = 27 / 39

                                                                  x = 9 / 13

              प्रतिस्थापन विधि :

                                      समीकरण (1 ) से ,

                                                            3x - 5y = 4  ------> (1)

x = (5y +4 ) / 3 --------> (3 )

x का यह मान समीकरण (2 ) में रखने पर ,

समीकरण (2) से ,

9x - 2y = 7 -------> (2)

9 * (5y +4 ) / 3 - 2y = 7

15y + 12 -2y = 7

13y = 7 - 12

13y = -5

y = -5 / 13

y का यह मान समीकरण (3 ) में रखने पर ,

x = (5y +4 ) / 3

x ={ 5 * (-5 / 13 ) + 4 } / 3

x = (-25 +52 ) / 39

x = 27 / 39

x = 9 / 13

x = 9 / 13 , y = -5 / 13 उत्तर

(iv) विलोपन विधि :

x / 2 + 2y / 3 = -1 => 3x + 4y = -6 ------>(1)

x - y / 3 = 3 => 3x - y = 9 -------> (2)

समीकरण ( 1 ) में से समीकरण (2 ) घटाने पर ,

5y = -15

y = -15 / 5

y = -3

y का मान समीकरण (2 ) में रखने पर ,

=> 3x - y = 9

=> 3x - (-3 ) = 9

=> 3x + 3 = 9

=> 3x = 9 - 3

=> 3x = 6

=> x = 6 / 3

=> x = 2

प्रतिस्थापन विधि :

समीकरण (2 ) से ,

3x - y = 9

x = (9 + y ) / 3 ------>(3)

x का यह मान समीकरण (1 ) में प्रतिस्थापित करने पर ,

3x + 4y = -6

3 (9 + y ) / 3 + 4y = -6

9 + y + 4y = -6

5y = -6 -9

5y = -15

y = -15 / 5

y = -3

y का मान समीकरण (3 ) में रखने पर ,

x = (9 + y ) / 3

x = 9 + (-3 ) / 3

x =9 -3 / 3

x = 6 / 3

x = 2

x = 2 , y = -3 उत्तर

ENGLISH TRANSLATION:

Class 10 Maths Exercise 3.4 answers NCERT Solutions Question number 1 all

EX 3.4 CLASS 10 QUESTION 1 PART 4

CLASS 10 MATHS CHAPTER 3 EXERCISE 3.4 QUESTION 2 IN HINDI

Question - ( 1 ) Solve the following pair of equations by elimination method and substitution method. Which method is more suitable.

( I ) x + y = 5 and 2x – 3y = 4

Deletion method:

x + y = 5 -----> ( 1 )

2x - 3y = 4 ------> ( 2 )

In equation ( 1 ) multiplied by 3 ,

Substituting this value of x in Equation (1),

x + y = 5 -----> ( 1 )

19 / 5 + y = 5

y = 5 - 19 / 5

y = (25 - 19 ) / 5

y = 6 / 5

Substitution Method : From Equation (1), x + y = 5 -----> ( 1 )

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

Substituting this value of x in Equation (2),

2x - 3y = 4 ------> ( 2 )

2 (5 - y ) - 3y = 4

10 - 2y -3y = 4

10 - 5y = 4

- 5y = 4 -10

-5y = -6

y = 6 / 5

Substituting this value of y in equation (3),

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

x = 5 - 6 / 5

x = (25 - 6) / 5

x = 19 / 5

Elimination Method : (ii) 3x + 4y = 10 -----> (1)

2 x - 2y = 2 ------> (2)

On multiplying equation ( 2 ) by 2 , ( so that the coefficient of y becomes equal ) .

4 x - 4y = 4 -------> (3)

Adding equation (1) and (3),

7 x = 14

x = 14/ 7

x = 2

Substituting the value of x in Equation (1),

3x + 4y = 10

3 * 2 + 4y = 10

6 + 4y = 10

4y = 10 - 6

4y = 4

y = 4 / 4 = 1

Substitution Method : 3x + 4y = 10 -----> (1)

2 x - 2y = 2 ------> (2)

From equation (2), 2x = 2 + 2y

x =( 2 + 2y ) / 2

x = 2 ( 1 + y ) / 2

x = 1 + y -------> (3)

Substituting the value of x from equation (3) in equation (1),

3x + 4y = 10

3 (1 + y ) + 4y = 10

3 + 3y + 4y = 10

3 + 7y = 10

7y = 10 - 3

7y = 7

y = 7 / 7

y = 1

Substituting the value of y in equation (3),

x = 1 + y -------> (3)

x = 1 + 1

x = 2

(iii) Deletion Method:(ELIMINATION METHOD )

3x - 5y -4 = 0 => 3x - 5y = 4 ------> (1)

9x = 2y + 7 => 9x - 2y = 7 -------> (2)

On multiplying by 3 in equation (1),

y = - 5 / 13

Substituting the value of y in Equation (1),

3x - 5y = 4

3 x - 5 * (-5 / 13 ) = 4

39 x +25 = 52

39 x = 52 -25

39 x = 27

x = 27 / 39

x = 9 / 13

Substitution Method:

From equation (1),

3x - 5y = 4 ------> (1)

x = (5y +4 ) / 3 --------> (3 )

Substituting this value of x in Equation (2),

From equation (2),

9x - 2y = 7 -------> (2)

9 * (5y +4 ) / 3 - 2y = 7

15y + 12 -2y = 7

13y = 7 - 12

13y = -5

y = -5 / 13

Substituting this value of y in Equation (3),

x = (5y +4 ) / 3

x ={ 5 * (-5 / 13 ) + 4 } / 3

x = (-25 +52 ) / 39

x = 27 / 39

x = 9 / 13

x = 9 / 13 , y = -5 / 13 Answer

(iv) Deletion Method:

x / 2 + 2y / 3 = -1 => 3x + 4y = -6 ------>(1)

x - y / 3 = 3 => 3x - y = 9 -------> (2)

On subtracting equation (2) from equation (1),

5y = -15

y = -15 / 5

y = -3

Substituting the value of y in Equation (2),

=> 3x - y = 9

=> 3x - (-3 ) = 9

=> 3x + 3 = 9

=> 3x = 9 - 3

=> 3x = 6

=> x = 6 / 3

=> x = 2

Substitution Method:

From equation (2),

3x - y = 9

x = (9 + y ) / 3 ------>(3)

Substituting this value of x in equation (1),

3x + 4y = -6

3 (9 + y ) / 3 + 4y = -6

9 + y + 4y = -6

5y = -6 -9

5y = -15

y = -15 / 5

y = -3

Substituting the value of y in Equation (3),

x = (9 + y ) / 3

x = 9 + (-3 ) / 3

x =9 -3 / 3

x = 6 / 3

x = 2

x = 2 , y = -3 Answer

NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLE EX . 3.4 QUESTION NO. 1 ALL SOLUTIONS

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