NCERT solution Class 10 pair of linear equations in two variables

NCERT Solutions

Class 10 Maths

Pair of Linear Equations in Two Variables

NCERT Questions

Ex. 3.1 Q.1
Ex.3.1 Q.2
Ex.3.1 Q.3
Ex.3.1 Q.4
Ex.3.1 Q.5
Ex.3.1 Q.6
Ex.3.1 Q.7
Ex.3.2 Q.1
Ex.3.2 Q.2
Ex.3.2 Q.3
Ex.3.3 Q.1
Ex.3.3 Q.2

Ex.3.3 Q.1

Solve the following pair of linear equations by the elimination method and the substitution method:

(1) x + y = 5 and 2x – 3y = 4

(2) 3x + 4y = 10 and 2x – 2y = 2

(3) 3x – 5y – 4 = 0 and 9x = 2y + 7

(4) + y = -1 and x – y = 3

x + y =5 and 2x –3y = 4

By elimination method

x + y =5 ... (1)

2x –3y = 4 ... (2)

Multiplying equation (1) by (2), we get

2x + 2y = 10 ... (3)

2x – 3y = 4 ... (2)

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

5y = 6

y =

Putting the value in equation (1), we get

x = 5 – ( ) =

Hence, x = and y =

By substitution method

x + y = 5 ... (1)

Subtracting y from both sides, we get

x = 5 – y ... (4)

Putting the value of x in equation (2) we get

2(5 – y) – 3y = 4

–5y = – 6

y = =

Putting the value of y in equation (4) we get

x = 5 –

x =

Hence, x = and y = again

(2) 3x + 4y = 10 and 2x – 2y = 2

By elimination method

3x + 4y = 10 .... (1)

2x – 2y = 2 ... (2)

Multiplying equation (2) by 2, we get

4x – 4y = 4 ... (3)

3x + 4y = 10 ... (1)

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

x + 0 = 14

Dividing both side by 7, we get

x = = 2

Putting in equation (1), we get

3x + 4y = 10

3(2) + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

y = = 1

Hence, answer is x = 2, y = 1

By substitution method

3x + 4y = 10 ... (1)

Subtract 3x both sides, we get

4y = 10 – 3x

Divide by 4 we get

y = (10 – 3x) ÷ 4

Putting this value in equation (2), we get

2x – 2y = 2 ... (1)

2x – 2(10 – 3x) ÷ 4) = 2

Multiply by 4 we get

8x – 2(10 – 3x) = 8

8x – 20 + 6x = 8

14x = 28

x = 28 ÷ 14 = 2

y = (10 – 3x) ÷ 4

y = 4 ÷ 4 = 1

Hence, answer is x = 2, y = 1 again.

(3) 3x – 5y – 4 = 0 and 9x = 2y + 7

By elimination method

3x – 5y – 4 = 0

3x – 5y = 4 ... (1)

9x = 2y + 7

9x – 2y = 7 ... (2)

Multiplying equation (1) by 3, we get

9 x – 15 y = 11 ... (3)

9x – 2y = 7 ... (2)

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

–13y = 5

y = –

Putting value in equation (1), we get

3x – 5y = 4 ... (1)

3x – 5(– ) = 4

Multiplying by 13 we get

39x + 25 = 52

39x = 27

x = =

Hence our answer is x = and y = –

By substitution method

3x – 5y = 4 ... (1)

Adding 5y on both sides we get

3x = 4 + 5y

Dividing by 3 we get

x = (4 + 5y) ÷ 3 ... (4)

Putting this value in equation (2) we get

9x – 2y = 7 ... (2)

9 ((4 + 5y) ÷ 3) – 2y = 7

On solve we get

3(4 + 5y) – 2y = 7

12 + 15y – 2y = 7

13y = – 5

y = –

x = {4 + 5 × (-5 ÷ 13)} ÷ 3

x = (4 – ) ÷ 3

x = {(52 - 25) ÷ 13} ÷ 3

x = 27 ÷ (13 × 3)

x =

Hence, we get x = and y = - again.

(4) + = – 1 and x – = 3

By elimination method

+ = – 1 ... (1)

x – = 3 ... (2)

Multiplying equation (1) by 2, we get

x + = – 2 ... (3)

x – = 3 ... (2)

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

= – 5

Dividing by 5 and multiplying by 3, we get

y = –

y = – 3

Putting this value in equation (2), we get

x – = 3 ... (2)

x – (–3) = 3

x + 1 = 3

x = 2

Therefore x = 2 and y = – 3.

By substitution method

x – = 3 ... (2)

Adding on both sides, we get

x = 3 + ... (4)

Putting the value in equation (1) we get

+ = – 1 ... (1)

(3 + ) ÷ 2 + = – 1

+ + = – 1

Multiplying by 6,

9 + y + 4y = – 6

5y = – 15

y = – 3

Therefore x = 2 and y = – 3

Video solution:

Pair of Linear Equations in Two Variables Class 10 Maths NCERT Chapter 3 NCERT Solutions

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