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.3.2 Q.1

Solve the following pair of linear equations by the substitution method.

(1) x + y = 14; x – y = 4

(2) s – t = 3; + = 6

(3) 3x – y = 3; 9x – 3y = 9

(4) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

(5) √2x + √3y = 0; √3x – √8y = 0

(6) x – y = -2; x + y =

(1) x + y = 14 ... (1)

x – y = 4 ... (2)

From equation (1), we get

x = 14 – y ... (3)

Putting this value in equation (2), we get

(14 – y) – y = 4

14 – 2y = 4

10 = 2y

y = 5 ... (4)

Putting this in equation (3), we get

x = 9

So, x = 9 and y = 5

(2) s – t = 3 ... (1)

+ = 6 ... (2)

From equation (1), we get s = t + 3

Putting this value in equation (2), we get

t + + = 6

2t + 6 + 3t = 36

5t = 30

t = ... (4)

Putting in equation (3), we get

s = 9

So, s = 9, t = 6

(3) 3x – y = 3 ... (1)

9x – 3y = 9 ... (2)

From equation (1), we get

y = 3x – 3 ... (3)

Putting this value in equation (2), we get

9x – 3(3x – 3) = 9

9x – 9x + 9 = 9

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation

between these variables can be given by y = 3x – 3

Therefore, one of its possible solutions is x = 1, y = 0.

(4) 0.2x + 0.3y = 1.3 ... (1)

0.4x + 0.5y = 2.3 ... (2)

0.2x + 0.3y = 1.3

Solving equation (1), we get

0.2x = 1.3 – 0.3y

Dividing by 0.2, we get

x = –

x = 6.5 – 1.5 y ... (3)

Putting the value in equation (2), we get

0.4x + 0.5y = 2.3

(6.5 – 1.5y) × 0.4x + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

–0.1y = 2.3 – 2.6

y =

y = 3

Putting this value in equation (3) we get,

x = 6.5 – 1.5 y

x = 6.5 – 1.5(3)

x = 6.5 – 4.5

x = 2

So, x = 2 and y = 3

(5) √2x + √3y = 0 …………... (1)

√3x – √8y = 0 …………. (2)

From equation (1), we get

y = -√2x ÷ √3 …………. (3)

Put value of y in equation (2), we get

√3x - √2(-√2x ÷ √3) = 0

3x – 2x = 0

x = 0

Put value of x in equation (3), we get

y = 0

Hence, x = 0, y = 0

(6) x – y = -2 ……… (1)

+ = ……………. (2)

From equation (1), we get

y = x + 2

y = (3x + 4) ÷ 2

y = 3(3x + 4) ÷ (2 × 5)

y = (9x + 12) ÷ 10 ……… (3)

Put value of y in equation (2), we get

+ ( ) × (9x + 12) ÷ 10 =

=> (20x + 27x + 36) ÷ 60 =

=> (47x + 36) ÷ 60 = 13 ÷ 6

=> (47x + 36) ÷ 10 = 13

=> 47x + 36 = 13 × 10

=> 47x + 36 = 130

=> 47x = 130 – 36

=> 47x = 94

=> x =

=> x = 2

Put value of x in equation (3), we get

y = (9 × 2 + 12) ÷ 10

=> y = (18 + 12) ÷ 10

=> y = 30 ÷ 10

=> y = 3

Hence, x = 2, y = 3

Video solution:

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

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