### Chapter 3: Pair of Linear Equations in Two Variables

Q
##### Pair of linear equations in two variables CBSE NCERT Solutions

Question:

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

(i) x + y = 14;   x – y = 4

(ii) s – t = 3;

(iii) 3x – y = 3;  9x − 3y = 9

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

(v)

(vi)   ;

(i) x + y = 14 …(1)

x – y = 4  … (2)

From equation (2),   x – y = 4 x = 4 + y

Putting  x = 4 + y  in equation (1), we get

4 + y + y = 14

2y = 10  = 5

Putting value of y in equation (1), we get

x + 5 = 14

= 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s – t = 3 … (1)

…(2)

Using equation (1), we can say that s = 3 + t

Putting s = 3 + t in equation (2), we get

6 + 2t + 3t = 36

5t + 6 = 36

5t = 30  = 6

Putting value of t = 6 in equation (1), we get

– 6 = 3  = 3 + 6 = 9

Therefore, t = 6 and s = 9

(iii) 3x – y = 3 … (1)

9x − 3y = 9 … (2)

Comparing equation 3x – y = 3 with   and equation 9x − 3y = 9 with  .

We get   and

Here,

Therefore, we have infinite many solutions for x and y

(iv) 0.2x + 0.3y = 1.3 … (1)

0.4x + 0.5y = 2.3 … (2)

Using equation (1), we can say that

0.2x = 1.3 − 0.3y

=

Putting this in equation (2), we get

2.6 0.6y + 0.5y = 2.3

0.1y = −0.3  = 3

Putting value of y in (1), we get

0.2x + 0.3 (3) = 1.3

0.2x + 0.9 = 1.3

0.2x = 0.4  = 2

Therefore, x = 2 and y = 3

(v)       ….(1)

….(2)

Using equation (1), we can say that

Putting this in equation (2), we get

= 0

Putting the value of y in (1), we get x = 0

Therefore, x = 0 and y = 0

(vi)   … (1)

… (2)

Using equation (2), we can say that

Putting this in equation (1), we get

= 3

Putting the value of y in equation (2), we get

x = 2

Therefore, = 2 and = 3

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