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## Class ten maths Pair of Linear Equn in Two Variables

#### TEXTBOOK’S EXERCISE 3.3

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

(i) x + y = 14; x– y = 4

Sol. (ii) Given equations are :

x + y = 14 ————(i)

x – y = 4 ————-(ii)

x = 4 + y ———-(iii)

From equation (2),

⇒ 4 + y + y = 14

Substituting the value of x in equation (1), we have,

⇒ 4 + y + y = 14 ⇒ 2y = 14 – 4

⇒ 2y = 10

Substituting y = 5 in equation (3), we have x = 4 + 5 = 9

Hence, x = 9 and y = 5

(ii) Given equations are :

S – t =3 or, S = 3 + t ———1

Substituting the value of s in equation (2), we have,

2 (3 + t) + 3t = 36

⇒ 6 + 2t + 3t = 36

⇒ 6 536 + = t

⇒ 5t = 36 – 6

Substituting t = 6 in equation (1), we have, s = 3 + 6 = 9

Hence, s = 9 and t = 6.

(iii) Given equations are :

3x – y = 3

⇒ y = 3x – 3 …(1)

And 9x – 3y = 9 …(2)

Substituting the value of y in equation (2), we have,

⇒ 93339 x x − − = ()

⇒ 9x – 9x + 9 = 9

⇒ 9 = 9

This statement is true for all values of x.

Hence, the given pair of linear equations has infinite number of solutions.

(iv) Given equations are :

0.2x + 0.3y = 1.3

0.2x + 0.3y + 1.3

⇒ 0.2 x = 1.3 – 0.3y

and 0.4x + 0.5y = 2.3

⇒ 0.5y = 2.3 – 0.4 x

Substituting the value of x in equation (2), we have

0.5y = 2.3 – 2.6 + 0.6y

⇒ – 0.1y = – 0.3

Substituting y = 3 in equation (1), we have,

⇒ x = 2

Hence, the required solution is x = 2 and y = 3.

(v) Given equations are :

From equation (2), we have

Substituting the value of x in equation (1), we have

Substituting y = 0 in equation (1), we have,

Hence, x = 0 and y = 0.

(vi) Given equations are :

From equation (1),

9x – 10y = –12 [Cross multiplication]

⇒ 9x = 10y – 12

From equation (2),

⇒ 2x + 3y = 13 …(4)

Substituting the value of x in equation (4), we have

Substituting y = 3 in equation (3), we have,

Hence, x = 2 and y = 3.

2. Solve 2x + 3y = 11 and 2x – 4y = –24 and hence find the value of m for which y = mx + 3.

Sol. Given linear equations are :

From equation (ii),

2x = 4y -24

⇒ 2x = 2 (2y – 12)

⇒ x = 2y – 12 …(iii)

Substituting the value of x in equation (i), we have

2 (2y – 12) +3y = 11

⇒ 4y – 24 + 3y = 11

⇒ 7y = 11 + 24

⇒ 7y = 35

Substituting y = 5 in equation (iii), we have,

x = 2 (5) – 12 =10 – 12 = -2

Putting the value of

x = -2, y = 5

in equation (iv), we have

⇒ 5 = m (–2) + 3

⇒ 5 – 3 = – 2m

⇒ 2 = – 2m

∴ m = –1

Hence, x = –2, y = 5 and m = –1.

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