CBSE Notes, Lectures

CBSE - Mathematics - Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables

NCERT Exercise Exercise 3.3

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

Solve the following pair of linear equations by the substitution method.
(i) = 14 ; – = 4

(ii) – = 3 ; s/3 + t/2 = 6
(iii) 3x – y = 3 ; 9x – 3y = 9 

(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
(v) √2x+ √3y = 0 ; √3x - √8y = 0 

(vi) 3/2x - 5/3y = -2 ; x/3 + y/2 = 13/6


Answer

(i) x + y = 14 ... (i)

x – y = 4 ... (ii)
From equation (i), we get

x = 14 - y ... (iii)
Putting this value in equation (ii), we get

(14 - y) - y = 4

14 - 2y = 4

10 = 2y

= 5 ... (iv)

Putting this in equation (iii), we get

= 9

∴ = 9 and y = 5


(ii) – = 3 ... (i)
s/3 + t/2 = 6 ... (ii)
From equation (i), we gett + 3
Putting this value in equation (ii), we get
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5 ... (iv)
Putting in equation (iii), we obtain
s = 9
∴ s = 9, t = 6

(iii) 3x - = 3 ... (i)
9x - 3y = 9 ... (ii)
From equation (i), we get
y = 3x - 3 ... (iii)
Putting this value in equation (ii), 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.

(iv) 0.2x + 0.3y = 1.3 ... (i) 
0.4x + 0.5y = 2.3 ... (ii)
0.2x + 0.3y = 1.3
Solving equation (i), we get
0.2x = 1.3 – 0.3y
Dividing by 0.2, we get
x = 1.3/0.2 - 0.3/0.2
x = 6.5 – 1.5 y …(iii)
Putting the value in equation (ii), 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
= -0.3/-0.1
y = 3
Putting this value in equation (iii) we get
x = 6.5 – 1.5 y
x = 6.5 – 1.5(3)
x = 6.5 - 4.5
x = 2
∴ = 2 and y = 3 

(vi) 3/2x - 5/3y = -2 ... (i)
x/3 + y/2 = 13/6 ... (ii)
From equation (i), we get
9x - 10y = -12
x = -12 + 10y/9 ... (iii)
Putting this value in equation (ii), we get

.