Class ten maths Pair of Linear Equn in Two Variables
Chapter 3: Pair of Linear Equations in Two Variables NCERT Solutions
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.