class ten maths Chapter 3 NCERT Solutions
Chapter 3: Linear equations in two variables NCERT solutions
TEXTBOOK’S EXERCISE 3.4
1. Solve the following pair of equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
Sol. (i) Given equations are :
x + y = 5 …(1)
2x – 3y = 4 …(2)
Elimination Method :
Multiplying both sides of (1) by 2, we have
2x + 2y = 10 …(3)
Subtracting equation (2) from equation (3), we have
Substituting the value of y in equation (1), we have
Substitution Method :
From equation (2), 2x = 4 + 3y
Substituting the value x in equation (1), we have
Substituting the value of y in equation (4), we have
(ii) Given equations are :
3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
Elimination Method :
Multiplying both sides of equation (2) by 2, we have
4x – 4y = 4 …(3)
Adding equations (3) and (2), we have
Substituting the value of x in equation (1), we have
3(2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 10 – 6
Hence, x = 2 and y = 1
Substitution Method :
From equation (2)
2x = 2 + 2y
⇒ x = y + 1
Substituting the value of x in equation (1), we have
3(y+1) + 4y = 10 ⇒ 3y + 3 + 4y = 10
⇒ 7y = 10 – 3 ⇒ 7y = 7
∴ y = 1
Substituting the value of y in equation (4), we have
X = 1 + 1 = 2
Hence, x = 2 and y = 1
(iii) Given equations are :
3x – 5y – 4 = 0 …(1)
9x = 2y + 7 ⇒ 9x – 2y – 7 = 0 …(2)
Elimination Method :
Multiplying both sides of equation (1) by 3, we have
9x – 15y – 12 = 0 …(3)
Subtracting equation (2) from equation (3), we have
y = – 5/ 13
Substituting the value of y in equation (1), we have
class ten maths Chapter 3 NCERT Solutions
Substitution Method :
From equation (2) we have
Substituting the value of x in equation (1), we have
– 13y – 5 = 0
⇒ – 13y = 5
y= − 5/13
Substituting the value of y in equation (4), we have
Hence, x = − 9/13 and y = 5/13
(iv) Given equations are :
3x + 4y = – 6 …(1)
And, x − y = 3 3
⇒ 3x – y = 9———2
Elimination Method :
Subtracting equation (2) from equation (1), we have
Elimination Method :
Subtracting equation (2) from equation (1), we have
Substituting the value of y in equation (1), we have
3x + 4 (–3) = – 6
⇒ 3x – 12 = – 6
⇒ 3x = – 6 + 12
⇒ 3x = 6
∴ x= = 63 2
Hence, x = 2, y = – 3.
Substitution Method :
From equation (2), y = 3x – 9 …(3)
Substituting the value of y in equation (1), we have
3x + 4 (3x–9) = –6
⇒ 3x + 12x – 36 = – 6
⇒ 15x = –6 + 36 = 30
∴ x = 30/15 = 2
Substituting the value of x in equation (3), we have
y = 3 (2) – 9 = 6 – 9 = – 3
Hence, x = 2, y = –3