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## class ten maths Chapter 3 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

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