Question

Solve the following pair of linear 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

(iv) \(\frac{x}{2} + \frac{2y}{3} = - 1\) and x - \(\frac{y}{3} = 3\)

Answer 1

(i) x + y = 5 and 2x – 3y = 4 

x + y = 5 ………….. (i) 

2x – 3y = 4 ………… (ii) 

Multiplying eqn. (i) by 3 

3x + 3y = 15 ………….. (iii) 

By adding eqn. (ii) to eqn. (iii) ‘y’ is eliminated.

Substituting the value of ‘x’ in eqn. (i), 

x + y = 5

(ii) 3x + 4y = 10 and 2x – 2y = 2 

3x + 4y = 10 ……….. (i) 

2x – 2y = 2 …………. (ii) 

Multiplying eqn. (ii) by 2, 

4x – 4y = 4 ………….. (iii) 

Adding eqn. (i) to eqn. (ii),

Substituting the value of ‘x’ in eqn. (i), 

3x + 4y = 10 

3 × 2 + 4y = 10 

6 + 4y = 10 

4y = 10 – 6 

4y = 4

∴ y = \(\frac{4}{4}\) = 1

∴ x = 2, y = 1. 

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 

3x – 5y – 4 = 0 ⇒ 3x – 5y = 4 ……… (i) 

9x = 2y + 7 ⇒ 9x – 2y = 7 ………. (ii) 

Multiplying eqn. (i) by 3, 

9x – 15y = 12 …………. (iii) 

Subtracting eqn. (iii) from eqn. (ii)

Substituting the value of ‘y’ in eqn. (i), 

3x – 5y = 4

3x + 4y = -6 ……….. (i) 

3x – y = 9 …………… (ii) 

Subtracting eqn. (ii) from eqn. (i),

Substituting the value of ‘y’ in eqn. (i), 

3x + 4y = -6 

3x + 4 (-3) = -6 

3x – 12 = -6 

3x = -6 + 12 

3x = 6 

∴ x = \(\frac{6}{3}\) = 2

∴ x = 2, y = 3