Question

By equating coefficients of variables, solve the following equations.

i. 3x – 4y = 7 ; 5x + 2y = 3 

ii. 5x + ly = 17 ; 3x – 2y = 4 

iii. x – 2y = -10 ; 3x – 3y = -12 

iv. 4x + y = 34 ; x + 4y = 16

Answer 1

i. 3x – 4y = 7 …(i) 

5x + 2y = 3 ….(ii) 

Multiplying equation (ii) by 2, 

10x + 4y = 6 …(iii)

Adding equations (i) and (iii),

∴ x = 1 

Substituting x = 1 in equation (i), 

3x – 4y = 7 

∴ 3(1) – 4y = 7 

∴ 3 – 4y = 7 

∴ 3 – 7 = 4y 

∴ -4 = 4y 

∴ y = -4/4

∴ y = -1 

∴ (1, -1) is the solution of the given equations. 

ii. 5x + 7y = 17 …(i) 

3x – 2y = 4 ….(ii) 

Multiplying equation (i) by 2, 

10x + 14y = 34 …(iii)

Multiplying equation (ii) by 7, 

21x – 14y = 28 …..(iv) 

Adding equations (iii) and (iv),

∴ x = 2 

Substituting x = 2 in equation (ii), 

3x – 2y = 4 

∴ 3(2) – 2y = 4 

∴ 6 – 2y = 4 

∴ 6 – 4 = 2y 

∴ 2 = 2y 

∴ y = 2/2

∴ y = 1 

∴ (2,1) is the solution of the given equations.

iii. x – 2y = -10 ….(i) 

3x – 5y = -12 …….(ii) 

Multiplying equation (i) by 3, 

3x – 6y = -30 …(iii) 

Subtracting equation (ii) from (iii),

∴ y = 18 

Substituting y = 18 in equation (i), 

x – 2y = -10 

∴ x – 2(18) = -10 

∴ x – 36 = -10 

∴ x = -10 + 36 = 26 

∴ (26, 18) is the solution of the given equations.

iv. 4x + y = 34 …(i) 

x + 4y = 16 …… (ii) 

Multiplying equation (i) by 4,

16x + 4y = 136 …(iii) 

Subtracting equation (ii) from (iii),

x = 8 

Substituting x = 8 in equation (i), 

4x + y = 34 

∴ 4(8) + y = 34 

∴ 32 + y = 34 

∴ y = 34 – 32 = 2 

∴ (8, 2) is the solution of the given equations.