Solve the following sets of simultaneous equations.

i. x + y = 4; 2x - 5y = 1

ii. 2x + y =5; 3x – y =5

iii. 3x – 5y =16; x - 3y = 8

iv. 2y – x =0; 10x + 15y =105

v. 2x +3y + 4 = 0; x – 5y = 11

vi. 2x – 7y =7; 3x + y =22

(i)

x + y = 4 eq.[1]

2x - 5y = 1 eq.[2]

We can write eq.[1] as,

x = 4 - y eq.[3]

Substituting eq.[3] in eq.[2],

⇒ 2(4 - y) - 5y = 1

⇒ 8 - 2y - 5y = 1

⇒ -7y = -7

⇒ y = 1

Substituting 'y' in eq.[3]

⇒ x = 4 - 1

⇒ x = 3

Hence, solution is x = 3 and y = 1.

(ii)

2x + y = 5 eq.[1]

3x - y = 5 eq.[2]

We can write eq.[1] as,

y = 5 - 2x eq.[3]

Substituting eq.[3] in eq.[2],

⇒ 3x - (5 - 2x) = 5

⇒ 3x - 5 + 2x = 5

⇒ 5x = 10

⇒ x = 2

Substituting 'x' in eq.[3]

⇒ y = 5 - 2(2)

⇒ y = 1

Hence, solution is x = 2 and y = 1.

(iii)

3x - 5y = 16 eq.[1]

x - 3y = 8 eq.[2]

We can write eq.[2] as,

x = 8 + 3y eq.[3]

Substituting eq.[3] in eq.[1],

⇒ 3(8 + 3y) - 5y = 16

⇒ 24 + 9y - 5y = 16

⇒ 4y = -8

⇒ y = -2

Substituting 'y' in eq.[3]

⇒ x = 8 + 3(-2)

⇒ x = 8 - 6 = 2

Hence, solution is x = 2 and y = -2

(iv)

2y - x = 0 eq.[1]

10x + 15y = 105 eq.[2]

We can write eq.[1] as,

x = 2y eq.[3]

Substituting eq.[3] in eq.[2],

⇒ 10(2y) + 15y = 105

⇒ 20y + 15y = 105

⇒ 35y = 105

⇒ y = 3

Substituting 'y' in eq.[3]

⇒ x = 2(3)

⇒ x = 6

Hence, solution is x = 6 and y = 3.

(v)

2x + 3y + 4 = 0 eq.[1]

x - 5y = 11 eq.[2]

We can write eq.[2] as,

x = 11 + 5y eq.[3]

Substituting eq.[3] in eq.[1],

⇒ 2(11 + 5y) + 3y + 4 = 0

⇒ 22 + 10y + 3y + 4 = 0

⇒ 13y + 26 = 0

⇒ 13y = -26

⇒ y = -2

Substituting 'y' in eq.[3]

⇒ x = 11 + 5(-2)

⇒ x = 11 - 10 = 1

Hence, solution is x = 1 and y = -2.

(vi)

2x - 7y = 7 eq.[1]

3x + y = 22 eq.[2]

We can write eq.[2] as,

y = 22 - 3x eq.[3]

Substituting eq.[3] in eq.[1],

⇒ 2x - 7(22- 3x) = 7

⇒ 2x - 154 + 21x = 7

⇒ 23x = 161

⇒ x = 7

Substituting 'x' in eq.[3]

⇒ y = 22 - 3(7)

⇒ y = 22 - 21 = 1

Hence, solution is x = 7 and y = 1.