# Solve the following pair of linear equations by the substitution method.

i) x + y = 14...............(i)

x - y = 4....................(ii)

From equation (i)
take x on one side and when we take y to the other side its sign changes and we get,

x = 14 - y .........(iii)

Putting value of x in equation (ii) we get,

(14 - y) - y = 4

14 - 2y = 4

2y = 10

Putting value of y in equation (iii) we get,

x = 14 - 5 = 9

Hence, x = 9 and y = 5

ii) s - t = 3......................(i)

and,

From equation (i) we get,

taking t to the other side, the sign of t changes to positive

s = t + 3...............(iii)

Putting value of x from (iii) to (ii)

⇒

⇒ 2t + 6 + 3t = 36

⇒ 5t = 30

⇒ t =

Putting value of t in equation (iii) , we get,

s = 6 + 3 = 9

Hence, s = 9, t = 6

iii) 3x - y = 3..................(i)

9x - 3y = 9......................(ii)

Comparing with general pair of equations i.e.

a1x + by + c1 = 0

a2x + b2y + c2 = 0, we jhave

a1 = 3, b1 = -1, c1 = -3

a2 = 9, b2 = -2 and c2 = -9

Here,

and In this case, the system of linear equation is consistent and has infinite solutions.

iv) 0.2 x + 0.3 y = 1.3 .....................(i)

0.4 x + 0.5 y = 2.3 ........................(ii)

From equation (i) . we get,

Putting value of x in equation (ii) we get,

(6.5 - 1.5 y) × 0.4 + 0.5 y = 2.3

6.5 x 0.4 - 1.5 y x 0.4 + 0.5 y = 2.3
2.6 - 0.6 y + 0.5 y = 2.3

- 0.1 y = -0.3

y =

Putting value of y in equation (iii) we get,

x = 6.5 - 1.5 × 3 = 6.5 - 4.5 = 2

Hence, x = 2 and y = 3

v)

From equation (i) , we get,

x =

Putting value of x in equation (ii). we get,

⇒

⇒
⇒

so, y = 0

Putting value of y in equation (iii) we get.

x = 0

Hence, x = 0 and y = 0

vi)
................(i)

and

From equation (i) we get,

By taking L.C.M and solving we get,

9 x - 10 y = -12

Putting this value of x in equation (ii), we get,

⇒

⇒

⇒ 47y = 117 + 24

⇒ 47y = 141

⇒ y =

Putting value of y in (iii)

⇒ x =

Hence, x = 2 and y = 3

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