Solve the following pair of linear equations by the substitution method.
(i) | (ii) |
(iii) | (iv) |
(v) | (vi) |
i) x + y = 14...............(i)
x - y = 4....................(ii)
From equation (i) we get,
x = 14 - y .........(iii)
Putting value of x in equation (ii) we get,
(14 - y) - y = 4
= 14 - 2y = 4
= 2y = 10
= y =
Putting value of y in equation (iii) we get,
x = 14 -5 = 9
Hence, x = 9 and y = 5
ii) s - t = 3......................(i)
=
From equation (i) we get,
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)
From equation (i) , we get,
y = 3x - 3...............(iii)
Putting value of y from equation (iii) to equation (ii)
9x - 3(3x - 3) = 9
= 9x - 9x + 9 = 9
= 9 = 9
This is always true, and pair of these equations have infinite possible solutions
Therefore one possible solution is = x = 1 and y = 0
iv) 0.2x + 0.3y = 1.3 .....................(i)
0.4 x + 0.5 y = 2.3 ........................(ii)
From equation (i) . we get,
x =
Putting value of x in equation (ii) we get,
(6.5 - 1.5y) × 0.4x + 0.5y = 2.3
2.6 - 0.6y + 0.5y = 2.3
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)
=
From equation (i) we get,
9x - 10y = -12
x =
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