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


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