Q. 13.8( 192 Votes )

Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x - 3y = 4 

(ii) 3x + 4y = 10 and 2x - 2y = 2 

(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7

(iv) x/2 + 2y/3 = - 1and x - y/3 = 3

Answer :

i) By elimination method

x + y = 5.............(i)


2x - 3y = 4..........(ii)


Multiplying equation (i) by 2 we get,


2x + 2y = 10 ..............(iii)


Subtracting equation (ii) from equation (iii) we get,


5y = 6


y =


Putting value of y in equation (i). we get,


x =


By substitution method


x + y = 5 ...........(i)


2x - 3y = 4 ......................(ii)


from equation (i)


x = 5 - y................(iii)


Putting value of x from equation (iii) to equation (ii) we get,


2(5 - y) - 3y = 4


= 10 - 2y - 3y = 4


= -5y = -6


= y =


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


x = .


ii) By elimination method


3x + 4y = 10 ..............(i)


2x - 2y = 2.................(ii)


Multiplying equation (ii) by 2 we get,


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


Adding equations (i) and (iii) we get,


7x = 14


= x =


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


3 (2) + 4y = 10


4y = 10 - 6 = 4


y =


By substitution method


3x + 4y = 10 ..............(i)


2x - 2y = 2.................(ii)


From equation (ii)

2x = 2 + 2y
Dividing both side by 2, we get
x = 1 + y

putting value of x in equation (i) we get,

3(1 + y) + 4y = 10
3 + 3y + 4y = 10
7y = 7
y = 1

and
x = 1 + y = 1 + 1 = 2

(iii) By elimination method


3x - 5y = 4................(i)


9x - 2y = 7..........(ii)


Multiplying equation (i) by 3 we get,


9x - 15y = 12.................(iii)


Subtracting equation (ii) from (iii) we get,


- 13y = 5


= y =


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


3x -


3x +

Taking LCM,
 

By substitution method


3x - 5y = 4................(i)


9x - 2y = 7..........(ii)


From equation (i)


x =


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


=


=


= 13y = -5


y =


Putting value of y in equation (i) we get


3x -


3x = 4 -


= x =


iv) By elimination method


.....(i)


=


Multiplying equation (i) by 2, we get,


x +


Subtracting equation (ii) from equation (iii) we get,


=


= y =


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


x +


= x - 4 = - 2


= x = 2


By substitution method


.....(i)


=


From equation (ii) we get,


x = 3 +      (iii)


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


=


=


= 5y = - 6 - 9 = - 15


= y =

Putting this value in (iii), we get
x = 3 - 1 = 2

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