Solve the following pair of linear equations by the elimination method and the substitution method :
(i)
and ![]()
(ii)
and ![]()
(iii)
and ![]()
(iv)
and ![]()
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 (1)
y = ![]()
putting value of y in equation (ii) we get,
2x - 2![]()
= 14x = 28
= x = ![]()
Putting value of x in equation (ii) we get,
4 - 2y = 2
= -2y = -2
= y= 1
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 + ![]()
39x = 27
= x = ![]()
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
= ![]()
= ![]()
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
= ![]()
= ![]()
From equation (ii) we get,
x = 3 + ![]()
Putting value of x in equation (i) we get,
= ![]()
= ![]()
= 5y = - 6 - 9 = - 15
= y = ![]()