Q. 13.8( 262 Votes )

# Solve the followi

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 = 103 + 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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