solve the following system of linear equations;

Question

2(ax-by)+(a+4b)=0,2(bx+ay)+(b-4a)=0

Answer 1

2(ax - by) + (a + 4b) =0

2ax - 2by + a + 4b =0          (1)

2(bx + ay) + (b - 4a) = 0

2bx + 2ay + b - 4a = 0          (2)

Multiply the first equation  by a and second equation by b

2a^2x - 2aby + a^2 + 4ab = 0     (3)

2b^2x + 2aby + b^2 - 4ab = 0     (4)

2(a^2 + b^2)x + (a^2 + b^2) = 0

Divide each side by (a^2 + b^2)

2x + 1 = 0

Subtract 1 from each side

2x + 1 - 1 = -1

2x = -1

x = -1 / 2

Substitute x = - 1 / 2 in the second equation

2b(-1 / 2) + 2ay + b - 4a = 0

-b + 2ay + b - 4a = 0

2ay = 4a

y = 2.

Answer 2

  System  of  linear  equations   are  2 ( a x - b y ) + ( a + 4b )  =  0    -> ( 1 )

                                                          2 ( bx + ay )  + ( b - 4a )   =  0     -> ( 2 )

   The  first  equation  Multiply  by  ' b'

 [ ( 2 ax - 2by )  + ( a + 4b ) ] b  =  0

 2abx - 2b^2y + ab + 4b^2  =  0    -> ( 3 )

 The  second  equation  Multiply  by  ' a '

 [ ( 2bx + 2ay ) + ( b - 4a ) ] a  = 0

 2abx + 2a^2y + ab - 4a^2   =  0    -> ( 4 )

Substract  equations  in  ( 3 ) & ( 4 )

 2abx - 2b^2y + ab + 4b^2   = 0

 2abx + 2a^2y + ab - 4a^2  = 0

___________________________

       -2b^2y - 2a^2y +4b^2 + 4a^2  = 0

     -2 y ( a^2 + b^2 ) +4 ( a^2 + b^2 )  = 0

    -2y + 4  = 0

   - 2y = -4  Therefore  y = 4/2 = 2

Substitute  the  'y'  value  in  equation   ( 3 )

2abx - 2b^2y + ab +4b^2  = 0

2abx -2 b^2 ( 2 ) + ab + 4b^2  = 0

2abx - 4b^2 + ab + 4b^2  = 0

2abx + ab  = 0

2abx = - ab

2x  =  -1

Therefore  x =  -1/ 2

The values  of x = - 1/2 , y = 2 .