Question

Solve each of the following system of equations by elimination method.

Answer 1

solution:-

given by:-

》 2 /x + 2/3y  = 1/6 , 3/x + 2/y = 0

》  2/x + 2/3y  = 1/6  ---- (1)

》  3/x + 2/y = 0  ---- (2)

 Let 1/x = a and  1/y = b

》    2 a + (2b/3) = 1/6

》    (6 a + 2 b)/3 = 1/6

》     6 (6 a + 2 b )  = 3

》    36 a + 12 b = 3

Dividing the whole equation by 3 we get

》     12 a + 4 b = 1   ------- (1)

》       3 a + 2 b = 0 ---------(2)

》     12 a + 4 b = 1

In the first equation we have + 4b and in the second equation also we have + 2b and the symbols are  same so we have to subtract them for eliminating the variable b.

Multiply the second equation by
》2 => 6 a + 4 b = 0
Subtracting the second equation from the first equation

》               12 a + 4 b = 1

》                6 a + 4 b = 0

 》               (-)     (-)   (-)

               --------------

》                       6 a  = 1

 》                       a = 1/6

now we have to apply the value of a in either given equations to get the value of another variable b

Substitute a = 1/6 in the first equation we get

》 12(1/6) + 4 b =1

》 2 + 4 b = 1

》  4 b =1-2
》 4 b = -1

》 b = -1/4

》(x = 6 ,y = -4) ans

☆i hope its help☆

Answer 2

HELLO DEAR,


let 1/x = p , 1/y = q

2p + 2q/3 = 1/6 ,

6p + 2q = 1/2

12p + 4q = 1------------( 1 )

3p + 2q = 0----------( 2 )

from------( 1 ) & ------( 2 )
[multiply by (2) in ----(2)]

12p + 4q = 1
6p + 4q = 0
———————
6p = 1

p = 1/6 [ put in ---( 1 ) ]

we get,

12/6 + 4q = 1

4q = 1 - 2

q = -1/4

q = -1/4 = 1/y

y = -4

AND,

p = 1/x = 1/6

x = 6 , y = -4


I HOPE ITS HELP YOU DEAR,. THANKS