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Asked by sm7319344 | 12 Jan, 2024, 09:02: PM
answered-by-expert Expert Answer

There are 8 questions. But only one question per posting is allowed.

Hence only first two questions are answered.

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qn, # 1

4x + i ( 3x-y) = 3 - i 6

equating real and imaginary parts , we get

4x = 3   and (3x-y) = -6

hence we get ,  x= 3/4

y = 3x+6 = 3 (3/4) + 6 = 33/4

x= 3/4  and y = 33/4

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qn. #2

( x4 +  2 x i ) - ( 3 x2 + i y )   = ( 3 - 5 i ) + [ 1 + (2 y) i ]

( x4 - 3 x2 ) + ( 2x - y ) i  = 4 + ( 2y - 5 ) i

we get

( x4 - 3 x2 ) = 4   ..............................(1)

(2x-y)  = ( 2y -5 ) ..........................(2)

From eqn.(1),

( x4 - 3 x2 ) = x2 ( x2 - 3 ) = 4

from above expression we get x2 = 4  ,  x = +2 or -2

From eqn.(2) ,

(2x-y)  = ( 2y -5 )

3y = 2x + 5

if x =2 , then  3y = 9  or y = 3

if x=-2 , then 3y = 1  or y = 1/3

hence we get

x=2 , y = 3   or    x=-2 , y = 1/3

 

 

 

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