Question

If 3x+2iy/5i-2=15/8x+3iy then x=-1,y=-3 or x=1 , y=3

Answer 1

Given :  (3x + 2yi)/(5i - 2)  = 15/(8x + 3yi)  

To find : show that  x=-1,y=-3 or x=1 , y=3

Solution:

(3x + 2yi)/(5i - 2)  = 15/(8x + 3yi)

=> (3x + 2yi)(8x + 3yi)  = 15(5i - 2)

=> 24x²  + i(9xy + 16xy)  - 6y²  = 75i  - 30

=> (24x² - 6y² ) + i (25xy)  = -30  + 75i

Equating Real & imaginary part

 25xy  = 75

=> xy  = 3

24x² - 6y² = - 30

=> 24x² - 6(3/x)² = - 30

=> 24x² -  54/x²  = - 30

=> 24x²  + 30  -  54/x² = 0

=> 24x²  -24 + 54  -  54/x² = 0

=> 24x(x  - 1/x)  + (54/x)(x  - 1/x)  = 0

=> (24x + 54/x)(x - 1/x)  =   0

24x + 54/x = 0  => x² = -54/24   ( -ve square value not possible )

x - 1/x = 0

=> x² = 1

=> x =  ±1

xy  = 3  

=> x = 1 , y = 3    

x  = - 1 , y = - 3

x=-1,y=-3 or x=1 , y=3

QED

Hence proved

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Answer 2

Step-by-step explanation:

Given : (3x + 2yi)/(5i - 2) = 15/(8x + 3yi)

To find : show that x=-1,y=-3 or x=1 , y=3

Solution:

(3x + 2yi)/(5i - 2) = 15/(8x + 3yi)

=> (3x + 2yi)(8x + 3yi) = 15(5i - 2)

=> 24x² + i(9xy + 16xy) - 6y² = 75i - 30

=> (24x² - 6y² ) + i (25xy) = -30 + 75i

Equating Real & imaginary part

25xy = 75

=> xy = 3

24x² - 6y² = - 30

=> 24x² - 6(3/x)² = - 30

=> 24x² - 54/x² = - 30

=> 24x² + 30 - 54/x² = 0

=> 24x² -24 + 54 - 54/x² = 0

=> 24x(x - 1/x) + (54/x)(x - 1/x) = 0

=> (24x + 54/x)(x - 1/x) = 0

24x + 54/x = 0 => x² = -54/24 ( -ve square value not possible )

x - 1/x = 0

=> x² = 1

=> x = ±1

xy = 3

=> x = 1 , y = 3

x = - 1 , y = - 3

x=-1,y=-3 or x=1 , y=3

QED

Hence proved