Question

If (x^2y-2)+i(x+2xy-5)=0 then

Answer 1

Given:

(x²y-2)+i(x+2xy-5) = 0

To find:

The value of x and y

Calculation:

The given equation can be rearranged as

   =>  (x²y-2)+i(x+2xy-5) = 0+0i

By comparing the values on the L.H,S and R.H.S side of the equation

   =>  x²y-2 = 0

         x²y = 2

         xy = 2/x    ……(1)

   =>  x+2xy-5 = 0

         x+2(2/x)-5 = 0          [from eq(1)]

         x²+4-5x = 0

         x²-4x-x+4 = 0

         x(x-4)-1(x-4) = 0

         x=1, x=4

By substituting the values in equation (1)    

For x=1      

         xy = 2/x

         1(y) = 2/1  => y = 2        

For x=4      

         xy = 2/x

         4(y) = 2/4  => y = 1/8

Final answer:

The possible values of (x,y) are (1,2) and (4,(1/8))  

Answer 2

Answer:

hope this help!

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