Question

find the value of x&y (1+ i)x-2i/3+i +(2-3i)y+i/3-i =i

Answer 1

((1+ i) x - 2 i)/(3 + i) + ((2 - 3i)y + i)/(3-i) = i

(x + xi - 2i)/(3+i)) + (2y- 3yi + i) / (3-i) = i

[(x +xi - 2i) (3-i) + (2y - 3yi + i) (3+i)] / ( 3² - i² ) = i

( 3x + 3xi - 6i - xi - xi² +2i² + 6y - 9yi +3i + 2yi -3yi² +i² ) / ( 9 + 1 ) = i

3x + 3xi - 6i - xi + x - 2 + 6y - 9yi + 3i + 2yi + 3y - 1 = 10 i

4 x + 9 y + 2xi - 7 yi - 3i - 3 = 10 i

4 x + 9 y + ( 2x - 7 y) i = 13 i + 3

The system:

4 x + 9 y = 3

2 x - 7 y = 13  / · ( - 2 )

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  4 x + 9 y = 3

+

- 4 x + 14 y = - 26

        23 y = - 23,   y = - 1  

4 x + 9 · ( - 1 ) = 3

4 x - 9 = 3

4 x = 12,       x = 12 : 4 = 3

Answer:  x = 3 and y = - 1