What are x and y if (8x + yi) / (1 + 4i)  = (2x + i)/ (3 + 2i)  

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We have to find x and y given that (8x + yi) / (1 + 4i) = (2x + i)/ (3 + 2i)

(8x + yi) / (1 + 4i) = (2x + i)/ (3 + 2i)

=> (8x + yi) (3 + 2i) = (2x +i) (1+ 4i)

=> 24x + 16xi + 3yi + 2yi^2 = 2x + i + 8xi + 4i^2

=> 24x + 16xi + 3yi – 2y = 2x + i + 8xi – 4

=> 24x – 2y + i (16x + 3y) = 2x – 4 + i (8x +1)

Equate the real and complex coefficients

We get 24x – 2y = 2x – 4

22x – 2y = -4

=> 11x – y = -2

=> y = 11x + 2

and 16x + 3y = 8x + 1

=> 8x + 3y – 1 = 0

substitute y = 11x + 2

=> 8x + 33x + 6 – 1 = 0

=> 41x = -5

=> x = -5/41

y = 11*(-5/41) + 2

=> (-55 + 82)/41

=> 27 / 41

Therefore x = -5/41 and y = 27/41

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