Find the real values of x and y for which:

(1 + i) y2 + (6 + i) = (2 + i)x


Given: (1 + i) y2 + (6 + i) = (2 + i)x

Consider, (1 + i) y2 + (6 + i) = (2 + i)x


y2 + iy2 + 6 + i = 2x + ix


(y2 + 6) + i(y2 + 1) = 2x + ix


Comparing the real parts, we get


y2 + 6 = 2x


2x – y2 – 6 = 0 …(i)


Comparing the imaginary parts, we get


y2 + 1 = x


x – y2 – 1 = 0 …(ii)


Subtracting the eq. (ii) from (i), we get


2x – y2 – 6 – (x – y2 – 1) = 0


2x – y2 – 6 – x + y2 + 1 = 0


x – 5 = 0


x = 5


Putting the value of x = 5 in eq. (i), we get


2(5) – y2 – 6 = 0


10 – y2 – 6 = 0


-y2 + 4 = 0


- y2 = -4


y2 = 4


y = √4


y = ± 2


Hence, the value of x = 5 and y = ± 2


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