Find the real values of x and y for which:

Solution:

Given: $(1+i) y^{2}+(6+i)=(2+i) x$

Consider, $(1+i) y^{2}+(6+i)=(2+i) x$

$\Rightarrow y^{2}+i y^{2}+6+i=2 x+i x$

$\Rightarrow\left(y^{2}+6\right)+i\left(y^{2}+1\right)=2 x+i x$

Comparing the real parts, we get

$y^{2}+6=2 x$

$\Rightarrow 2 x-y^{2}-6=0 \ldots(i)$

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

$2 x-y^{2}-6-\left(x-y^{2}-1\right)=0$

$\Rightarrow 2 x-y^{2}-6-x+y^{2}+1=0$

$\Rightarrow x-5=0$

$\Rightarrow x=5$

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

$2(5)-y^{2}-6=0$

$\Rightarrow 10-y^{2}-6=0$

$\Rightarrow-y^{2}+4=0$

$\Rightarrow-y^{2}=-4$

$\Rightarrow y^{2}=4$

$\Rightarrow y=\sqrt{4}$

$\Rightarrow y=\pm 2$

Hence, the value of x = 5 and y = ± 2