Complex number exercise to find x ,y in the given expression

Exercise:

Find $x,y$ such that $\left( 3x-i \right)\left( 2y+i \right)+11=7i$

Solution: we have $\left( 3x-i \right)\left( 2y+i \right)=-11+7i$

So $\left| 3x-i \right|\left| 2y+i \right|=\left| -11+7i \right|$ $\Leftrightarrow \sqrt{9{{x}^{2}}+{{1}^{2}}}\times \sqrt{4{{y}^{2}}+{{1}^{2}}}=\sqrt{{{11}^{2}}+{{7}^{2}}}$

$\Rightarrow \sqrt{\left( 9{{x}^{2}}+1 \right)\left( 4{{y}^{2}}+1 \right)}=\sqrt{170}$ $\Leftrightarrow \left( 9{{x}^{2}}+1 \right)\left( 4{{y}^{2}}+1 \right)=170$

So $9{{x}^{2}}+4{{y}^{2}}+36{{x}^{2}}{{y}^{2}}=169$

$\Rightarrow 6xy+i3x-i2y-{{i}^{2}}=-11+7i\Rightarrow \left( 6xy+1 \right)+i\left( 3x-2y \right)=-11+7i$

So $\left\{ \begin{array}{c}
  & 6xy+1=-11 \\
 & 3x-2y=7 \\
 & \And  \\
 & 9{{x}^{2}}+4{{y}^{2}}+36{{x}^{2}}{{y}^{2}}=169 \\
\end{array} \right.$ $\Rightarrow \left\{ \begin{array}{c}
  & 6xy=-12 \\
 & 3x-2y=7 \\
 & 9{{x}^{2}}+4{{y}^{2}}+36{{x}^{2}}{{y}^{2}}=169 \\
\end{array} \right.$

So ${{x}^{2}}{{y}^{2}}=\frac{144}{36}$

Hence $9{{x}^{2}}+4{{y}^{2}}+144=169\Leftrightarrow 9{{x}^{2}}+4{{y}^{2}}=25$

but $3x=7+2y\Leftrightarrow x=\frac{7+2y}{3}\Leftrightarrow {{x}^{2}}=\frac{1}{9}\left( 49+4{{y}^{2}}+28y \right)$

hence $49+4{{y}^{2}}+28y+4{{y}^{2}}=25\Leftrightarrow 8{{y}^{2}}+28y+24=0$

thus $y=-2\,\,or\,\,\frac{-3}{2}$ that is $x=1\,\,or\,\,\frac{4}{3}$

if $\left( x,y \right)=\left( 1,-2 \right)\Leftrightarrow \left( 3-i \right)\left( -4+i \right)+11\overset{??}{\mathop{=}}\,7i\,\Leftrightarrow 7i=7i$ accepted

so $\left( x,y \right)=\left( \frac{4}{3},-\frac{3}{2} \right)\Leftrightarrow \left( 4-i \right)\left( -3+i \right)+11\overset{??}{\mathop{=}}\,7i\Leftrightarrow 7i=7i$accepted

hence $\left( x,y \right)\in \left\{ \left( 1,-2 \right),\left( \frac{4}{3},-\frac{3}{2} \right) \right\}$