Line parallel to x axis and passing through point of intersection of lines ax^2 + 2by + 3b =0 and bx-2ay-3a=0 is?

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Line parallel to x axis and passing through point of intersection of lines ax + 2by + 3b =0 and bx-2ay-3a=0 is?
$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{ax}+2\mathrm{by}+3\mathrm{b}=0 \\ \Rightarrow \mathrm{x}=\frac{-3\mathrm{b}-2\mathrm{by}}{\mathrm{a}} ...\left(1\right) \\ \mathrm{and} \mathrm{bx}-2\mathrm{ay}-3\mathrm{a}=0 \\ \Rightarrow \mathrm{b}\left(\frac{-3\mathrm{b}-2\mathrm{by}}{\mathrm{a}}\right)-2\mathrm{ay}-3\mathrm{a}=0 \\ \Rightarrow \frac{-3{\mathrm{b}}^2}{\mathrm{a}}-\frac{2{\mathrm{b}}^2\mathrm{y}}{\mathrm{a}}-2\mathrm{ay}-3\mathrm{a}=0 \\ \Rightarrow \frac{3{\mathrm{b}}^2+2{\mathrm{b}}^2\mathrm{y}+2{\mathrm{a}}^2\mathrm{y}+3{\mathrm{a}}^2}{2}=0 \\ \Rightarrow 3\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)+2\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)\mathrm{y}=0 \\ \Rightarrow \mathrm{y}=\frac{-3}{2} \\ \mathrm{and} \mathrm{x}=\frac{-3\mathrm{b}-2\mathrm{b}\left({\displaystyle \frac{-3}{2}}\right)}{\mathrm{a}} \\ \Rightarrow \mathrm{x}=\frac{-3\mathrm{b}+3\mathrm{b}}{\mathrm{a}}=0 \\ \mathrm{So}, \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{is} \left(0,\frac{-3}{2}\right) \end{gathered}$$
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