+4 $${\mathtt{6}}{\mathtt{\,\times\,}}\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right) = {\mathtt{3}} = {\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{a}}\right) = \tiny\text{Error: }$$6(a+2)=3=3(10-3a)
