\(\dfrac{93}{17}.\dfrac{1}{x}+\dfrac{-4}{7}.\dfrac{1}{x}+\dfrac{13}{70}=\dfrac{4}{11}\)
=>\(\dfrac{1}{x}.\left(\dfrac{93}{17}+\dfrac{-4}{7}\right)=\dfrac{137}{770}\)
=>\(\dfrac{1}{x}.\dfrac{583}{119}=\dfrac{137}{770}\)
=>\(\dfrac{1}{x}=\dfrac{2329}{64130}=>x=\dfrac{64130}{2329}\)