Đáp án:
$\begin{array}{l}
\dfrac{{{x_1}}}{{19300}} + \dfrac{{{x_2}}}{{10500}} = 2,{6.10^{ - 5}}\\
\Rightarrow \dfrac{{{x_1}}}{{193.100}} + \dfrac{{{x_2}}}{{105.100}} = \dfrac{{2,6}}{{100000}}\\
\Rightarrow \dfrac{{{x_1}}}{{193}} + \dfrac{{{x_2}}}{{105}} = \dfrac{{2,6}}{{1000}}\\
\Rightarrow \dfrac{{{x_1}}}{{193}} + \dfrac{{{x_2}}}{{105}} = \dfrac{{26}}{{100}}\\
\Rightarrow \dfrac{{{x_1}}}{{193}} + \dfrac{{{x_2}}}{{105}} = \dfrac{{13}}{{50}}\\
\Rightarrow \dfrac{{{x_1}}}{{193}} = \dfrac{{13}}{{50}} - \dfrac{{{x_2}}}{{105}}\\
\Rightarrow {x_1} = \dfrac{{2509}}{{50}} - \dfrac{{193{x_2}}}{{105}}\\
\Rightarrow \dfrac{{2509}}{{50}} - \dfrac{{193{x_2}}}{{105}} + {x_2} = 0,3 = \dfrac{3}{{10}}\\
\Rightarrow \dfrac{{88}}{{105}}{x_2} = \dfrac{{1247}}{{25}}\\
\Rightarrow {x_2} = \dfrac{{26187}}{{440}}\\
\Rightarrow {x_1} = \dfrac{{2509}}{{50}} - \dfrac{{193{x_2}}}{{105}} = \dfrac{{ - 5211}}{{88}}\\
Vậy\,\left( {{x_1};{x_2}} \right) = \left( {\dfrac{{ - 5211}}{{88}};\dfrac{{26187}}{{440}}} \right)
\end{array}$
