Đáp án:
Áp dụng tính chất dãy tỉ số bằng nhau ta có
$\begin{array}{l}
a)3x = 2y\\
\Rightarrow \dfrac{x}{2} = \dfrac{y}{3} = \dfrac{{x + y}}{{2 + 3}} = \dfrac{{10}}{5} = 2\\
\Rightarrow \left\{ \begin{array}{l}
x = 2.2 = 4\\
y = 3.2 = 6
\end{array} \right.\\
Vậy\,x = 4;y = 6\\
b)\dfrac{{x - 2}}{{y + 3}} = \dfrac{4}{6} = \dfrac{2}{3}\\
\Rightarrow \dfrac{{x - 2}}{2} = \dfrac{{y + 3}}{3} = \dfrac{{y + 3 - \left( {x - 2} \right)}}{{3 - 2}}\\
= \dfrac{{y - x + 3 + 2}}{1} = \dfrac{{ - 4 + 5}}{1} = 1\\
\Rightarrow \left\{ \begin{array}{l}
x - 2 = 2\\
y + 3 = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 4\\
y = 0
\end{array} \right.\\
Vậy\,x = 4;y = 0\\
c)\dfrac{x}{4} = \dfrac{y}{{ - 10}} = \dfrac{{2y}}{{ - 20}}\\
= \dfrac{{x + 2y}}{{4 + \left( { - 20} \right)}} = \dfrac{{12}}{{ - 16}} = \dfrac{{ - 3}}{4}\\
\Rightarrow \left\{ \begin{array}{l}
x = - \dfrac{3}{4}.4 = - 3\\
y = \dfrac{{ - 3}}{4}.\left( { - 10} \right) = \dfrac{{15}}{2}
\end{array} \right.\\
Vậy\,x = - 3;y = \dfrac{{15}}{2}
\end{array}$
