Đáp án:

a) \(\left\{ \begin{array}{l}
x = 4\\
y =  - 1
\end{array} \right.\) 

Giải thích các bước giải:

\(\begin{array}{l}
a)\left\{ \begin{array}{l}
\dfrac{7}{{20}}x + 4y =  - \dfrac{{13}}{5}\\
\dfrac{3}{4}x - 6y = 9
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x - 9} \right):6\\
\dfrac{7}{{20}}x + 4.\dfrac{{\dfrac{3}{4}x - 9}}{6} =  - \dfrac{{13}}{5}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x - 9} \right):6\\
\dfrac{7}{{20}}x + \dfrac{{3x - 36}}{6} =  - \dfrac{{13}}{5}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x - 9} \right):6\\
\dfrac{7}{{20}}x + \dfrac{{x - 12}}{2} =  - \dfrac{{13}}{5}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x - 9} \right):6\\
\dfrac{{7x + 10x - 120 + 13.4}}{{20}} = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x - 9} \right):6\\
17x = 68
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 4\\
y =  - 1
\end{array} \right.\\
b)\left\{ \begin{array}{l}
10x - 9y = 8\\
15x + 21y = \dfrac{1}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{10x - 8}}{9}\\
15x + 21.\dfrac{{10x - 8}}{9} = \dfrac{1}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{10x - 8}}{9}\\
15x + \dfrac{{70x - 56}}{3} = \dfrac{1}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{10x - 8}}{9}\\
\dfrac{{45x + 70x - 56}}{3} = \dfrac{1}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{10x - 8}}{9}\\
230x - 112 = 3
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y =  - \dfrac{1}{3}
\end{array} \right.\\
c)\left\{ \begin{array}{l}
\dfrac{{33}}{{10}}x + \dfrac{{42}}{{10}}y = 1\\
9x + 14y = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
33x + 42y = 10\\
9x + 14y = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x =  - \dfrac{1}{3}\\
y = \dfrac{1}{2}
\end{array} \right.
\end{array}\)