Giải các hệ phương trình
a) \(\left\{ \begin{array}{l}
- 0,5x + 0,4y = 0,7\\
0,3x - 0,2y = 0,4
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
\frac{3}{5}x - \frac{4}{3}y = \frac{2}{5}\\
- \frac{2}{3}x - \frac{5}{9}y = \frac{4}{3}
\end{array} \right.\)
a) \(\left\{ \begin{array}{l}
- 0,5x + 0,4y = 0,7\\
0,3x - 0,2y = 0,4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 0,5x + 0,4y = 0,7\\
0,6x - 0,4y = 0,8
\end{array} \right.\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x = 15\\
0,3x - 0,2y = 0,4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 15\\
y = 20,5
\end{array} \right.
\end{array}\)
Vậy hệ có 1 nghiệm là (x;y) = (15;20,5)
b) \(\left\{ \begin{array}{l}
\frac{3}{5}x - \frac{4}{3}y = \frac{2}{5}\\
- \frac{2}{3}x - \frac{5}{9}y = \frac{4}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{3}{5}x - \frac{4}{3}y = \frac{2}{5}\\
- \frac{3}{5}x - \frac{1}{2}y = \frac{6}{5}
\end{array} \right.\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
- \frac{{11}}{6}y = \frac{8}{5}\\
\frac{3}{5}x - \frac{4}{3}y = \frac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - \frac{{14}}{{11}}\\
y = - \frac{{48}}{{55}}
\end{array} \right.
\end{array}\)
Vậy hệ có 1 nghiệm là \(\left( {x;y} \right) = \left( { - \frac{{14}}{{11}}; - \frac{{48}}{{55}}} \right)\)
-- Mod Toán 10