Вопрос:

Решите систему уравнений: (3a+1)/5+(2b-1)/3=2/5; (a-5)/2+(b-3)/4=1.

Ответ:


\[\left\{ \begin{matrix}
\frac{3a + 1}{5} + \frac{2b - 1}{3} = \frac{2}{5}\ \ \ | \cdot 15 \\
\frac{3a - 2}{2} + \frac{b - 3}{4} = 1\ \ \ \ \ \ | \cdot 4\ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]


\[\left\{ \begin{matrix}
3 \cdot (3a + 1) + (2b - 1) = 3 \cdot 2 \\
2 \cdot (3a - 2) + b - 3 = 4\ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \]


\[\left\{ \begin{matrix}
9a + 3 + 10b - 5 = 6 \\
6a - 4 + b - 3 = 4\ \ \ \ \ \\
\end{matrix}\text{\ \ } \right.\ \text{\ \ \ \ }\]


\[\left\{ \begin{matrix}
9a + 10b = 6 + 2 \\
6a + b = 4 + 7\ \ \ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ }\]


\[\left\{ \begin{matrix}
9a + 10b = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
6a + b = 11\ \ \ | \cdot ( - 10) \\
\end{matrix} \right.\ \]


\[\left\{ \begin{matrix}
9a + 10b = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\
- 60a - 10b = - 110\ \ \ (2) \\
\end{matrix} \right.\ \]


\[- 51a = - 102\]


\[a = 2\]


\[\left\{ \begin{matrix}
a = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\
6a + b = 11 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]


\[\left\{ \begin{matrix}
a = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\
b = 11 - 6a \\
\end{matrix} \right.\ \text{\ \ \ \ \ }\]


\[\left\{ \begin{matrix}
a = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
b = 11 - 6 \cdot 2 \\
\end{matrix} \right.\ \text{\ \ \ \ }\]


\[\left\{ \begin{matrix}
a = 2\ \ \ \ \\
b = - 1 \\
\end{matrix} \right.\ \]


\[Ответ:(2;\ - 1).\]





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