Решите систему уравнений: 1/x+2/(y+1)=3; 3/x+4/(y+1)=7.

\[\left\{ \begin{matrix} \frac{1}{x} + \frac{2}{y + 1} = 3 \\ \frac{3}{x} + \frac{4}{y + 1} = 7\ \\ \end{matrix}\text{\ \ \ } \right.\ \]

\[\left\{ \begin{matrix} a + b = 3\ \ \ \ \ \\ 3a + 2b = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} a = 3 - b\ \ \ \ \ \\ 3a + 2b = 7 \\ \end{matrix} \right.\ \]

\[3 \bullet (3 - b) + 2b = 7\]

\[9 - 3b + 2b = 7\]

\[- b = - 2\]

\[b = 2.\]

\[b = 2 \Longrightarrow a = 3 - 2 = 1.\]

\[\frac{1}{x} = 1 \Longrightarrow x = 1.\]

\[\frac{2}{y + 1} = 2\]

\[2 \bullet (y + 1) = 2\]

\[y + 1 = 1\]

\[y = 0\]

\[Ответ:(1;0).\]