Задача

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[tex]\begin{array}{|l} \frac {2y} {x+3} - \frac {3}{y} = -1 \\ \frac {3y}{x+3} - \frac {1}{y} = \frac 12 \end{array}[/tex][tex]DM: x \ne -3; y \ne 0[/tex]
Полагаме [tex]\frac {y}{x+3} = u ; \frac {1}{y} = v[/tex]
[tex]\begin{array}{|l} 2u -3v = -1 \\ 3u - v = \frac 12 \end{array} \Rightarrow -7u = \frac {-5}{2} \Rightarrow u = \frac {5}{14} \Rightarrow v = 3u - \frac 12 = \frac {15}{14} - \frac {7}{14} = \frac {4}{7} = - \frac 32[/tex]
[tex]\begin{array}{|l} \frac {y}{x+3} = \frac {5}{14} \\ \frac {1}{y} = \frac {4}{7} \Rightarrow y = \frac {7}{4} \end{array}[/tex]
[tex]\frac {\frac {7}{4}}{x+3} = \frac {5}{14} \Rightarrow 5x + 15 = \frac {49}{2} \Rightarrow 5x = \frac {19}{2} \Rightarrow x = \frac {19}{10}[/tex]
[tex]( \frac {19}{10};\frac {7}{4})[/tex]

[Гост]

Благодаря!:)