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

\[3 \bullet (7 - x) + 4x = 2x(7 - x)\]

\[21 - 3x + 4x = 14x - 2x^{2}\]

\[2x^{2} + x - 14x + 21 = 0\]

\[2x^{2} - 13x + 21 = 0\]

\[D = ( - 13)^{2} - 4 \cdot 2 \cdot 21 =\]

\[= 169 - 168 = 1\]

\[x_{1} = \frac{13 + \sqrt{1}}{2 \cdot 2} = \frac{13 + 1}{4} = \frac{14}{4} =\]

\[= 3\frac{2}{4} = 3,5\]

\[x_{2} = \frac{13 - \sqrt{1}}{2 \cdot 2} = \frac{13 - 1}{4} = \frac{12}{4} =\]

\[= 3\]

\[x_{1} = 3,5 \Longrightarrow \ \ \ \ \ \ \ \ y_{1} = 7 - 3,5 =\]

\[= 3,5.\]

\[x_{2} = 3 \Longrightarrow \ \ \ \ \ \ \ \ \ \ y_{2} = 7 - 3 = 4.\]

\[Ответ:(3,5;3,5),\ (3;4).\]