Решите систему уравнений: y+x=5; 4/x+3/y=3.

\[4 \bullet (5 - x) + 3x = 3x(5 - x)\]

\[20 - 4x + 3x = 15x - 3x^{2}\]

\[20 - x - 15x + 3x^{2} = 0\]

\[3x^{2} - 16x + 20 = 0\]

\[D = ( - 16)^{2} - 4 \bullet 3 \bullet 20 =\]

\[= 256 - 240 = 16\]

\[x_{1} = \frac{16 + \sqrt{16}}{2 \cdot 3} = \frac{16 + 4}{6} = \frac{20}{6} =\]

\[= 3\frac{2}{6} = 3\frac{1}{3}\]

\[x_{2} = \frac{16 - \sqrt{16}}{2 \cdot 3} = \frac{16 - 4}{6} = \frac{12}{6} =\]

\[= 2\]

\[x_{1} = 3\frac{1}{3} \Longrightarrow \text{\ \ \ \ \ \ \ }y_{1} = 5 - 3\frac{1}{3} =\]

\[= 1\frac{2}{3}.\]

\[x_{2} = 2 \Longrightarrow \text{\ \ \ \ \ \ \ \ \ \ }y_{2} = 5 - 2 = 3.\]

\[Ответ:\left( 3\frac{1}{3};1\frac{2}{3} \right),\ (2;3).\]