Question

помагити с решением только надо " Б" решат пажалоста прошу ​

Answer 1

Ответ:

$\left \{ {{y+x = 5} \atop {\frac{4}{x} + \frac{3}{y} = 3 }} \right. \\\left \{ {{y= 5-x} \atop {\frac{4}{x} + \frac{3}{y} = 3 }} \right. \\\frac{4}{x} + \frac{3}{5-x} = 3 \\\frac{4(5-x)}{x(5-x)} + \frac{3x}{x(5-x)} = \frac{3x(5-x)}{x(5-x)} \\4(5-x) + 3x = 3x(5-x)\\20-4x+3x = 15x -3x^{2} \\3x^{2} -16x+20 = 0\\D = 256 - 240 = 16\\x_{1} = (16+4):6 = \frac{20}{6} = \frac{10}{3} = 3\frac{1}{3}\\y_{1} = 5-3\frac{1}{3} = 1 \frac{2}{3}\\ x_{2} = (16-4):6 = 12:6 = 2 \\y_{2} = 5-2 = 3$