Question

$\frac{1}{x} + \frac{1}{y} = \frac{4}{9} \\ \\ x + y = 12$
Решите систему ​

Answer 1

методом подстановки:

х1,у1=9,3

х2,у2=3,9

Answer 2

Ответ:

$x_1 = 9; y_1 = 3\\x_2 = 3; y_2 = 9$

Объяснение:

$\left[\begin{array}{ccc}\frac{1}{x} +\frac{1}{y} =\frac{4}{9} \\x+y=12\end{array}\right\\\left[\begin{array}{ccc}\frac{1}{x} +\frac{1}{y} =\frac{4}{9} \\y=12-x\end{array}\right\\\left[\begin{array}{ccc}\frac{1}{x} +\frac{1}{12-x} =\frac{4}{9} \\x+y=12\end{array}\right\\\left[\begin{array}{ccc}\frac{12-x + x}{x(12-x)} =\frac{4}{9}\\x+y=12\end{array}\right\\\left[\begin{array}{ccc}\frac{12}{x(12-x)} =\frac{4}{9}\\x+y=12\end{array}\right\\\left[\begin{array}{ccc}9*12=4x(12-x)\\x+y=12\end{array}\right$

$12x - x^2 = 9 * 3\\x^2 - 12x + 27 = 0\\x_1 = 9; y_1 = 3\\x_2 = 3; y_2 = 9$