[tex] log_{3}(x) - log_{3}(y) = 1 \\ x + y= 12 \\ \\ log_{3}(12 - y) - log_{3}(y) = 1 \\ x = 12 - y \\ \\ log_{3}(12 - y) - log_{3}(y) = 1 \\ log_{3}( \frac{12 - y}{y} ) = 1 \\ log_{3}( \frac{12 - y}{y} ) = log_{3}(3) \\ \frac{12 - y}{y} = 3 \\ 3y = 12 - y \\ 3y + y = 12 \\ 4y = 12 \\ y = 12 \div 4 \\ y = 3 \\ \\ x = 12 - 3 = 9[/tex]

Ответ: ( 9 ; 3 )