Решить систему уравнений [latex] \left \{ {{(x - 4)(y+3)=0} \atop {4y - 3x=12}} \right. [/latex] В бланк ответов внести наибольшее возможное знчаение дроби [latex] \frac{ x_{0} }{ y_{0} } [/latex] где пара чисел ([latex] x_{0} ...

[latex] \left \{ {{(x-4)(y + 3) = 0} \atop {4y - 3x = 12}} \right. [/latex] Разойдёмся на две системы: [latex] \left \{ {{x - 4 = 0} \atop {4y - 3x = 12}} \right. [/latex] [latex] \left \{ {{y + 3 = 0} \atop {4y - 3x = 12}} \right. [/latex] [latex] \left \{ {{x =4 } \atop {4y - 3x = 12}} \right. [/latex] [latex]\left \{ {{y = - 3} \atop {4y - 3x = 12}} \right. [/latex] [latex]\left \{ {{x =4 } \atop {4y - 12 = 12}} \right.[/latex] [latex]\left \{ {{y = - 3} \atop {-12 - 3x = 12}} \right. [/latex] [latex]\left \{ {{x =4 } \atop {y = 6}} \right.[/latex] [latex]\left \{ {{y = - 3} \atop {x = -8}} \right. [/latex] Наибольшее будет [latex] \frac{-8}{-3} = \frac{8}{3} [/latex]. Ответ: [latex] \frac{8}{3}.[/latex]