Заранее спасибо! решите систему {x/2=y/3 {x+y=4

х/2-у/3=0 х+у=4 3х-2у=0   I х+у=4     I 2 3х-2у=0 2х+2у=8 сложим 5х=8 х=1,6 у=4-х=2,4

[latex] \left \{ {{ \frac{x}{2}= \frac{y}{3} } \atop {x+y=4}} \right. \\ \\ \left \{ {{3x=2y} \atop {x+y=4}} \right. \\ \\ \left \{ {{x=4-y} \atop {3(4-y)=2y}} \right. \\ \\ \left \{ {{x=4-y} \atop {12-3y=2y}} \right. \\ \\ \left \{ {{x=4-y} \atop {5y=12}} \right. \\ \\ \left \{ {{x=4-y} \atop {y= \frac{12}{5} }} \right. \\ \\ \left \{ {{y= \frac{12}{5} } \atop {x= \frac{8}{5} }} \right. [/latex] Ответ: [latex]( \frac{8}{5} ; \frac{12}{5} )[/latex]