Question

Помогите пожалуйста: "Решите систему уравнений методом подстановки" а) у=х-1; { х+3у=9 б) 3х-2у=12; { х+2у=-4 в) 2х+у=1; { 2х+у=3 Умоляю!!!

Answer 1

$\left \{ {y=x-1} \atop {x+3y=9}} \right. \left \{ {y=x-1} \atop {x+3(x-1)=9}} \right. \left \{ {y=x-1} \atop {x+3x-3=9}} \right. \left \{ {y=x-1} \atop {4x=12}} \right. \left \{ {y=3-1} \atop {x=3}} \right \left \{ {y=2} \atop {x=3}}$$\left \{ {3x-2y=12} \atop {x+2y=-4}} \right. \left \{ {3*(-4-2y)-2y=12} \atop {x=-4-2y}} \right. \left \{ {-12-6y-2y=12} \atop {x=-4-2y}} \right. \left \{ {-8y=24} \atop {x=-4-2y}} \right. \left \{ {y=-3} \atop {x=-4-2*(-3)}} \right. \\ \left \{ {y=-3} \atop {x=-4-(-6)}} \right. \left \{ {y=-3} \atop {x=-4+6}} \right. \left \{ {y=-3} \atop {x=2}} \right.$
$\left \{ {{2x+y=1} \atop {2x+y=3}} \right. \left \{ {{y=1-2x} \atop {2x+1-2x=3}} \right. \left \{ {{y=1-2x} \atop {0=4}} \right. \left \{ {{y=1-2x} \atop {0 \neq 4}} \right.$ Значит, система не имеет решений.