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.