Question

Решите методом подстановки систему х-2у=3 и 5х+у=4

Answer 1

$\left \{ {{x-2y=3,} \atop {5x+y=4;}} \right. \ \ \left \{ {{x=2y+3,} \atop {5x+y=4;}} \right.\\\\ 5x+y=4\\ 5*(2y+3)+y=4\\ 10y+15+y=4\\ 11y=-11\\ y=-1;\\ x=2y+3=-2+3=1.\\ OTBET:\ 1;\ -1$