Помогите решить системи (х+5у=13 (у=3х-1 (3х-у=-9 (2х+3у=8

Ответ:

$\left \{ {{x = -2} \atop {y = 3}} \right$ $\left \{ {{y=2} \atop {x=1}} \right.$

Объяснение:

$\left \{ {{x + 5y = 13} \atop {3x - y = -9}} \right.$

$\left \{ {{x = 13 - 5y} \atop {3(13 - 5y) - y = -9}} \right.$

$\left \{ {{x + 5y = 13} \atop {39 - 15y - y = -9}} \right.$

$\left \{ {{x + 5y = 13} \atop {-16y = -48}} \right$

$\left \{ {{x + 5y = 13} \atop {y = 3}} \right.$

$\left \{ {{x + 5*3 = 13} \atop {y = 3}} \right.$

$\left \{ {{x = -2} \atop {y = 3}} \right$

$\left \{ {{y=3x-1} \atop {2x+3y=8}} \right.$

$\left \{ {{y=3x-1} \atop {2x + 3(3x-1)= 8}} \right.$

$\left \{ {{y=3x-1} \atop {2x + 9x - 3 = 8}} \right.$

$\left \{ {{y=3x-1} \atop {11x = 11}} \right.$

$\left \{ {{y=3x-1} \atop {x=1}} \right.$

$\left \{ {{y=2} \atop {x=1}} \right.$