Question

решите систему 3у+х=12; у-4х+9=0

Answer 1

$\left \{ {{3y+x=12} \atop {y-4x+9=0}} \right.\\\\\left \{ {{x+3y=12}|*4 \atop {-4x+y=-9}} \right.\\\\+\left \{ {{4x+12y=48} \atop {-4x+y=-9}} \right.\\-------\\13y=39\\y=3\\x=12-3y\\x=12-3*3\\x=3$

Ответ : (3 ; 3)