Question

решите систему уравнений {х/3+у/2=1 {х+3у=0

Answer 1

{x/3+y/2=1

{x+3y=0

{x/3+y/2=1

{x=-3y

{-3y/3+y/2=1

{x=-3y

{-y+y/2=1

Домножаем(-у на два) в -у+у/2=1

Получаем -2у/2+у/2=1

{-2у/2+у/2=1

{х=-3у

{-у/2=1

{х=-3у

{-у=2×1=2

{х=-3у

Отсюда

у=-2

х=6

Answer 2

$\left \{ {{\frac{x}{3}+\frac{y}{2}=1\; |\cdot 6} \atop {x+3y=0}} \right. \; \left \{ {{2x+3y=6} \atop {x=-3y}} \right. \; \left \{ {{2(-3y)+3y=6} \atop {x=-3y}} \right. \; \left \{ {{-3y=6} \atop {x=-3y}} \right. \; \left \{ {{y=-2} \atop {x=6}} \right. \; \Rightarrow \; (6,-2)$