Răspuns:
Explicație pas cu pas:
[tex]\left \{ {{x+3(y-1)=-10} \atop {3(x-1)+y=-8}} \right. ~\left \{ {{x+3y-3=-10} \atop {3x-3+y=-8}} \right. ~\left \{ {{x+3y=-7~|*(-3)} \atop {3x+y=-5}} \right. ~\left \{ {{-3x-9y=21} \atop {3x+y=-5}} \right.~[/tex]
Deoarece acum aplicam metoda reducerii (sau adunarii), adunam ecuatiile parte cu parte, obtinem -8y=16, ⇒y=-2. Inlocuim in careva ecuatie a sistemului, fie in x+3y=-7, ⇒x+3·(-2)=-7, ⇒x-6=-7, ⇒x=-7+6=-1.
Raspuns: S={(-1; -2)}.
Metoda substitutiei. vom porni de la sistemul deacum obtinut,
[tex]\left \{ {{x+3y=-7} \atop {3x+y=-5}} \right. ~\left \{ {{x=-7-3y} \atop {3*(-7-3y)+y=-5}} \right. ~\left \{ {{x=-7-3y} \atop {-21-9y+y=-5}} \right. ~\left \{ {{x=-7-3y} \atop {-8y=-5+21}} \right.\\\left \{ {{x=-7-3y} \atop {-8y=16}} \right.~\left \{ {{x=-7-3y} \atop {y=16:(-8)}} \right.~\left \{ {{x=-7-3y} \atop {y=-2}} \right.[/tex]
Inlocuim in prima, x=-7-3·(-2)=-7+6=-1.
Raspuns: S={(-1;-2)}
