Question

A de la 6 dau 30 de p

Answer 1

Răspuns:

Explicație pas cu pas:

$\left \{ {{x+3(y-1)=-10} \atop {3(x-1)+y=-8}} \right. ~\left \{ {{x+3y-3=-10} \atop {3x-3+y=-8}} \right. ~\left \{ {{x+3y=-7~|*(-3)} \atop {3x+y=-5}} \right. ~\left \{ {{-3x-9y=21} \atop {3x+y=-5}} \right.~$

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,

$\left \{ {{x+3y=-7} \atop {3x+y=-5}} \right. ~\left \{ {{x=-7-3y} \atop {3*(-7-3y)+y=-5}} \right. ~\left \{ {{x=-7-3y} \atop {-21-9y+y=-5}} \right. ~\left \{ {{x=-7-3y} \atop {-8y=-5+21}} \right.\\\left \{ {{x=-7-3y} \atop {-8y=16}} \right.~\left \{ {{x=-7-3y} \atop {y=16:(-8)}} \right.~\left \{ {{x=-7-3y} \atop {y=-2}} \right.$

Inlocuim in prima, x=-7-3·(-2)=-7+6=-1.

Raspuns: S={(-1;-2)}

Answer 2

Răspuns:

Explicație pas cu pas:

a)

• Metoda substitutiei

{  x+3(y-1) = -10

{ 3(x-1)+y= -8

___________

{ x + 3 y - 3 = - 10 =>   {  x + 3 y = - 7    =>   x = - 7 - 3 y

{ 3 x - 3 + y = - 8   =>   {  3 x + y = - 5

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3 x + y = - 5

3 ( - 7 - 3 y ) + y = - 5

- 21 - 9 y + y = - 5

- 8 y = -5 + 21

y = 16 : ( - 8 )                       =>       y = - 2

x = - 7 - 3 × ( - 2 ) = - 7 + 6   =>      x = - 1

Solutii:   ( x;  y)  = ( - 1;  - 2 )

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• Metoda reducerii

{   x + 3 y = - 7     [ × 3

{ 3 x + y = - 5

___________________

{ 3 x + 9 y = - 21

{ 3 x + y = - 5  

____________scad relatiile

/         8 y = - 21 - ( - 5)

8 y = - 21 + 5

y = ( - 16 ) : 8   =>     y = - 2

_____________________

x + 3 y = - 7

x + 3 × ( - 2 ) = - 7

x - 6 = - 7

x = - 7 + 6        =>    x = - 1

Solutii (  x;  y) = (  - 1;  - 2 )