Question

rezolvați prin metoda reducerii urmatoarele sisteme:

A){4x-2(y+1)=2.
{3x-2y=1

B){4x+y=-16
{2(x-2)-3y=2

C){3(x+4)=4(y+3).
{2(x-2)=y+1

D){4(x+7)=4(y+5)
{3(x+7)=4(y+3) ajutoooor dau ​

Answer 1

Răspuns:

a)4x-2(y+1)=2

3x-2y=1

b)4x+y=-16

2(x-2)-3y=2

c)3(x+4)=4(y+3)

2(x-2)=Y+1

2)rezolvati prin metoda reducerii

sistemele:

a)4(x+2)+3(y+1)=15

5(x+1)+4y=11

b)2(x+1)-3(y-1)=-1

5(x-2)-4(y-3)=1

c)3(x+y)+2(y-x)=-5

4(x-y)-2(2x-y)=4