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Volki2020
@Volki2020
April 2021
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Решите систему управлений
{0,2x-0,3(2y+1)=1,5
3(x+1)+3y=2y-2
1
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No03
No03
{0,2x-0,3(2y+1)=1,5
{3(x+1)+3y=2y-2
{0.2х-0,6у-0,3-1,5=0
{3х+3+3у-2у+2=0
{0,2х-0,6у=1,8
{
3х+у=-5
у
=-5-3х
0,2х-0,6(-5-3х)=1,8
0,2х+3+1,8х-1,8=0
2х=-1,2
х=-0,6
у=-5-3(-0,6)=-5+1,8=-3,2
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