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3(x-y)-4(y+1)=x
2x-2-3y-6=0
3x-3y-4y-4=x
2x-3y=8
2x-7y=4
odejmuję stronami
2x-3y-(2x-7y)=8-4
2x-3y-2x+7y=4
4y=4
y=1
2x-3=8
2x=11
x=5,5
y=1
x=5,5
II. 2⅓x - 0,6=4
(1,2-x)/5 = (3,4y+1)/6
na pewno w 1. równaniu nie ma nigdzie y?