a)
substitutie
x-y=3
y=x-1
x-x+1=3
y=x-1
1=3 (fals)
=> sistemul nu are solutii
reducere
x-y=3
y=x-1
x-y=3
y-x=-1
x-y+y-x=2
0=2 (fals)
=> sistemul nu are solutii
b)
substitutie
y=x-3
3x+y=-3
3x+x-3=-3
4x-3=-3
4x=0
x=0
y=0-3
y=-3
reducere
y=x-3
3x+y=-3
y-x=-3 | ×3
3x+y=-3
3y-3x=-9
y+3x=-3
4y=-12
y=-3
-3-x=-3
-x=0
x=0
