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Mam dwa rozwiązania:
(2a-b)/(b-a)=3 //×(b-a)
2a-b=3b-3a
5a=4b
b=5/4×a
(3a-4b)/(2+2a)=(3a-4×((5/4)a))/(2+2a)=(3a-5a)/(2+2a)
-2a/(2+2a)= -1-a=0 więc a=1, a b=5/4
lub:
2a-b/b-a=3
2a-1-a=3
a=4
3×4-2b+2×3=0
2b=18
b=9
b-a=6a-3b
4b=5a
b=1¼a
podkladm
(3a-4*1¼a)/(2+2a)=(3a-5a)/(2+2a)=-2a/(2+2a)=-a-1
-a-1=0
a=1
b=1¼