Question

metoda mersului invers: 8-[6400:16-(360-a×2)×5]:20=3 si [(36-46:2×0)+(a×7-3×12)]:7=6 va rog imi trebuie urgent...

Answer 1

8-[6400:16-(360-a×2)×5]:20=3
[400-(360-a×2)×5]:20=8-3
[400-(360-a×2)×5]:20=5
400-(360-a×2)×5=5×20
400-(360-a×2)×5=100
(360-a×2)×5=400-100
(360-a×2)×5=300
360-a×2=300:5
360-a×2=60
a×2=360-60
a×2=300
a=300:2
a=150


[(36-46:2×0)+(a×7-3×12)]:7=6
(36-23×0)+(a×7-36)=6 ×7
(36-0)+(a×7-36)=42
36+(a×7-36)=42
a×7-36=42-36
a×7-36=6
a×7=6+36
a×7=42
a=42:7
a=6