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stiind ca x+3y+2z+16=50 determinati 2x+6x+4z+12
va rog frumos ajutatima


Răspuns :

x+3y+2z+16=50
x+3y+2z=50 -16
x+3y+2z=34

2x+6y+4z+12=2
(x+3y+2z)+12=2×34+12=68+12=80