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Szekelymariana
a fost răspuns

stiind ca3x+6y+19=43determinati 6y+38+3x stiind ca 2x+2y+38=62 determinati 6y+38+3x stiind ca 5x+18+20y=43 determinati 15x+38+60y stiind ca 3x+2y+4z=30 determinati 4z+43+2y+27+3x stiind ca x+3+y+4+z+5=12 determinati 2012+x+y+z

Răspuns :

Infrumusetare
6y+38+3xdivide 3x+6y+19=12y +57+6x=12y:2=6x=6y=6y+6y+6y+57=18y+57=43=18Y=57-43=14=18Y=14=Y=18:14=NU CONVINE

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