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Feliciamate11
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xy+x-3y+3=0 determinati perechi (x,y) de numere intregi nenule

Răspuns :

Iria2001cristiniub
xy+x−3y+3=0
xy+x−3y+3+−3=0+−3
xy+x−3y=−3
xy+x−3y+3y=−3+3y
xy+x=3y−3
x(y+1)=3y−3
[tex] \frac{x(y+1)}{y+1}= \frac{3y-3}{y+1} [/tex]
[tex]x= \frac{3y-3}{y+1} [/tex]

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