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Sa se rezolve urmatorul sistem: [tex] \left \{ {{x+[y]=13,9} \atop {[x]+2y=26,3}} \right. [/tex]

Răspuns :

Pentru orice a numar real are loc (identitatea lui Hermite):
[tex][a]+[a+\frac{1}{2}]=[2a]\\ [/tex]
Aplicand partea intreaga celor doua ecuatii ale sistemului obtinem:
[tex][x]+[y]=13\\ \ [x]+[2y]=26 [/tex]
Scadem relatiile si avem:[tex][2y]-[y]=13[/tex]
Dar
[tex][2y]=[y]+[y+\frac{1}{2}][/tex]
De unde rezulta
[tex][y+\frac{1}{2}]=13 \Rightarrow y+\frac{1}{2}\in[13,14)\Rightarrow y\in[12.5,13.5)\\ \\Cazul ~I\\ ~ [y]=12\Rightarrow x=13.9-12=1.9\\ 1+2y=26.3\Rightarrow y=\frac{25.3}{2}=12.65\\ \\Cazul ~II\\ ~ [y]=13\Rightarrow x=13.9-13=0.9\\ 0+2y=26.3\Rightarrow y=\frac{26.3}{2}=13.15\\ \\ S=\{(1.9;12.65),(0.9;13.15) \}[/tex]