Svolgimento:
Per risolvere questo sistema basta operare una sostituzione. Sostituiamo:Sistemi: {(2/x-1/y=3),(1/x+2/y=4):}
Svolgimento: Per risolvere questo sistema basta operare una sostituzione. Sostituiamo: t=1/x q=1/y Il sistema diventa: {(2t-q=3),(t+2q=4):} {(2t-q=3),(-2t-4q=-8):} {(2t-q=3),(-5q=-5):} {(2t-q=3),(q=1):} {(2t-1=3),(q=1):} {(t=2),(q=1)
1 min. di lettura
Vota
4
/
5
[math]t=1/x[/math]
[math]q=1/y[/math]
Il sistema diventa:[math]\egin{cases} 2t-q=3 \\ t+2q=4 \ \end{cases}[/math]
[math]\egin{cases} 2t-q=3 \\ -2t-4q=-8 \ \end{cases}[/math]
[math]\egin{cases} 2t-q=3 \\ -5q=-5 \ \end{cases}[/math]
[math]\egin{cases} 2t-q=3 \\ q=1 \ \end{cases}[/math]
[math]\egin{cases} 2t-1=3 \\ q=1 \ \end{cases}[/math]
[math]\egin{cases} t=2 \\ q=1 \ \end{cases}[/math]
e quindi[math]\egin{cases} x=1/2 \\ y=1 \ \end{cases}[/math]
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26 Maggio 2014
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