Да се решат системите

[Гост]

1) [tex]\begin{array}{|l} x^{2 } + xy = 15 \\ y^{2 } + xy = 10 \end{array}[/tex]

2) [tex]\begin{array}{|l} x^{2 } + xy = 6 \\ y^{2 } + xy = 3 \end{array}[/tex]

[mail_dinko]

[tex]\begin{array}{|l} x^2 + xy = 15 \rightarrow x(x+y) = 15 \\ y^2 + xy = 10 \rightarrow y(x+y) = 10 \end{array}[/tex]
[tex]\frac {x \cancel {(x+y)}}{y \cancel {(x+y)}} = \frac {15}{10} \Leftrightarrow \frac {x}{y} = \frac {3}{2} \Rightarrow y = \frac {2}{3} x[/tex]
[tex]\frac {4}{9} x^2 + x . \frac {2}{3}. \frac {3}{3} x = 10[/tex]
[tex]\frac {10}{9} x^2 = 10 \Leftrightarrow x^2 = 9 \Rightarrow x _ {1,2} = \pm 3 \Rightarrow y _ {1,2} = \pm 3 .\frac {2}{3} = \pm 2[/tex]
[tex](\pm 3 ; \pm 2)[/tex]

[tex]\begin{array}{|l} x^2 + xy = 6 \rightarrow x(x+y) = 6 \\ y^2 + xy = 3 \rightarrow y(x+y) = 3 \end{array}[/tex]
[tex]\frac {x \cancel {(x+y)}}{y \cancel {(x+y)}} = \frac {6}{3 } \Leftrightarrow \frac {x}{y} = 2 \Rightarrow x = 2y[/tex]
[tex]4y^2 + 2y^2 = 6 \Leftrightarrow 6y^2 = 6 \Rightarrow y = \pm 1[/tex]
[tex]x = 2y = \pm 2[/tex]
[tex](\pm 2; \pm 1)[/tex]

[pal702004]

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