👤

rezolva prin metoda substitutiei xradical din 3+y=2 si 2x-yradical din 3=1

Răspuns :


[tex]\left\{\begin{matrix}
x \sqrt{3} + y = 2 \\

2x - y \sqrt{3} = 1\end{matrix}\right.[/tex]

[tex]y = 2 - x \sqrt{3} [/tex]

[tex]2x - y \sqrt{3} = 1[/tex]

[tex]2x - (2 - x \sqrt{3} ) \times \sqrt{3} = 1[/tex]

[tex]2x - 2 \sqrt{3} + 3x = 1[/tex]

[tex]2x + 3x = 1 + 2 \sqrt{3} [/tex]

[tex]5x = 1 + 2 \sqrt{3} [/tex]

[tex]x = \frac{1 + 2 \sqrt{3} }{5} [/tex]

[tex]y = 2 - x \sqrt{3} [/tex]

[tex]y = 2 - \frac{ \sqrt{3} (1 + 2 \sqrt{3} )}{5} [/tex]

[tex]y = 2 - \frac{ \sqrt{3} + 6}{5} [/tex]

[tex]y = \frac{10}{5} - \frac{ \sqrt{3} + 6 }{5} [/tex]

[tex]y = \frac{10 - \sqrt{3} - 6 }{5} [/tex]

[tex]y = \frac{4 - \sqrt{3} }{5} [/tex]