Question

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

Answer 1

$\left\{\begin{matrix} x \sqrt{3} + y = 2 \\ 2x - y \sqrt{3} = 1\end{matrix}\right.$

$y = 2 - x \sqrt{3}$

$2x - y \sqrt{3} = 1$

$2x - (2 - x \sqrt{3} ) \times \sqrt{3} = 1$

$2x - 2 \sqrt{3} + 3x = 1$

$2x + 3x = 1 + 2 \sqrt{3}$

$5x = 1 + 2 \sqrt{3}$

$x = \frac{1 + 2 \sqrt{3} }{5}$

$y = 2 - x \sqrt{3}$

$y = 2 - \frac{ \sqrt{3} (1 + 2 \sqrt{3} )}{5}$

$y = 2 - \frac{ \sqrt{3} + 6}{5}$

$y = \frac{10}{5} - \frac{ \sqrt{3} + 6 }{5}$

$y = \frac{10 - \sqrt{3} - 6 }{5}$

$y = \frac{4 - \sqrt{3} }{5}$