Question

Rezolvati sistemul de equatii :
Rad6 x + rad2 y = 2rad2 (3 + rad3)
Rad2 x + rad3 y = 6 + 2rad6

Answer 1

$\left\{\begin{matrix} \sqrt{6} x + \sqrt{2}y = 2 \sqrt{2}(3 + \sqrt{3} ) \\ \sqrt{2}x + \sqrt{3} y = 6 + 2 \sqrt{6} \end{matrix}\right.$

$\left\{\begin{matrix} \sqrt{6} x + \sqrt{2}y = 6 \sqrt{2} + 2 \sqrt{6} \\ \sqrt{2}x + \sqrt{3} y = 6 + 2 \sqrt{6} \: | \times ( - \sqrt{3} ) \end{matrix}\right.$

$\left\{\begin{matrix} \sqrt{6} x + \sqrt{2} y = 6 \sqrt{2} + 2 \sqrt{6} \\ - \sqrt{6} x - 3y = - 6 \sqrt{3} - 2 \sqrt{18} \end{matrix}\right.$

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

$\sqrt{2} y - 3y = 2 \sqrt{6} - 6 \sqrt{3}$

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

$y = \frac{( \sqrt{2} + 3)(2 \sqrt{6} - 6 \sqrt{3} )}{2 - 9}$

$y = - \frac{2 \sqrt{12} - 6 \sqrt{6} + 6 \sqrt{6} - 18 \sqrt{3} }{7}$

$y = - \frac{4 \sqrt{3} - 18 \sqrt{3} }{7}$

$y = \frac{14 \sqrt{3} }{7}$

$y = 2 \sqrt{3}$

$\sqrt{2} x + \sqrt{3} y = 6 + 2 \sqrt{6}$

$\sqrt{2} x + 2 \sqrt{3} \times \sqrt{3} = 6 + 2 \sqrt{6}$

$\sqrt{2} x + 6 = 6 + 2 \sqrt{6}$

$\sqrt{2} x = 2 \sqrt{6}$

$x = \frac{2 \sqrt{6} }{ \sqrt{2} }$

$x = 2 \sqrt{3}$