Întrebare sistem , clasa a ix-a

Se da sistemul:
2x^2 - y^2 - xy =14 / *2
x^2 + 2y^2 - 5xy = - 4 /*7

Am facut asa: 4x^2 - 2y^2 - 2xy=28
7x^2 + 14y^2 - 35xy = - 28
11x^2 + 12y^2 -37xy = 0 / :y^2
11x^2/y^2 + 12y^2/y^2 - 37xy/y^2 = 0
11(x/y)^2 - 37x/y + 12 = 0
Notam cu t = x/y
11t^2 - 37t + 12 = 0
Delta = 841
x1 = 3 ; x2 = 4/11
Caz I: x/y = 3
2(3y)^2 - 3y*y - y^2 =14
18y^2 - 3y^2 - y^2 = 14
14y^2 = 14 ; y = +1; -1
y=1 ; x=3
y=-1; x=-3
Caz II: x/y = 4/11 ; x = 4y/11
2(4y/11)^2 - 4y/11 *y - y^2 = 14
32y^2/121 - 4y^2/11 - y^2 = 14
32y^2 - 44y^2 - 121y^2 = 1694
-133y^2 = 1694
y^2 = -1694/133

Spuneti-mi, va rog , daca am facut bine, si cum rezolv mai departe cazul II, daca nu am gresit cumva.
Va multumesc!

Sistemul arata asa: \(\left\{\begin{matrix} 2x^{2}-y^{2}- xy=\frac{14}{2} \\ x^{2}+2y^{2}-5xy=-\frac{4}{7} \end{matrix}\right.\) ?

Buna ziua! La egal este14, respective -4, am aplificat cu 2 si 7

Este bine ce ati facut. Mai departe (daca se cer si rezolvari complexe):

\(y^{2}=-\frac{1694}{133}\Leftrightarrow y^{2}=-\frac{2\cdot 121}{19}\), adica \(y\pm 11i\sqrt{\frac{2}{19}}\).
Cum \(x=\frac{4}{11}y\Rightarrow x=\pm 4i\sqrt{\frac{2}{19}}\).

Rezolvarile complexe sunt: \(\left\{\begin{matrix} x=4i\sqrt{\frac{2}{19}}\\ y=11i\sqrt{\frac{2}{19}} \end{matrix}\right.\) si \(\left\{\begin{matrix} x=-4i\sqrt{\frac{2}{19}}\\ y=-11i\sqrt{\frac{2}{19}} \end{matrix}\right.\).