Odpowiedź:
[tex]\huge\boxed{x=\dfrac{20}{19};\ y=-\dfrac{13}{38}}[/tex]
Szczegółowe wyjaśnienie:
[tex]\left\{\begin{array}{ccc}6y+x+1=0&|-6y-1\\2y-6x+7=0\end{array}\right\\\\\left\{\begin{array}{ccc}x=-6y-1&(1)\\2y-6x+7=0&(2)\end{array}\right[/tex]
Podstawiamy (1) do (2):
[tex]2y-6(-6y-1)+7=0\\2y+36y+6+7=0\\38y+13=0\qquad|-13\\38y=-13\qquad|:38\\\boxed{y=-\dfrac{13}{38}}[/tex]
Podstawiamy wartość y do (1):
[tex]x=-6\!\!\!\!\diagup^3\cdot\left(-\dfrac{13}{38\!\!\!\!\!\diagup_{19}}\right)-1\\x=\dfrac{39}{19}-1\\\\x=\dfrac{39}{19}-\dfrac{19}{19}\\\\\boxed{x=\dfrac{20}{19}}[/tex]
