Odpowiedź:
[tex]\huge\boxed{\left\{\begin{array}{ccc}x=4\\y=1\end{array}\right}\\\boxed{\left\{\begin{array}{ccc}x=-6\\y=12\end{array}\right}[/tex]
Szczegółowe wyjaśnienie:
[tex]\left\{\begin{array}{ccc}\dfrac{2x-2y}{3}-\dfrac{x+2y}{6}=x-3y&|\cdot6\\5x-7y=13\end{array}\right\\\left\{\begin{array}{ccc}6\!\!\!\!\diagup^2\cdot\dfrac{2x-2y}{3\!\!\!\!\diagup_1}-6\!\!\!\!\diagup^1\cdot\dfrac{x+2y}{6\!\!\!\!\diagup_1}=6\cdot x-6\cdot3y\\5x-7y=13\end{array}\right\\\left\{\begin{array}{ccc}2(2x-2y)-(x+2y)=6x-18y\\5x-7y=13\end{array}\right[/tex]
[tex]\left\{\begin{array}{ccc}4x-4y-x-2y=6x-18y\\5x-7y=13\end{array}\right\\\left\{\begin{array}{ccc}3x-6y=6x-18y&|+6y-6x\\5x-7y=13\end{array}\right\\\left\{\begin{array}{ccc}-3x=-12y&|:(-3)\\5x-7y=13\end{array}\right\\\left\{\begin{array}{ccc}x=4y&(1)\\5x-7y=13&(2)\end{array}\right[/tex]
podstawiamy (1) do (2):
[tex]5\cdot4y-7y=13\\20y-7y=13\\13y=13\qquad|:13\\\boxed{y=1}[/tex]
podstawiamy wartość y do (1):
[tex]x=4\cdot1\\\boxed{x=4}\\\\\left\{\begin{array}{ccc}x=4\\y=1\end{array}\right[/tex]
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[tex]\left\{\begin{array}{ccc}4(3x+y)-(x-3y)=4y+5x\\3(2x+3y)+5(x-y)=-18\end{array}\right\\\left\{\begin{array}{ccc}12x+4y-x+3y=4y+5x\\6x+9y+5x-5y=-18\end{array}\right\\\left\{\begin{array}{ccc}11x+7y=4y+5x&|-4y-5x\\11x+4y=-18\end{array}\right\\\left\{\begin{array}{ccc}6x+3y=0&|\cdot4\\11x+4y=-18&|\cdot(-3)\end{array}\right[/tex]
[tex]\underline{+\left\{\begin{array}{ccc}24x+12y=0\\-33x-12y=54\end{array}\right}\\.\qquad-9x=54\qquad|:(-9)\\.\qquad\boxed{x=-6}[/tex]
podstawiamy wartość x do pierwszego równania:
[tex]6\cdot(-6)+3y=0\\-36+3y=0\qquad|+36\\3y=36\qquad|:3\\\boxed{y=12}\\\\\left\{\begin{array}{ccc}x=-6\\y=12\end{array}\right[/tex]
