Odpowiedź:
[tex]\begin{cases}\frac{x+2y}{4}+\frac{x-3y}{3}=-5\frac{2}{3}\\\frac{2x-y}{4}-\frac{3x-2y}{2}=9\frac{1}{2}\end{cases}\\\\\\\begin{cases}\frac{x+2y}{4}+\frac{x-3y}{3}=-\frac{17}{3}/\cdot12\\\frac{2x-y}{4}-\frac{3x-2y}{2}=\frac{19}{2}/\cdot4\end{cases}\\\\\\\begin{cases}3(x+2y)+4(x-3y)=-68\\2x-y-2(3x-2y)=38\end{cases}\\\\\\\begin{cases}3x+6y+4x-12y=-68\\2x-y-6x+4y=38\end{cases}\\\\\\\begin{cases}7x-6y=-68\\-4x+3y=38/\cdot2\end{cases}[/tex]
[tex]+\begin{cases}7x-6y=-68\\-8x+6y=76\end{cases}\\---------\\-x=8\ \ /\cdot(-1)\\x=-8\\\\\\-4\cdot(-8)+3y=38\\\\32+3y=38\\\\3y=38-32\\\\3y=6\ \ /:3\\\\y=2\\\\\begin{cases}x=-8\\y=2\end{cases}[/tex]
