Odpowiedź:
[tex]\begin{cases}5x+4y+1=0\\3x+7y-4=0\end{cases}\\\\\\\begin{cases}5x+4y=-1\\3x+7y=4\end{cases}\\\\\\\begin{cases}5x=-4y-1\ \ /:5\\3x+7y=4\end{cases}\\\\\\\begin{cases}x=-\frac{4}{5}y-\frac{1}{5}\\3x+7y=4\end{cases}\\\\\\3(-\frac{4}{5}y-\frac{1}{5})+7y=4\\\\-\frac{12}{5}y-\frac{3}{5}+7y=4\ \ /\cdot5\\\\-12y-3+35y=20\\\\23y=20+3\\\\23y=23\ \ /:23\\\\y=1[/tex]
[tex]x=-\frac{4}{5}\cdot1-\frac{1}{5}=-\frac{4}{5}-\frac{1}{5}=-\frac{5}{5}=-1\\\\\\\begin{cases}x=-1\\y=1\end{cases}[/tex]
