Odpowiedź:
[tex]\begin{cases}\frac{5x+2y}{3}-\frac{2x-1}{2}=1\ \ /\cdot6\\\frac{3x-2}{2}-2y=3\ \ /\cdot2\end{cases}\\\\\\\begin{cases}2(5x+2y)-3(2x-1)=6\\3x-2-4y=6\end{cases}\\\\\\\begin{cases}10x+4y-6x+3=6\\3x-4y=6+2\end{cases}\\\\\\\begin{cases}4x+4y=6-3\\3x-4y=8\end{cases}\\\\\\\begin{cases}4x+4y=3\ \ /:4\\3x-4y=8\end{cases}\\\\\\\begin{cases}x+y=\frac{3}{4}\\3x-4y=8\end{cases}\\\\\\\begin{cases}x=\frac{3}{4}-y\\3x-4y=8\end{cases}[/tex]
[tex]3(\frac{3}{4}-y)-4y=8\\\\\frac{9}{4}-3y-4y=8\\\\-7y=8-\frac{9}{4}\\\\-7y=\frac{32}{4}-\frac{9}{4}\\\\-7y=\frac{23}{4}\ \ /:(-7)\\\\y=-\frac{23}{4}\cdot\frac{1}{7}\\\\y=-\frac{23}{28}\\\\\\x=\frac{3}{4}-(-\frac{23}{28})=\frac{3}{4}+\frac{23}{28}=\frac{21}{28}+\frac{23}{28}=\frac{44}{28}=\frac{11}{7}\\\\\\\begin{cases}x=\frac{11}{7}\\y=-\frac{23}{28}\end{cases}[/tex]
