{1/2х-1/3у=1 1/4х+2/3у=3​

$\left \{ {{\frac{1}{2}x-\frac{1}{3}y=1|*(-0,5)} \atop {\frac{1}{4}x+\frac{2}{3}y=3}} \right. \Longrightarrow\ \left \{{{-\frac{1}{4}x+\frac{1}{6}y=-0.5} \atop {\frac{1}{4}x+\frac{2}{3}y=3}} \right.$

$+ \left \{{{-\frac{1}{4}x+\frac{1}{6}y=-0.5} \atop {\frac{1}{4}x+\frac{2}{3}y=3}} \right. \\\\\frac{5}{6}y=2.5\\y=3\\\\ \frac{1}{4}x+\frac{2}{3}*3=3\\\\\frac{1}{4}x+2=3\\\\\frac{1}{4}x=1\\\\x=4$

Ответ: y=3; x=4

{1/2х-1/3у=1 1/4х+2/3у=3​

{1/2х-1/3у=1 1/4х+2/3у=3