[tex]\displaystyle\bf\\\left \{ {{3x-4y=18} \atop {5x+2y=6} \ |\cdot 2} \right. \\\\\\+\left \{ {{3x-4y=18} \atop {10x+4y=12}} \right. \\----------\\13x=30\\\\x=30:13=2\frac{4}{13} \\\\\\2y=6-5x=6-5\cdot2\frac{4}{13} =6-5\cdot \frac{30}{13} =6-\frac{150}{13}=6-11\frac{7}{13} =-5\frac{7}{13}\\\\\\y=-5\frac{7}{13} :2=-\frac{72}{13} \cdot \frac{1}{2} =-\frac{72}{26} =-\frac{36}{13}=-2\frac{10}{13} \\\\\\Otvet \ : \ \Big(2\frac{4}{13} \ ; \ -2\frac{10}{13} \Big)[/tex]