Найдите решение системы: 4(3х - 4у) + 14 = 3(5х + 2у) -2(х - 6у) = 5(3х - 4у) - 1

$\left \{ {{4(3x - 4y) + 14 = 3(5x + 2y)} \atop {-2(x - 6y) = 5(3x - 4y) - 1}} \right. \\ \left \{ {{12x-16y+14=15x+6y} \atop {-2x+12y=15x-20y-1}} \right. \\ \left \{ {{15x+6y+16y-12x=14} \atop {15x+2x-20y-12y=1}} \right. \\ \left \{ {{3x+22y=14} \atop {17x-32y=1}} \right. \\3x=14-22y \\x= \frac{14-22y}{3} \\17*\frac{14-22y}{3} -32y=1 \\17(14-22y)-96y=3 \\238-374y-96y=3 \\-470y=-235 \\y= \frac{235}{470} = \frac{1}{2}=0,5 \\x= \frac{14-22*0,5}{3} = \frac{14-11}{3} =1$
Ответ: (1;0,5)