Алгебра: 4А.03(б), 4А.04(б),4А.06(б) пожалуйста

Решите задачу:

$5.4(3y-2)-7.2(2y-3)=1.2 \\\ 16.2y-10.8-14.4y+21.6=1.2 \\\ 1.8y=-9.6 \\\ y=- \frac{9.6}{1.8} =- \frac{96}{18} =- \frac{16}{3} =-5 \frac{1}{3}$

$5+ \frac{7x-12}{3} =x+13 \\\ \frac{7x-12}{3} =x+8 \\\ 7x-12=3x+24 \\\ 4x=36 \\\ x=9$

$\frac{10.5}{y-3.6} = \frac{5.1}{y+1.8} \\\ 10.5(y+1.8) = 5.1(y-3.6) \\\ 105(y+1.8) =51(y-3.6) \\\ 35(y+1.8) =17(y-3.6) \\\ 35y+63=17y-61.2 \\\ 18y=-124.2 \\\ y=-6.9$