5(х-1)/6-1≥2*(х+1) / 3
5(x-1)/5≥2*(x+1)/3
x-1≥2*(x+1)/3
x-1-2*(x+1)/3≥0
3x-3-2x-2/3≥0
x-5/3≥0 | *3
x-5≥0
x≥5
ответ: x∈{5;+∞)
