Решите неравенство:х+3<5,7(х+10)+2 целых 2/5_______________________4 целых...

-12 \\ " alt="5)x+3<5,7(x+10)+2 \frac{2}{5} \\ x+3<5,7(x+10)+2,4 \\ x+3<5,7x+57+2,4 \\ x-5,7x<57+2,4-3 \\ -4,7x<56,4 \\ x>-12 \\ " align="absmiddle" class="latex-formula">

x∈(-12;+oo)

$6)4 \frac{1}{6} y+(2-y) \leq 4y-3 \\ 4 \frac{1}{6} y+2-y \leq 4y-3 \\ 4 \frac{1}{6} y-y- 4y \leq -3-2 \\ -\frac{5}{6}y \leq -5 \\ y \geq 6 \\$

y∈[6;+oo)