Question

Решите неравенство:

х+3<5,7(х+10)+2 целых 2/5

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4 целых 1/6у+(2-у)<_4у-3

Answer 1

$5)x+3<5,7(x+10)+2 \frac{2}{5} \\ x+3<5,7(x+10)+2,4 \\ x+3<5,7x+57+2,4 \\ x-5,7x<57+2,4-3 \\ -4,7x<56,4 \\ x>-12$

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)