If |x+5|≥10, then
(3)
We know that |x-a|≥r
⇔ x≤a-r or x≥a+r
∴ |x+5|≥10⇔x≤-5-10, or x≥-5+10
⇔ x≤-15 or x≥5⇔x∈(-∞,-15] or x∈[5,∞)
⇔ x∈(-∞,-15]∪[5,∞)