What interval consists of all *w* which satisfy **neither **-2(6+2w) ≤ -16 nor -3*w *≥ 18?

Guest Aug 23, 2023

#1**0 **

We can solve the first inequality as follows:

-2(6+2w) ≤ -16 -12-4w ≤ -16 -4w ≤ -4 w ≥ 1

We can solve the second inequality as follows:

-3w ≥ 18 w ≤ -6

The solutions to the first inequality are all real numbers greater than or equal to 1. The solutions to the second inequality are all real numbers less than or equal to -6.

The interval that consists of all w which satisfy neither of these inequalities is the union of the intervals (-infinity, -6) and (1, infinity).

In interval notation, this is the interval (-infinity, -6) U (1, infinity).

Guest Aug 23, 2023

#2**0 **

I am confused with this answer. Why is it that the interval that satisfies neither inequality is the union of the intervals found? I would expect the interval that satisfies neither of the inequalities to be the intervals of real numbers not contained within the original interval. That would be \((-1, 6)\). This is the interval of real numbers that is greater than -1 and less than 6.

The3Mathketeers
Aug 24, 2023