3x+5 > 14
3x+5-5 > 14-5 ... subtract 5 from both sides
3x > 9
3x/3 > 9/3 .... divide both sides by 3
x > 3
So x can be any number larger than 3
We're also told that [tex]2 \le x \le 8[/tex] meaning x is between 2 and 8 (including both endpoints).
Overlap these two intervals, and you'll get this new compound inequality: [tex]3 < x \le 8[/tex] so x can be larger than 3, and smaller than 8. It cannot be equal to 3, but it can equal 8. It can equal any number between 3 and 8.
If x is a whole number, then the following values are solutions: {4, 5, 6, 7, 8} so basically anything from 3 to 8, excluding 3 but including 8.
If x is a real number, then you simply write [tex]3 < x \le 8[/tex]
note: if x is a real number, then the answer in interval notation is (3, 8]. The curved parenthesis says "exclude this endpoint" and the square bracket means "include this endpoint".
