14. A compound inequality with 'or' is true if one or
both statements are true. The solution can be a
combination of both inequalities or only one.
2x+ 1 < 11 or x > 3x +
5
2
2x < 10; x < 5
-2x > 2; x < -1 5
Since the second inequality is part of the
first inequality, ( x < 5 includes x <-1), the
solution is the first inequality. {x| x < 5}
15. -2 > x + 1 or x + 2 > 3
-3 > x; x < - 3 or, x > 1
There isn't a number that is both less than
- 3 and greater than 1. But with inequalities
using 'or', it doesn't have to be both. Our
solution, therefore looks like this:
