Hey everyone- can anyone pls assist with how to solve these:
This is nowhere in my book and I am not technically required to know this for my class but I am interested in how this would be solved, I do know how to do it if it were either/or:
i.e. |x|<2 you would just do this: -2< x <2
and obviously |x|> -5 would be all real #s
but to I have no idea how to solve when its already compounded with two different values at once!
Any help greatly appreciated!
Hey thanks plato!
So what you're saying is the left side gives (-∞,∞), and the right side gives (-2,2), but since the right side is more specific it overrules the left?
What if the problem were something like: 2 ≤ |x| ≤ 5
should be treated as two inequalities: and . Notice the "and"- they must both be true which means that the solution set of the "double" inequality is the intersection of the two separate inequalities.
Further, itself means if x is non-negative or if x is negative. Here, we have "or" so the solution set is the union of the two separate solution sets.
gives the two inequalities and .
(a) as long as x is non-negative which it certainly is if it is larger than or equal to 2: [tex]2\le x.
(b) which is the same as as long as x is negative, which it is as long as x is less than -2: .
The union of those two sets is the two separate intervals,
(a) as long as x is non-negative:
(b) which is the same as as long as x is negative:
The union of those two sets is the single interval, [-5, 5].
Now take the intersection of those two sets. The interval intersect [-5, 5] is [-5, -2]. The interval [2, \infty) intersect [-5, 5] is [2, 5]. The solution set for [tex]2\le x\le 5[/itex] is .
You probably could have done that more simply by reducing the absolute value first: and