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Question 1013286: If the left side of an inequality is an absolute value and the right side is not and the sides are separated by a sign indicating "less than" or "less than or equal to", then any solution that is a compound inequality involves two single inequalities separated by "or".
True or false
Found 2 solutions by fractalier, ikleyn:
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! False.
Look at an example: |x + 1| < 5
This breaks into
x+1 < 5 and x+1 > -5
x < 4 and x > -6
These overlap...a logical AND case, not an OR case.
Answer by ikleyn(52893)  (Show Source):
You can put this solution on YOUR website! .
If the left side of an inequality is an absolute value and the right side is not and the sides are separated by a sign indicating "less than" or "less than or equal to", then any solution that is a compound inequality involves two single inequalities separated by "or".
True or false?
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True.
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If f(x) is a function or an expression containing an unknown variable x, then the inequality
|f(x)| < c, where c is a positive constant, has the solution set which is the  of solutions of these two systems of inequalities:
1) f(x) >= 0 and f(x) < c,  <--- This AND is the same as OR. In other words, it is not exclusive AND; in opposite, it is inclusive AND.
2) f(x) <= 0 and f(x) > -c.
Again: {x: |f(x)| < c} == {x: f(x) >= 0 and f(x) < c} U {x: f(x) <= 0 and f(x) > -c}.
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