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# Confusing modulus question Watch

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1. Hi everyone,

I have been given the following problem:

| x + 2 | < 3 => |x| <1

According to the worksheet there's a counterexample to this statement, but I can't find one - can anyone else find one, or is it me being silly?
2. Have you tried drawing a number line and thinking about lengths? Try not to think about |x| < 1 => |x+2| < 3 because that is a different statement.
3. (Original post by Electrogeek)
Hi everyone,

I have been given the following problem:

| x + 2 | < 3 => |x| <1

According to the worksheet there's a counterexample to this statement, but I can't find one - can anyone else find one, or is it me being silly?
Minus 2 does not work
4. | x + 2 | < 3 --> -3 < x + 2 < 3 --> -5 < x < 1
x in (-5, -1] would be a counter example to the statement, if the problem stated that | x + 2 | < 3 is only true for |x| < 1.
.
5. AH... So the counter example is to show that the rule exceeds the |x| <1 (because the domain where the equation is true is -5< x < 1)

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