Truth set of $-|x| \lt 2$? An exercise in my Algebra I book (Pearson and Allen, 1970, p. 261) asks for the graph of the truth set for $-\left|x\right| \lt 2; x \in \mathbb{R}$.
I've re-stated the inequality in the equivalent form of $\left|x\right| \gt -2$. I know that the truth set of $\left|x\right| = -2$ is $\emptyset$, but I'm not certain how to handle the inequality in conjunction with the absolute value.
I suspect the truth set is $\{x \mid x \ge 0\}$, but I am not certain whether this is correct, or how to prove it using the algebraic concepts I've learned thus far. (I suspect this is a flaw with the book as this is not the first time it assumes knowledge that hasn't yet been presented.)
Is the truth set I arrived at correct? Is there a simple proof of the solution using Algebra I concepts (i.e. the field axioms, basic order properties, etc.)?
Bibliography
Pearson, H. R and Allen, F. B., 1970. Modern Algebra - A Logical Approach (Book I). Boston: Ginn and Company.

$$-|x|<2\stackrel{\text{multiplication by}\,\,(-1)}\Longleftrightarrow |x|>-2$$
So: for what values of (real) $\,x\,$ it is true that $\,|x|>-2\,$ ? Hint: this is a rather huge subset of the reals...
