# For every real number $x$, $\lvert 2x - 6 \rvert \gt x \iff \lvert x-4 \rvert \gt 2$

I was hoping to get a bit of feedback on a proof I've done involving absolute values. This problem is taken from D. Velleman's How to Prove it (#3.5.11). I've only written on one side of the biconditional, but the I believe the other conditional can be structured similarly.

Problem Statement:
Prove that for every real number $$x$$, $$\lvert 2x - 6 \rvert \gt x$$ iff $$\lvert x - 4 \rvert \gt 2$$.

My question:
Do the statements below constitute a valid proof or do they require supplemental explanation?

($$\rightarrow$$) suppose $$2x - 6 \gt 0$$. Per the definition of $$\lvert 2x - 6 \rvert$$ we proceed by cases.

Case 1: $$2x - 6 \geq 0$$. Since $$2x - 6 \geq 0$$, $$\lvert 2x - 6 \rvert = 2x - 6.$$ Then $$\lvert 2x - 6 \rvert \gt x$$ $$= 2x - 6 \gt x$$ $$= -6 \gt -x$$ $$= 6 \lt x$$ $$= 4+2 \lt x$$ $$= 2 \lt x-4$$ $$= 2 \lt \lvert x - 4 \rvert$$ $$= \lvert x-4 \rvert \gt 2$$
Case 2: $$2x - 6 \lt 0$$. Since $$2x - 6 \lt 0$$, $$\lvert 2x - 6 \rvert = 6 - 2x$$.
Then
$$\lvert 2x - 6 \rvert \gt x$$ $$= 6 - 2x \gt x$$ $$= 6 \gt 3x$$ $$= 2 \gt x$$ $$= 4 > x + 2$$ $$= 4 - x \gt 2$$ $$= \lvert x - 4 \rvert \gt 2$$
EDIT
I removed the word 'equivalence' from the problem because it might not be the most appropriate word to use. I'm mostly just asking whether the series of statements I've given provide a valid proof of the conclusion.

• Tom do your = denote the equivalence? – user376343 Dec 1 '18 at 20:25