Solving simple modulus with a variable but use inegality operator

The question is very simple but my brain forgot some theory for 5-6 years. Please help me

I have this simple |4x-2| >= -1

Sorry if I took your time for explanation.

Thank you

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This is definitely not a question about modules. Perhaps you refer to $|\cdot |$ as modulus? Also, the inequality you give is true for all $x$, since $|\cdot |$ always gives a nonnegative number. – Alex Becker Sep 25 '12 at 19:45
@AlexBecker: True, sorry for my bad english – Snake Eyes Sep 25 '12 at 19:49

All $x$ numbers are solutions, since $|z| \ge 0$ for any (real or complex) number, and $0\ge -1$ and $\ge$ is transitive.

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I think 'module' is a mistranslation into English from your native language. This is an inequality which is true for any $x$, because absolute value always returns a nonnegative number.

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I got it but I neeed to calculate the x . I corrected the title – Snake Eyes Sep 25 '12 at 19:47
Sure. But $x$ can be any complex number and your inequality will still be satisfied. There are infinitely many solutions. – sourisse Sep 25 '12 at 19:54

Here is an example on how to solve it, in the case where not all $x$'s are solutions. Say you had to solve $x$ such that $$|4x-2|\leq 1.$$ What $| x |\leq n$ means is that $x$ is comprised between $-n$ and $n$, and so you have $-n\leq x\leq n$. In your (my) case, it would become \begin{array}{rll} -1\leq& 4x-2&\leq 1 \\ 1\leq& 4x&\leq 3\\ \frac{1}{4}\leq& x&\leq \frac{3}{4} \end{array} so for $\geq$ inequality, you take the complement of that set in $\mathbb{R}$.

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I don't know if you meant $1$ or $-1$, but in any case $|4x-2| \ge 1$ is equivalent to $4x-2 \le -1$ or $4x-2 \ge 1$. – Javier Sep 25 '12 at 20:39
for some reason I did the case $|4x-2|\leq 1$ – Jean-Sébastien Sep 25 '12 at 21:09