To solve: \(\displaystyle{\left|{3}{x}-{1}\right|}\succ{2}\)

The given equation is of the form of \(\displaystyle{\left|{x}\right|}{>}{k}\). However, the standard method for solving inequalities that involve absolute value cannot be applied to \(\displaystyle{\left|{3}{x}-{1}\right|}\succ{2}\) because k is negative.

We will solve the given equation \(\displaystyle{\left|{3}{x}-{1}\right|}\succ{2}\) by inspection.

\(\displaystyle{\left|{3}{x}-{1}\right|}\succ{2}\) is satisfied by all real numbers because the absolute value of \(\displaystyle{\left({3}{x}-{1}\right)}\), regardless of what number is substituted for x, will always be greater than -1.

The solution set is the set of all real numbers, which we can express in interval notation as \(\displaystyle{\left(-\infty,\infty\right)}\).

Conclusion:

Absolute values are always greater or equal to zero. So \(\displaystyle{\left|{3}{x}-{1}\right|}\succ{2}\) is true for all x and the interval notation is \(\displaystyle{\left(-\infty,\infty\right)}\).