Calculation:

To solve: \(\displaystyle{\left|{3}{x}+{1}\right|}\leq{13}\)

Let's continue to think in terms of distance on a number line. The number, \(\displaystyle{3}{x}+{1}\), must be less than or equal to 13 units away from zero.

\(\displaystyle{\left|{3}{x}+{1}\right|}\leq{13}\) is equivalent to \(\displaystyle-{13}\leq{3}{x}+{1}\leq{13}\)

By using the property \(\displaystyle{\left|{x}\right|}{<}{k}\) is equivalent to \(\displaystyle{x}\succ{k}\ {\quad\text{and}\quad}\ {x}{<}{k}\), where k is positive number,

We can write \(\displaystyle-{13}\leq{3}{x}+{1}\leq{13}\ {a}{s}\ {3}{x}+{1}\geq-{13}\ {\quad\text{and}\quad}\ {3}{x}+{1}\leq{13}\).

Now we have to solve this conjunction.

First we have to isolate the absolute value expression on one side of the inequality before solving the inequality, so we have to subtract 1 from both sides.

\(\displaystyle{3}{x}+{1}\geq-{13}\)

\(\displaystyle{3}{x}+{1}-{1}\geq-{13}-{1}\)

\(\displaystyle{3}{x}\geq-{14}\)

Divide both sides by 3, we get

\(\displaystyle{\frac{{{3}{x}}}{{{3}}}}\geq-{\frac{{{14}}}{{{3}}}}\)

\(\displaystyle{x}\geq-{\frac{{{14}}}{{{3}}}}\)

And \(\displaystyle{3}{x}+{1}\leq{13}\)

\(\displaystyle{3}{x}+{1}-{1}\leq{13}-{1}\)

\(\displaystyle{3}{x}\leq{12}\)

Divide both sides by 3, we get

\(\displaystyle{\frac{{{3}{x}}}{{{3}}}}\leq{\frac{{{12}}}{{{3}}}}\)

\(\displaystyle{x}\leq{4}\)

We can write an absolute value inequality as a compound inequality (i.e.)

\(\displaystyle{\frac{{-{14}}}{{{3}}}}\leq{x}\leq{4}\)

The solution set is \(\displaystyle{\left[{\frac{{-{14}}}{{{3}}}},{4}\right]}\)

Conclusion: The solution set is \(\displaystyle{\left[{\frac{{-{14}}}{{{3}}}},{4}\right]}\)