# To solve:|3x+1| \leq 13.General strategy to solve the inequaliti

To solve:
$$\displaystyle{\left|{3}{x}+{1}\right|}\leq{13}$$.
General strategy to solve the inequalities that involve absolute value:
Absolute value inequalities deal with the inequalities $$\displaystyle{\left({<},{>},\leq,{\quad\text{and}\quad}\ \geq\right)}$$ on the expressions with absolute sign.
We can use the property $$\displaystyle{\left|{x}\right|}{<}{k}$$ is equivalent to $$\displaystyle{x}\succ{k}\ {\quad\text{and}\quad}\ {x}{<}{k}$$, where k is a positive number and we can write a conjuction such as $$\displaystyle{x}\succ{k}\ {\quad\text{and}\quad}\ {x}{<}{k}$$ in the compact form.
$$\displaystyle-{k}{<}{x}{<}{k}$$.
For example, $$\displaystyle{\left|{x}\right|}{<}{2}\ {\quad\text{and}\quad}\ {\left|{x}\right|}{>}{2}$$.
$$\displaystyle{\left|{x}\right|}{<}{2}$$, represents the distance between x and 0 that is less than 2.
Whereas $$\displaystyle{\left|{x}\right|}{>}{2}$$, represents the distance between x and 0 that is greater than 2.
We can write an absolute value inequality as a compound inequality $$\displaystyle{\left({i}.{e}.\right)}-{2}{<}{x}{<}{2}$$.
When solving an absolute value inequality it's necessary to first isolate the absolute value expression on one side of the inequality before solving the inequality.
$$\displaystyle{\left|{a}{x}+{b}\right|}{<}{c}$$, where $$\displaystyle{c}{>}{0}$$
$$\displaystyle=-{c}{<}{a}{x}+{b}{<}{c}$$
$$\displaystyle{\left|{a}{x}+{b}\right|}{>}{c}$$, where $$\displaystyle{c}{>}{0}$$
$$\displaystyle={a}{x}+{b}{<}-{c}\ {\quad\text{or}\quad}\ {a}{x}+{b}{>}{c}$$
We can replace > above with $$\displaystyle\geq\ {\quad\text{and}\quad}\ {<}\ {w}{i}{t}{h}\ \leq$$.

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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 $$\displaystyle{\left[{\frac{{-{14}}}{{{3}}}},{4}\right]}$$