Step 1

We have to solve the absolute value inequality:

\(\displaystyle{\left|{2}{\left({x}-{1}\right)}+{4}\right|}\leq{8}\)

We know for any modulus function,

\(|f(x)|=\begin{cases}f(x) & if\ f(x)\geq 0\\-f(x) & if\ f(x)<0\end{cases}\)

We also know that if \(\displaystyle{\left|{f{{\left({x}\right)}}}\right|}\leq\) a then it must satisfy that

\(\displaystyle-{a}\leq{f{{\left({x}\right)}}}\leq{a}\)

Applying above condition for given inequality,

\(\displaystyle-{a}\leq{f{{\left({x}\right)}}}\leq{a}\)

\(\displaystyle-{8}\leq{2}{\left({x}-{1}\right)}+{4}\leq{8}\)

Step 2

Simplifying further,

\(-8-4 \leq 2(x-1)+4-4 \leq 8-4\) (substracted 4)

\(\displaystyle-{12}\leq{2}{\left({x}-{1}\right)}\leq{4}\)

\(\displaystyle{\frac{{-{12}}}{{{2}}}}\leq{\frac{{{2}{\left({x}-{1}\right)}}}{{{2}}}}\leq{\frac{{{4}}}{{{2}}}}\) (divided all sides by 2)

\(\displaystyle-{6}\leq{\left({x}-{1}\right)}\leq{2}\)

\(\displaystyle-{6}+{1}\leq{x}-{1}+{1}\leq{2}+{1}\)

\(\displaystyle-{5}\leq{x}\leq{3}\)

In interval notation we can write it as \(\displaystyle{x}\in{\left[-{5},{3}\right]}\).

Hence, solution of the absolute value inequality is \(\displaystyle{x}\in{\left[-{5},{3}\right]}\).