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,

\(\displaystyle{\left|{f{{\left({x}\right)}}}\right|}={b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{f{{\left({x}\right)}}}&{\quad\text{if}\quad}\ {f{{\left({x}\right)}}}\geq{0}\backslash-{f{{\left({x}\right)}}}&{\quad\text{if}\quad}\ {f{{\left({x}\right)}}}{<}{0}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\)</span>

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,

\(\displaystyle-{8}-{4}\leq{2}{\left({x}-{1}\right)}+{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 \(\displaystyle{x}\in{\left[-{5},{3}\right]}\).

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,

\(\displaystyle{\left|{f{{\left({x}\right)}}}\right|}={b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{f{{\left({x}\right)}}}&{\quad\text{if}\quad}\ {f{{\left({x}\right)}}}\geq{0}\backslash-{f{{\left({x}\right)}}}&{\quad\text{if}\quad}\ {f{{\left({x}\right)}}}{<}{0}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\)</span>

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,

\(\displaystyle-{8}-{4}\leq{2}{\left({x}-{1}\right)}+{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 \(\displaystyle{x}\in{\left[-{5},{3}\right]}\).