Question

# Solve absolute value inequality. |2(x-1)+4|\leq8

Piecewise-Defined Functions
Solve absolute value inequality.
$$\displaystyle{\left|{2}{\left({x}-{1}\right)}+{4}\right|}\leq{8}$$

2021-02-26

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 $$\displaystyle{x}\in{\left[-{5},{3}\right]}$$.