Solve absolute value inequality. |2(x−1)+4|≤8

Step 1
We have to solve the absolute value inequality:
${\displaystyle |2(x-1)+4|\le 8}$
We know for any modulus function,
$|f(x)|=\{\begin{array}{ll}f(x)& iff(x)\ge 0\\ -f(x)& iff(x)<0\end{array}$
We also know that if ${\displaystyle \left|f\left(x\right)\right|\le }$ a then it must satisfy that
${\displaystyle -a\le f\left(x\right)\le a}$
Applying above condition for given inequality,
${\displaystyle -a\le f\left(x\right)\le a}$
${\displaystyle -8\le 2(x-1)+4\le 8}$
Step 2
Simplifying further,
$-8-4\le 2(x-1)+4-4\le 8-4$ (substracted 4)
${\displaystyle -12\le 2(x-1)\le 4}$
${\displaystyle \frac{-12}{2}\le \frac{2(x-1)}{2}\le \frac{4}{2}}$ (divided all sides by 2)
${\displaystyle -6\le (x-1)\le 2}$
${\displaystyle -6+1\le x-1+1\le 2+1}$
${\displaystyle -5\le x\le 3}$
In interval notation we can write it as ${\displaystyle x\in [-5,3]}$.
Hence, solution of the absolute value inequality is ${\displaystyle x\in [-5,3]}$.