Q:

# Solve the inequality involving absolute value. |x−2|+4≥10 Enter the exact answer in interval notation. To enter ∞ , type infinity. To enter ∪ , type U. Preview Show your work and explain, in your own words, how you arrived at your answers

Accepted Solution

A:
$|x-2| \geq 10-4$
$|x-2| \geq 6$
$\begin{array} { l }\begin{array} { l }x-2 \geq 6,& x-2 \geq 0\end{array},\\\begin{array} { l }-\left( x-2 \right) \geq 6,& x-2 < 0\end{array}\end{array}$
$\begin{array} { l }\begin{array} { l }x \geq 8,& x-2 \geq 0\end{array},\\\begin{array} { l }-\left( x-2 \right) \geq 6,& x-2 < 0\end{array}\end{array}$
$\begin{array} { l }\begin{array} { l }x \geq 8,& x \geq 2\end{array},\\\begin{array} { l }-\left( x-2 \right) \geq 6,& x-2 < 0\end{array}\end{array}$
$\begin{array} { l }\begin{array} { l }x \geq 8,& x \geq 2\end{array},\\\begin{array} { l }x \leq -4,& x-2 < 0\end{array}\end{array}$
$\begin{array} { l }\begin{array} { l }x \geq 8,& x \geq 2\end{array},\\\begin{array} { l }x \leq -4,& x < 2\end{array}\end{array}$
$\begin{array} { l }x \in \left[ 8, +\infty\right\rangle,\\\begin{array} { l }x \leq -4,& x < 2\end{array}\end{array}$
$\begin{array} { l }x \in \left[ 8, +\infty\right\rangle,\\x \in \left\langle-\infty, -4\right]\end{array}$
$x \in \left\langle-\infty, -4\right] \cup \left[ 8, +\infty\right\rangle$