$\begin{array}{cc}\text{E}\text{n}\text{t}\text{e}\text{r}\text{e}\text{d}& \text{A}\text{n}\text{s}\text{w}\text{e}\text{r}\text{P}\text{r}\text{e}\text{v}\text{i}\text{e}\text{w}\\ \text{C}\text{h}\text{o}\text{i}\text{c}\text{e}2& \text{C}\text{h}\text{o}\text{i}\text{c}\text{e}2\\ x-8& x-8\\ x+x-8& x+x-8\\ 2& 2\\ 2& 2\\ 2x16& 2x16\\ 8& 8\\ 2x-88& 2x-88\end{array}881616168816161616$Prove the following statement by cases: For any real number $x,x+|x-8|8.$

Part $1$

Part $2$

We will prove by two cases: $x-80$ and In this step, complete the proof of

the first case.

Proof.

Case $1$: Let $x$ be a real number such that $x-80.$

Since $x-80,|x-8|=$

It follows that $x+|x-8|=$

Since $x-80,x8.$

Multiply both sides of $x8$ by yields

Subtracts both sides of

by

yields

Hence, $x+|x-8|8.$