Property 6
Equations with absolute values are divided into parts according to the critical value of the expression and solution is performed.
\(|x|=\begin{cases}x&x\ge0\\-x&x<0\end{cases}\)
\(\text{(critical point is 0)}\)
\(|x-2|=\begin{cases}x-2&x\ge2\\-x+2&x<2\end{cases}\)
\(\text{(critical point is 2)}\)
1.
\(|x-6|=x-2\)
\(\Rightarrow\text{S.S.}=?\)
2.
\(|x-6|=2x-3\)
\(\Rightarrow\text{S.S.}=?\)
3.
\(|2x+6|=x-4\)
\(\Rightarrow\text{S.S.}=?\)
4.
\(|x-8|=2x\)
\(\Rightarrow\text{S.S.}=?\)
5.
\(|2x-4|=x+6\)
\(\Rightarrow\text{S.S.}=?\)
6.
\(|x-6|=2x+10\)
\(\Rightarrow\text{S.S.}=?\)
7.
\(|x|=2x-4\)
\(\Rightarrow\text{S.S.}=?\)
8.
\(|x+2|+|x-4|=6\)
\(\Rightarrow\text{S.S.}=?\)
9.
\(|x-1|+|x+6|=7\)
\(\Rightarrow x=?\)
10.
\(|x-4|+|x+8|=12\)
\(\Rightarrow\text{S.S.}=?\)
11.
\(|x-3|+|x+5|=12\)
\(\Rightarrow\text{S.S.}=?\)
12.
\(|x-2|+|x+6|=10\)
\(\Rightarrow\text{S.S.}=?\)
13.
\(|x-2|+|x+4|=4\)
\(\Rightarrow\text{S.S.}=?\)