Solve the following systems of linear in equations:
$1 \leq|x-2| \leq 3$
$1 \leq|x-2|$ and $|x-2| \leq 3$
When,
$|x-2| \geq 1$
Then
$x-2 \leq-1$ and $x-2 \geq 1$$x-2 \leq-1$ and $x-2 \geq 1$
Now when,
$x-2 \leq-1$
Adding 2 to both the sides in above equation
$x-2+2 \leq-1+2$
$x \leq 1$
Now when,
$x-2 \geq 1$
Adding 2 to both the sides in above equation
$x-2+2 \geq 1+2$
$x \geq 3$
For $|x-2| \geq 1: x \leq 1$ or $x \geq 3$
When,
$|x-2| \leq 3$
Then,
$x-2 \geq-3$ and $x-2 \leq 3$
Now when,
$x-2 \geq-3$
Adding 2 to both the sides in above equation
$x-2+2 \geq-3+2$
$x \geq-1$
Now when
$x-2 \leq 3$
Adding 2 to both the sides in above equation
$x-2+2 \leq 3+2$
$x \leq 5$
For $|x-2| \leq 3: x \geq-1$ or $x \leq 5$
Combining the intervals:
$x \leq 1$ or $x \geq 3$ and $x \geq-1$ or $x \leq 5$
Merging the overlapping intervals:
$-1 \leq x \leq 1$ and $3 \leq x \leq 5$
Therefore,
$x \in[-1,1] \cup[3,5]$
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