Solve the following systems of linear in equations:

Solution:

$1 \leq|x-2|$ 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,

$x \in[-1,1] \cup[3,5]$