Solve system of linear inequation: 1 < lx - 2l < 3 - Vidyadip

Question

Solve system of linear inequation: 1 < lx - 2l < 3

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Answer

When,
$|x-2| \leq 1$
Then,
$x-2 \leq-1$ 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:
$-1 \leq x \leq 1 \text { and } 3 \leq x \leq 5$
Therefore,
$x \in[-1,1] \cup[3,5]$

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