Question:
Solve $|4-x|+1<3$
Solution:
As, $|4-x|+1<3$
$\Rightarrow|4-x|<3-1$
$\Rightarrow|4-x|<2$
$\Rightarrow-2<4-x<2 \quad($ As,$|x|
$\Rightarrow-2-4<-x<2-4$ $\Rightarrow-6<-x<-2$ $\Rightarrow 2 $\therefore x \in(2,6)$
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