Solve the following
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,$|x|<a \Rightarrow-a<x<a)$

$\Rightarrow-2-4<-x<2-4$

$\Rightarrow-6<-x<-2$

$\Rightarrow 2<x<6$

$\therefore x \in(2,6)$

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