−(x − 3) + 4 < 5 − 2x

$-(x-3)+4<5-2 x$

$\Rightarrow-x+3+4<5-2 x$

$\Rightarrow-x+7<5-2 x$

$\Rightarrow-x+2 x<5-7 \quad[$ Transposing $-2 \mathrm{x}$ to LHS and 7 to RHS $]$

$\Rightarrow x<-2$

Hence, the solution set of the given inequation is $(-\infty,-2)$.