In order to simplify the notation consider: $$ x=[x_1,x_2,...x_n] $$
Consider the following inequation:
$$ s(x)(u(x)+A(x))<0 $$
My goal is to choose the function u(x) such that the inequation holds. s(x) and A(x) are known.
For instance: if s>0: $$u(x)+A(x)<0 $$ if s<0: $$ u(x)+A(x)>0 $$
This problem seems quite easy to solve.
if s>0: $$u(x)<-A(x) $$
and if s<0: $$u(x)>-A(x) $$
However, the textbook solution is quite different and I would like to know why:
Here is the solution:
if s>0 $$u(x)<-\begin{vmatrix} A(x) \end{vmatrix} $$
if s<0 $$u(x)>\begin{vmatrix} A(x) \end{vmatrix} $$
When $s(x)\ne 0$, there is a lot of freedom to choose the function $u$. Both choices are correct, but your is done in a most general framework, which also shows that the textbook choice is correct.