Solving Simple Rational Inequalities

Solving the inequality
$f(x)=\frac{4x-3}{4-x} \lt 2$
is not as simple as makling
$x$
the subject. We could naively multiply both sides by
$4-x$
, obtaining
$4x-3 \lt 2(4-x)=8-2x$
.
Solving this inequality gives
$4x+2x \lt 3+8 \rightarrow 6x \lt 11 \rightarrow x \lt \frac{11}{6}$

This ignores the possibility that when we multiply both sides by
$4-x$
we may be multiplying by a negative number. When we multiply by a negative numbe the inequality sign has to point the other way. This will only happen in this case if
$4-x \lt 0 \rightarrow 4 \lt x \rightarrow x \gt 4$
. With this condition we have
$4x-3 \gt 2(4-x)=8-2x$
,
$x \gt 4$

$4x-3 \gt 8-2x$
,
$x \gt 4$

$4x+2x \gt 3+8$
,
$x \gt 4$

$6x \gt \rightarrow x \gt \frac{11}{6}$
,
$x \gt 4$

We have two inequalitiers here
$x \gt \frac{11}{6} , \: x \gt 4$
.
The first of these is redundant, so
$x \gt 4$
.
The complete solution is
$x \lt \frac{11}{6}$
or
$x \gt 4$
.