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  
\[x \gt 4\]
.