## Solving Simple Rational Inequalities

\[f(x)=\frac{4x-3}{4-x} \lt 2\]

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 \[x \gt 4\]

.