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