Answer: \(x\leqslant1\) or \(x\geqslant3\).
We start with the main method. Split the modulus expression into its equivalent linear cases, or compare the two sides graphically where a graph is requested.
The graph of \(y=|4-x|\) is V-shaped with vertex \((4,0)\) and \(y\)-intercept \((0,4)\).
The graph of \(y=|2x-5|\) is V-shaped with vertex \(\left(\frac52,0\right)\) and \(y\)-intercept \((0,5)\).
For the inequality, first find the intersection points:
\(|4-x|=|2x-5|.\)
Squaring both sides gives
\((4-x)^2=(2x-5)^2.\)
So
\(16-8x+x^2=4x^2-20x+25.\)
Hence \(3x^2-12x+9=0\), so
\(x^2-4x+3=0.\)
Therefore \(x=1\) or \(x=3\).
Testing a value between the intersections, for example \(x=2\), gives
\(|4-2|=2\) and \(|2(2)-5|=1\), so the inequality is false between \(1\) and \(3\).
Hence the solution is
\(x\leqslant1\) or \(x\geqslant3.\)
This completes the solution and gives the required result.
