first off let's keep in mind that an absolute value is really a ± equation in disguise, so |5x -4| can pretty much be rewritten as ±(5x - 4), so is a dual equation, let's solve both for "x".
[tex]|5x-4|\leqslant 3x\implies \pm(5x-4)\leqslant 3x \\\\[-0.35em] ~\dotfill\\\\ +(5x-4)\leqslant 3x\implies 5x-4\leqslant 3x\implies 2x-4\leqslant 0\implies 2x\leqslant 4 \\\\\\ x\leqslant \cfrac{4}{2}\implies {\LARGE \begin{array}{llll} x\leqslant 2 \end{array}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]-(5x-4)\leqslant 3x\implies (5x-4)\stackrel{\stackrel{notice}{\downarrow }}{\geqslant} -3x\implies 8x-4\geqslant 0 \\\\\\ 8x\geqslant 4\implies x\geqslant \cfrac{4}{8}\implies {\LARGE \begin{array}{llll} x\geqslant \cfrac{1}{2} \end{array}}[/tex]
Check the picture below.
