I sposób

$\left|3x-1\right|=\left|4-x\right|$
$3x-1=4-x\vee 3x-1=-4+x$
$4x=52x=-3$
   $x=\frac{5}{4}x=-\frac{3}{2}$

$x\in \left\{-\frac{3}{2},\frac{5}{4}\right\}$

II sposób

$\left|3x-1\right|=\left|4-x\right||{\left(\right)}^2$
${\left(3x-1\right)}^2={\left(4-x\right)}^2$
$9x^2-6x+1=16-8x+x^2$
$8x^2+2x-15=0$
$\Delta =2^2-4\cdot 8\cdot \left(-15\right)=484,\sqrt{\Delta }=22$
$x_1=\frac{-2-22}{16}=-\frac{24}{16}=-\frac{3}{2}$

$x_2=\frac{-2+22}{16}=\frac{20}{16}=\frac{5}{4}$

$x\in \left\{-\frac{3}{2},\frac{5}{4}\right\}$