[tex]6|6-4x|=8x+4\\\\Formula\ for\ absolute\ value\\\\
|a|=a,\ for\ a \geq 0\\ or\\|a|=-a,\ for\ a<0\\\\I\\
for\ 6-4x \geq 0\ \ \ --> \ \ -4x \geq -6\ \ -->\ \ \ x \leq \frac{3}{2}\\\\
6(6-4x)=8x+4\\
36-24x=8x+4\ \ |subtract\ 4\\
32-24x=8x\ \ | add\ 24x\\
32=32x\ \ | divide\ by\ 32\\
x=1\ \ \ and\ 1 \leq \frac{3}{2}\\\\
II\\
for\ 6-4x <0\ \ \ --> \ \ -4x < -6\ \ -->\ \ \ x > \frac{3}{2}\\\\
6(-6+4x)=8x+4\\
-36+24x=8x+4\ \ | subtract4\\
-40+24x=8x\ \ \ | subtract\ 24x\\
-40=-16x\ \ | divide\ by\ -16[/tex][tex]x=2,5\ \ and\ 2,5>\frac{3}{2}\\\\Solutions:\\
x\in \{1;2,5\}[/tex]
