Answer :

luana
[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]