Answer :
[tex]|1-3x| < 5\\\\x=-1\\\\|1-3\cdot(-1)| < 5\\\\|1+3| < 5\\\\|4| < 5\\\\4 < 5-the\ true[/tex]
[tex]x=1\\\\|1-3\cdot1| < 5\\\\|1-3| < 5\\\\|-2| < 5\\\\2 < 5-the\ true[/tex]
[tex]x=2\\\\|1-3\cdot2| < 5\\\\|1-6| < 5\\\\|-5| < 5\\\\5 < 5-the\ false[/tex]
[tex]|1-3x| < 5\\\\1-3x < 5\ \wedge\ 1-3x > -5\\\\-3x < 5-1\ \wedge\ -3x > -5-1\\\\-3x < 4\ /:(-3)\ \wedge\ -3x > -6\ /:(-3)\\\\x > -\frac{4}{3}\ \wedge\ x < 2\\\\x\in(-\frac{4}{3};\ 2)\\\\I\ x=-1\in(-\frac{4}{3};\ 2)\\\\II\ x=1\in(-\frac{4}{3};\ 2)\\\\III\ x=2\notin(-\frac{4}{3};\ 2)[/tex]
[tex]x=1\\\\|1-3\cdot1| < 5\\\\|1-3| < 5\\\\|-2| < 5\\\\2 < 5-the\ true[/tex]
[tex]x=2\\\\|1-3\cdot2| < 5\\\\|1-6| < 5\\\\|-5| < 5\\\\5 < 5-the\ false[/tex]
[tex]|1-3x| < 5\\\\1-3x < 5\ \wedge\ 1-3x > -5\\\\-3x < 5-1\ \wedge\ -3x > -5-1\\\\-3x < 4\ /:(-3)\ \wedge\ -3x > -6\ /:(-3)\\\\x > -\frac{4}{3}\ \wedge\ x < 2\\\\x\in(-\frac{4}{3};\ 2)\\\\I\ x=-1\in(-\frac{4}{3};\ 2)\\\\II\ x=1\in(-\frac{4}{3};\ 2)\\\\III\ x=2\notin(-\frac{4}{3};\ 2)[/tex]
