a,
$ | 2x - 8 | = 3x + 1 $
\(\left[ \begin{array}{l}2x - 8 = 3x + 1 \\ 2x - 8 = - ( 3x +1 ) \end{array} \right.\) $\\$ \(\left[ \begin{array}{l}x = -9\\ 2x - 8 = - 3x - 1 \end{array} \right.\) $\\$ \(\left[ \begin{array}{l} x = -9 \\ 5x = 7 \end{array} \right.\) $\\$ \(\left[ \begin{array}{l} x = -9 \\ x = \dfrac{7}{5} \end{array} \right.\)
b,
$ | x - 4 | = 4 - x $
\(\left[ \begin{array}{l} x - 4 = 4 - x \\ x - 4 = -4 + x \end{array} \right.\) $\\$ \(\left[ \begin{array}{l} 2x = 8 \\ 0 = 0 \end{array} \right.\) $\\$ \(\left[ \begin{array}{l}x = 4 \\0=0\end{array} \right.\)
$\to x = 4 $
c,
$ | 2x - 6 | = 2x - 6 $
\(\left[ \begin{array}{l}2x - 6 = 2x - 6\\2x - 6 = -2x + 6 \end{array} \right.\) $\\$ \(\left[ \begin{array}{l}0 =0 \\ 4x = 12 \end{array} \right.\) $\\$ \(\left[ \begin{array}{l}0=0\\x=3\end{array} \right.\) $\\$ $\to x = 3 $
