## Precalculus (6th Edition)

The solutions are $x=2$ and $x=4$
$|6-2x|+1=3$ Take $1$ to the right side of the equation: $|6-2x|=3-1$ $|6-2x|=2$ Solving this absolute value equation is equivalent to solving two separate equations, which are: $6-2x=2$ and $6-2x=-2$ $\textbf{Solve the first equation:}$ $6-2x=2$ Take $6$ to the right side: $-2x=2-6$ $-2x=-4$ Solve for $x$: $x=\dfrac{-4}{-2}$ $x=2$ $\textbf{Solve the second equation:}$ $6-2x=-2$ Take $6$ to the right side: $-2x=-2-6$ $-2x=-8$ Solve for $x$: $x=\dfrac{-8}{-2}$ $x=4$ The solutions are $x=2$ and $x=4$