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------------------------------------------------ Solve [tex]|x-6|+3=16[/tex].

A. [tex]x=19, 7[/tex]
B. [tex]x=19, -7[/tex]
C. [tex]x=19, -19[/tex]
D. [tex]x=19[/tex]

We are given the equation
[tex]$$
|x-6| + 3 = 16.
$$[/tex]

Step 1. Isolate the absolute value

Subtract 3 from both sides:
[tex]$$
|x-6| = 16 - 3 = 13.
$$[/tex]

Step 2. Split into two cases

The definition of absolute value tells us that if
[tex]$$
|A| = B \quad (\text{with } B \ge 0),
$$[/tex]
then
[tex]$$
A = B \quad \text{or} \quad A = -B.
$$[/tex]

Here, with [tex]$A = x-6$[/tex] and [tex]$B = 13$[/tex], we have:
[tex]\[
\begin{cases}
x - 6 = 13, \\
x - 6 = -13.
\end{cases}
\][/tex]

Step 3. Solve both equations

1. For the first case:
[tex]$$
x - 6 = 13 \quad \Longrightarrow \quad x = 13 + 6 = 19.
$$[/tex]
2. For the second case:
[tex]$$
x - 6 = -13 \quad \Longrightarrow \quad x = -13 + 6 = -7.
$$[/tex]

Final Answer

The solutions are:
[tex]$$
x = 19 \quad \text{or} \quad x = -7.
$$[/tex]

Thus, the answer is [tex]$x = 19, -7$[/tex].