The problem asks us to solve the linear equation $10 + 3(x - 2) = mx + 3$ for $x$, where $m$ is a real parameter.

1. Problem Description

The problem asks us to solve the linear equation $10 + 3(x - 2) = mx + 3$ for $x$, where $m$ is a real parameter.

2. Solution Steps

First, we simplify the left-hand side of the equation:

$10 + 3(x - 2) = 10 + 3x - 6 = 3x + 4$.

So the equation becomes $3x + 4 = mx + 3$.

Now, we want to isolate $x$. We can rearrange the terms to group the $x$ terms together:

$3x - mx = 3 - 4$.

$(3 - m)x = -1$.

Now, if $3 - m \neq 0$, we can divide both sides by $(3 - m)$ to solve for $x$:

$x = \frac{-1}{3 - m} = \frac{1}{m - 3}$.

If $3 - m = 0$, then $m = 3$, and the equation becomes:

$(3 - 3)x = -1$,

$0x = -1$,

$0 = -1$.

This equation has no solution.

3. Final Answer

If $m \neq 3$, then $x = \frac{1}{m - 3}$.

If $m = 3$, there is no solution.