Example  3.  Suppose $2x + 1 = 2$. Find $x$.
Solution: The equation can be solved using the common algebraic steps:
 $2x + 1 = 2$ $\ \ \ \ -1 = -1$ $2x\ \ \ \ \ = 1$ $\frac 12 (2x) = \frac 12 (1)$ $\ \ \ \ \ x = \frac 12$

This conforms with the symbolic result,  $x = \frac {C- B} A$  where $A=2, B = 1$, and $C = 2$. So the solution is $x = \frac {2- 1} 2 = \frac {1} 2.$

Comment: We can consider the expression on the left hand side of the equation as a linear function of $x$ giving $f(x) = 2x + 1 = s(m(x))$ where $m(x) =2x$ and $s(x)=x+1$
The problem can be restated now: To find a value for $x$ where $f(x) = 2$.
This problem and its solution can be visualized with mapping diagrams for the linear functions $f$ treated as a composition of $m$ followed by $s$.
 Check the box in the GeoGebra sketch to see a visualization of the solution. On the GeoGebra mapping diagram for the composition, find $2$ on the final target axis. Undo $s$ by subtracting $1$ from $2$. This corresponds to going back on the diagram from $2$ on the final axis to $2-1=1$ on the middle axis, reversing the arrow that would go from $1$ to $2$ visualizing $s$. Thus if $m(x) = 1$ then $f(x) = s(m(x))=s(1)=2$  Undo $m$ by multiplying $1$ by $\frac 12$. This corresponds to going back on the diagram from $1$ on the middle axis to $\frac 12$ on the source axis, reversing the arrow that would go from $\frac 12$ to $1$ visualizing $m$. Thus when $x=\frac 12$, $f(x) = s(m(\frac 12))=s(1)=2$ On the GeoGebra sketch you can change  the values of $m=A$ and $b=B$ (with the sliders) and $C$ (currently $2$ on the target axis) to see other linear equations $Ax + B = C$ are solved visually with the mapping diagram.