Consider the linear function $f(x) = mx + b$.

A standard form of a linear equation with variable $x$ is an equation of the form $Ax + B = C$ with $A \ne 0$.
Beginning algebra spends a considerable amount of time solving this equation without any reference to functions.
When $B = 0$ this equation has the solution $x = \frac C A$.
When $B \ne 0$ the equation can be solved so that $x = \frac {C- B} A$.

Example  LEQ.1 : Suppose $5x - 10 = 20$. Find $x$.
A slightly more ambitious form of a linear equation with variable $x$ is an equation of the form $A_1x + B = A_2x +C$ with $A_1 \ne A_2$. Algebraically this is solved by solving the related equation $Ax + B = C$ where $A = A_1 - A_2$.

Example LEQ.2 : Suppose $5x - 10 = 3x + 20$. Find $x$..
You can use this next dynamic example to solve linear equations like those in Examples LEQ.1 and LEQ.2 visually with a mapping diagram of $f$ (and $g$) and the lines in the graph of $f$ (and $g$).

Example LF.DLEQ.0 Dynamic Views for solving an equation $f(x) = mx+b = c$ on Graphs and Mapping Diagrams

LINEAR Equations with 2 Variables

A linear equation with variables $x$ and $y$ is an equation of the form $Ax + By = C$ with $A^2 + B^2 \ne 0$.
When $B = 0$ this equation has the solution $x = - \frac C A$.
When $B \ne 0$ the equation can be solved so that $y = - \frac A B x + \frac C B$. So $y$ is a linear function of $f(x)$ where $m = -\frac A B$ and $b = \frac C B$.

Questions involving linear equations usually involve a second piece of information to have a unique solution.

Example  3.  Suppose $5x - 10y = 20$ (i) and $x=8$ (ii). Find $y$.
Solution: A. Replace $x$ in the equation (i) by $8$ to give the equation $5*8 -10 y = 40- 10 y = 20$ (iii).  This leads to the equation $10 y = 20$ (iv) and thus the solution to the solution $y = 2$.
B.  Express $y$ as a function of $x$ giving $y = f(x) = \frac 5 {10} x - \frac {20} {10}$ (v). Now evaluate $f(8) = \frac 1 2 8 - 2 =4$.

In Solution B we find the linear function being used to solve the problem merely by evaluation  of the function. This evaluation can be visualized both on the graph and the mapping diagram.

For the graph of $f$
: Find $x=8$ on the X axis , then find the point on the graph of $f$ directly above (below) that point, $(8,2)$, determine it's second coordinate, $2$, and that is the desired value for $y$.

For the mapping diagram of $f$
: Find $x=8$ on the X axis , then find the focus point of $f$, $F= [\frac 1 2, -2]$ on the mapping diagram. Draw the line through F and the point $x=8$, to find the point of intersection of this line with the Y axis, $y=2$, which is the desired value for $y$.

Example  4.  Suppose $5x - 10y = 20$ (i) and $y=8$ (ii). Find $x$.
Solution: A. Replace $y$ in the equation (i) by $8$ to give the equation $5*x -10*8 = 5x- 80 = 20$ (iii).  This leads to the equation $5x = 100$ (iv) and thus the solution to the solution $x = 20$.
B.  Express $y$ as a function of $x$ giving $y = f(x) = \frac 5 {10} x - \frac {20} {10} = \frac 1 2 x - 2$ (v). Now find x where $f(x) = 8 = \frac 1 2 x - 2$, which leads to $x=20$.

In Solution B we find the linear function being used to solve the problem solving a related equation based on the linear function $f$. This solution can be visualized both on the graph and the mapping diagram.

For the graph of $f$
: Find $y=8$ on the Y axis , then find the point on the graph of $f$ directly right (left)  ofthat point, $(20,8)$, determine it's first coordinate, $20$, and that is the desired value for $x$.

For the mapping diagram of $f$
: Find $y=8$ on the Y axis , then find the focus point of $f$, $F= [\frac 1 2, -2]$ on the mapping diagram. Draw the line through F and the point $y=8$, to find the point of intersection of this line with the X axis, $x=20$, which is the desired value for $x$.