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 $x = \frac {C- B} A$.

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 $A = A_1 - A_2$.

You can use this next dynamic example to solve linear equations like
those in Examples LEQ.1 and LEQ.2 visually with focus points on a mapping diagram of
$f$ (and $g$) and the lines in the graph of $f$ (and $g$).

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 $f$

For the mapping diagram of $f$

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 $f$

For the mapping diagram of $f$

Example 5. Suppose $5x - 10y = 20$ (i) and $x+y=1$ (ii). Find $x$ and $y$.

Solution: A. Using (ii) we see that $y=1-x$. Replace $y$ in the equation (i) by $1-x$ to give the equation $ 5*x -10*(1-x) = 15x- 10 = 20$ (iii). This leads to the equation $15x = 30$ (iv) and thus the solution to the solution $x = 2$ and consequently $y=1-2=-1$..

B. From (i) express $y$ as a function of $x$ giving $ y = f(x) = \frac 5 {10} x - \frac {20} {10} = \frac 1 2 x - 2 $ (v). From (ii) express $y$ as function of $x$ giving $y=g(x) =1-x .

Now find x where $f(x) = g(x)$ or $1-x = \frac 1 2 x - 2 $, which leads to $1= \frac 32 x -2$ and thus $x=2$ and $y=f(2)=g(2)= -1$.

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

For the graph of $f$

For the mapping diagram of $f$