Systems of linear equations: Basic concepts and methods
Solving a linear equation with a single unknown
Any linear equation can be reduced to a basic form. Given such a basic form, solving the equation is not so difficult anymore. Here we discuss how this is done for a linear equation with a single unknown.
Solving a linear equation with a single unknown In general, the solutions of the linear equation \(a\, x+b=0\) with unknown \(x\) and real numbers \(a\) and \(b\) can be found as follows.
\(\,\)case

\(\,\)solutions

\(\,a\ne0\phantom{x}\)

\(\,\) exactly one: \(x=−\dfrac{b}{a}\,\)

\(\,a=0\) and \(b\ne0\,\)

\(\,\) none

\(\,a=0\) and \(b=0\,\)

\(\,\) any number \(x\,\)

\(\phantom{abc}\)
There is no need to remember these rules, because the solutions are easy to find by reductions (it is not strictly necessary to reduce the equation to a basic form first). The three cases are also identified in terms of geometric lines. For each case we give an example.
\(x=6\)
In order to see this, we reduce the equation as follows: \[\begin{array}{rclcl}8 x+36&=& 6 x48&\phantom{x}&\blue{\text{the original equation }}\\ 14 x+36&=&48&\phantom{x}&\blue{\text{the term }6 x\text{ moved to the left hand side}}\\ 14 x &=&84&\phantom{x}&\blue{\text{the term }36\text{ moved to the right hand side}} \\ x &=&6&\phantom{x}&\blue{\text{dividing by }14\text{}}\tiny.\end{array}\]
The only solution of the equation is \(x=6\).
In order to see this, we reduce the equation as follows: \[\begin{array}{rclcl}8 x+36&=& 6 x48&\phantom{x}&\blue{\text{the original equation }}\\ 14 x+36&=&48&\phantom{x}&\blue{\text{the term }6 x\text{ moved to the left hand side}}\\ 14 x &=&84&\phantom{x}&\blue{\text{the term }36\text{ moved to the right hand side}} \\ x &=&6&\phantom{x}&\blue{\text{dividing by }14\text{}}\tiny.\end{array}\]
The only solution of the equation is \(x=6\).