Systems of linear equations and matrices: Linear equations
Solving a linear equation with a single unknown
Each linear equation can be reduced to a basic form. Given such a basic form, solving the equation is not so difficult anymore. Here we recall how this is done for a linear equation with a single unknown.
Solving a linear equation in a single unknown In general, the solutions of the linear equation \(a\cdot x+b=0\) with unknown \(x\) and real numbers \(a\) and \(b\) 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 \(x\,\)

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 can also be identified geometrically in terms of lines, as we will see later. For each case we give an example.
#x=4#
To see this, we reduce the equation as follows.
\[\begin{array}{rclcl}3 x+12&=&24&\phantom{x}&\color{blue}{\text{the term }3 x\text{ moved to the left hand side}}\\ 3 x &=&12&\phantom{x}&\color{blue}{\text{the term }12\text{ moved to the right hand side}} \\ x &=&4&\phantom{x}&\color{blue}{\text{dividing by }3\text{}}\tiny.\end{array}\]
Hence, the only solution to the equation is #x=4#.
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