Solving a linear equation with a single unknown

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 \(x\,\)

\(\phantom{abc}\)

The linear equation \(3x+8=0\) has solution \(x=-\frac{8}{3}\).
The equation \(0\cdot x + 8 = 0\) does not essentially contain the unknown: it is equivalent to \(8=0\), an equation which is not satisfied. There is no solution.
The equation \(0\cdot x + 0 = 0\) also does not essentially contain the unknown, but it is equivalent to \(0=0\), an equation which is always satisfied. Each value of \(x\) is a solution.

We explain why. (The equation is \(ax+b=0\) )

\(\,\)case	\(\,\)solutions	\(\,\)statement
\(\,a\ne0\,\)	\(\,\) exactly one: \(x=−\dfrac{b}{a}\,\)	\(\,\) subtract the left- and right-hand side \(b\) \(\,\) and divide then both sides by \(a.\,\)
\(\,a=0\) and \(b\ne0\,\)	\(\,\) no	\(\,\) The equation is \(b=0\) and that is not true, \(\,\) regardless of the choice of \(x.\,\)
\(\,a=0\) and \(b=0\,\)	\(\,\) any number \(x\,\)	\(\,\) The equation is \(b=0\); it is true \(\,\) for each choice of \(x.\,\)

In the above rule for the solution of the linear equation \(a\cdot x+b=0\) with unknown \(x\), the symbols \(a\) and \(b\) represent real numbers for which specific values can be entered. But in fact we have written the equation and the solution using \(a\) and \(b\) without writing down specific values. In other words, \(a\) and \(b\) are parameters.

Actually, all statements also apply to complex numbers instead of real numbers.

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.

Find the exact solution of the equation \(\;-8 x+36=6 x-48\;\) in \(x\).

\(x=6\)

In order to see this, we reduce the equation as follows: \[\begin{array}{rclcl}-8 x+36&=& 6 x-48&\phantom{x}&\blue{\text{the original equation }}\\ -14 x+36&=&-48&\phantom{x}&\blue{\text{the term }6 x\text{ moved to the left hand side}}\\ -14 x &=&-84&\phantom{x}&\blue{\text{the term }36\text{ moved to the right hand side}} \\ x &=&6&\phantom{x}&\blue{\text{dividing by }-14\text{}}\tiny.\end{array}\]
The only solution of the equation is \(x=6\).

New example