Solving a linear equation with several unknowns

Systems of linear equations: Basic concepts and methods

Solving a linear equation with several unknowns

In the same way as we previously described the general solution of a linear equation with a single unknown, we can also solve a linear equation with two or more unknowns; it is only a matter of selecting unknowns for their roles as temporary parameters.

First, we discuss equations with two unknowns.

Solving the linear equation with two unknowns The linear equation \(a\, x+b\, y=c\), with unknown \(x\) and \(y\) where \(a\), \(b\), and \(c\) are numbers of parameters, are numbers or parameters, can be solved in the following two ways:

Viewing \(y\) as a temporary parameter we solve the linear equation with unknown \(x\), and conclude that \(x=-\frac{b}{a}\, y+\frac{c}{a}\). This is possible if only if \(a\ne0\); therefore the solutions are all pairs \([x,y]\) of the form \([-\frac{b}{a}\, y+\frac{c}{a},y]\). Here, the role of \(y\) as a parameter becomes clear: for each value of \(y\) there is exactly one solution.
Viewing \(x\) as a temporary parameter we solve the linear equation with unknown \(y\), and conclude that \(y=-\frac{a}{b}\cdot x+\frac{c}{b}\). This is possible if and only if \(b\ne0\); therefore, the solutions are all pairs \([x,y]\) of the form \([x,-\frac{a}{b}\, x+\frac{c}{b}]\). Here, the role of \(x\) as a parameter becomes clear: for each value of \(x\) there is exactly one solution.

The linear equation \(2x+3y + 4 =0\) can be used to express

\(x\) terms of \(y\): \(x=-\frac{3y+4}{2}\)
\(y\) terms of \(x\): \(y=-\frac{2x+4}{3}\)

In equations like \(0\cdot x+3y+4=0\) and \(2x+0\cdot y+4 = 0\) not all unknowns are actually present: the equation can be rewritten in such a way that there appear fewer unknowns.

If both \(a\ne0\) and \(b\ne0\), then there is overlap between the first and second case. Each of the two provides a way to describe the set of solutions. The former does so by viewing \(x\) as a function of \(y\), the latter by viewing \(y\) as a function of \(x\). Exclusively for the first case, the vertical line for \(a=0\) occurs, and for the second case, the horizontal line for \(b=0\) occurs.

By now we have identified the possible solutions in the cases \(a\ne0\) or \(b\ne0\). The results constitute the set of points on a line in the plane. If \(a=0\) and \(b=0\), then the equation becomes \(0=c\) and two cases remain:

if \(c=0\), then each pair \(\rv{x,y}\) is a solution;
if \(c\ne0\), then there is no solution.

Reduce \(\;6 x+8 y=1\;\) to an equation of the form \(\;y=a\,x+b\).

\(y = -{{3}\over{4}}x + {{1}\over{8}}\)

We proceed as in solving a linear equation with unknown \(y\). Thus, we consider \(x\) as a parameter.

\[ \begin{array}{rclcl} 6 x+8 y&=&1&\phantom{xxxxx}&\blue{\text{the given equation}}\\
8\cdot y &=& 1- 6 x&\phantom{xxxxx}&\blue{\text{terms without }y\text{ moved to the right}}\\
y &=& \displaystyle 6 x+8 y=1&\phantom{xxxxx}&\blue{\text{divided by the coefficient }8\text{ of }y} \\
y &=&\displaystyle -{{3}\over{4}} x + {{1}\over{8}}&\phantom{xxxxx}&\blue{\text{simplified order of terms changed}}
\end {array}\]

New example

Remains the general case.

Isolating an unknown in a linear equation with \(n\) unknowns Consider the linear equation \[a_1x_1 + \cdots + a_nx_n + b = 0\] where \(a_1, \ldots, a_n\) and \(b\) are real or complex numbers and \(x_1, \ldots, x_n\) unknowns.

Suppose that the coefficient of one of the unknowns is distinct from zero, say \(a_1\neq 0\); then the unknown \(x_1\) can be isolated by viewing all the other unknowns as temporary parameters: \[ x_1=\frac{-b - a_2x_2 -\cdots - a_nx_n}{a_1}\]
If all coefficients are zero, there remain two cases:
- if \(b=0\), then each list of \(n\) numbers \(\rv{x_1,\ldots,x_n}\) is a solution;
- if \(b\ne0\), then there is no solution

The solution described for \(a_1\ne 0\) can be obtained from the case of one unknown by viewing the equation as \[a_1x_1+c = 0\] where \(c = a_2x_2 + \cdots + a_nx_n + b\).