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.

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 $\;6 x+6 y=1\;$ to an equation of the form $\;y=a\,x+b$ .

$y = -x + {{1}\over{6}}$

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

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

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 $c = a_2x_2 + \cdots + a_nx_n + b$ .