Systems of linear equations and matrices: Linear equations
Solving a linear equation with several unknowns
In the same way 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 a linear equation with two unknowns The linear equation \(a\cdot x+b\cdot y=c\), with unkowns \(x\) and \(y\), where \(a\), \(b\), and \(c\) 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 \(x=-\frac{b}{a}\cdot y+\frac{c}{a}\). This is possible if only if \(a\ne0\); therefore the solutions are all pairs \(\rv{x,y}\) of the form \([-\frac{b}{a}\cdot y+\frac{c}{a},y]\). 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 \(\rv{x,y}\) of the form\(\rv{x,-\frac{a}{b}\cdot x+\frac{c}{b}}\). Here, the role of \(x\) as a parameter becomes clear: for each value of \(x\) there is exactly one solution.
If \(a\ne0\) 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.
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, and
- if \(b\ne0\), then there is no solution.
We proceed as in solving a linear equation with unknown #y#. Thus, we view #x# as a parameter.
\[ \begin{array}{rclcl} 4 x+6 y&=&4&\phantom{xxxxx}&\color{blue}{\text{the original equation}}\\
6\cdot y &=& 4- 4 x&\phantom{xxxxx}&\color{blue}{\text{terms without }y\text{ moved to the right}}\\
y &=& \displaystyle \frac{4-4x}{6}&\phantom{xxxxx}&\color{blue}{\text{divided by the coefficient }6\text{ of }y} \\
y &=&\displaystyle -{{2}\over{3}} x + {{2}\over{3}}&\phantom{xxxxx}&\color{blue}{\text{simplified order of terms changed}}
\end {array}\]
Or visit omptest.org if jou are taking an OMPT exam.
