Real Linear Equation Solutions

Why

We want to solve linear equations.

Example

Suppose we want to find $x_1, x_2 \in \R $ to satisfy the system of linear equations

\[ \begin{aligned} 3x_1 + 2x_2 &= 10, \text{ and} \\ 6x_1 + 5x_2 &= 20, \\ \end{aligned} \]

with constants $((3, 2), 10)$ and $((6, 4), 20)$.

We can associate to the first equation an equation for $x_1$ in terms of $x_2$. We call this solving the first equation for $x_1$.

\[ 3x_1 + 2x_2 = 10 \iff 3x_1 = 10 - 2x_2 \iff x_1 = (1/3)(10 - 2x_2). \]

Define $f_1: \R \to \R $ as $f_1(y) = (1/3)(10 - 2y)$. Then $3x_1 + 2x_2 = 10$ if and only if $x_1 = f_1(x_2)$. We have written $x_1$ as a function of $x_2$ and obtained a new equation. The equation is not linear, however, as $f_1$ is not linear.

Using the equation for $x_1$ in terms of $x_2$, we can substitute this equation into our second linear equation. The two linear equations hold if and only if

\[ 6f(x_2) + 5x_2 = 20 \iff 20 - 4x_2 + 5x_2 = 20 \iff x_2 = 0. \]

So the equations are satisifed if and only if $x_2$ is $0$. If $x_2 = 0$,then $3x_1 = 10$ and $6x_1 = 20$. Both of these are equivalent to $x_1 = 10/3$. So we have that $x_1$ must be $10/3$ and $x_2$ must be 0.

Clearly this is a solution. Is it the only one?¹

Future editions will expand. ↩︎