My math book says, a Linear equation has exactly one solution. Because $ax + b = 0$; $x =-\frac{b}{a}$. But I've solved many linear equations with multiple solutions before. (I'm not very good in math. Need help...)

  • 2
    $\begingroup$ A linear equation has exactly one of three possible amounts of solutions: 1, 0, or infinite. $\endgroup$
    – Jared
    May 17 '16 at 2:52
  • $\begingroup$ Think of this as two lines $x \mapsto 0$ and $x \mapsto ax+b$, then the question is whether or not the lines intersect at exactly one point. It is easy to visualise that if they cross that they intersect at exactly one point. It is easy to see that if they are parallel but separate then they never intersect and have no crossing point (solution). If they are parallel and lie on top of each other then all points are crossing points. $\endgroup$
    – copper.hat
    May 17 '16 at 3:00
  • $\begingroup$ Are there equations that have more than one possible amount of solution? $\endgroup$
    – Joana
    May 17 '16 at 3:02
  • 1
    $\begingroup$ @Joana Yes--these are called non-linear equations, such as $x^2 + 5x + 4 = 0$. These are solved (algebraically) by transforming them into linear equations: $x^2 + 5x + 4 = (x + 1)(x + 4)$ therefore if $(x + 1)(x + 4) = 0$ then it's solved if $x + 1 = 0 \rightarrow x = -1$ or if $x + 4 = 0\rightarrow x = -4$. Therefore for the equation $x^2 + 5x + 4 = 0$ there are two possible solutions: $x = \{-1, -4\}$--both found by transforming this problem into two linear problems! $\endgroup$
    – Jared
    May 17 '16 at 3:05
  • $\begingroup$ Can you show an example of a linear equation you've solved that has multiple solutions? $\endgroup$
    – user856
    May 17 '16 at 3:21

If we assume that all linear equations have the form:

$$ ax + b= 0 $$

(which is completely valid and should be how we view linear equations)

then linear equations have either 1, 0, or infinite solutions. It's quite simple if $a \neq 0$ then they have exactly one solution: $x = -\frac{b}{a}$.

On the other hand, if $a = 0$ then if $b = 0$ we have infinite solutions (any value of $x$ solves $0x + 0 = 0$) and if $a = 0$ and $b \neq 0$ then there are no solutions (there is no value of $x$ that makes $0x + 1 = 0$, for example).


If we have $2$ unknowns, then the linear system

$$a_1 x_1 + a_2 x_2 = b$$

has, in general, infinitely many solutions. Why is that? Assuming that $a_1 \neq 0$, we write

$$x_1 = \frac{b}{a_1} - \left(\frac{a_2}{a_1}\right) x_2$$

Let $x_2 = \gamma$, where $\gamma \in \mathbb{R}$. Then, the solution set is a line parameterized as follows

$$\begin{bmatrix}x_1\\ x_2\end{bmatrix} = \begin{bmatrix} \frac{b}{a_1}\\ 0\end{bmatrix} + \gamma \begin{bmatrix} - \frac{a_2}{a_1}\\ 1\end{bmatrix}$$

  • $\begingroup$ But in a linear system (in the sense of the OP's question) we have a single unknown--not two. I get the want to extrapolate to many systems (which this definitely has an application)...once you start talking about a matrix with a zero determinant and being set to $\begin{bmatrix}0\end{bmatrix}$. $\endgroup$
    – Jared
    May 17 '16 at 3:17
  • $\begingroup$ @Jared With one unknown, only $0 x = 0$ has more than one solution, which is boring. I didn't want to obscure things, but who hasn't seen a linear system with two unknowns when plotting the graph of $y = m x + b$? $\endgroup$ May 17 '16 at 3:22
  • $\begingroup$ Whys is it "more boring" to have $0x = 0$ therefore $x$ can be anything rather than $x + y = 2$ and $ 2x + 2y = 4$ where $x$ can be anything? $\endgroup$
    – Jared
    May 17 '16 at 3:27
  • $\begingroup$ @Jared Because $\mathbb{R}$ only has one direction, whereas $\mathbb{R}^2$ has infinitely many. $\endgroup$ May 17 '16 at 3:29
  • $\begingroup$ I don't understand why having "infinitely many directions" makes a problem more interesting--I don't even know what that means...it would seem that a 3D problem is "more interesting" than a 2D problem and thus a "4D" problem more interesting than a "3D" problem, etc. $\endgroup$
    – Jared
    May 17 '16 at 3:32

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