$\begingroup$ can you show an example of a direct equation you've resolved that has multiple solutions? $\endgroup$

If us assume that

*all*linear equations have the form:

$$ax + b= 0$$

(which is fully valid and also *should* be exactly how we view direct equations)

then straight equations have actually either 1, 0, or limitless solutions. It"s quite straightforward if $a \neq 0$ climate they have *exactly* one solution: $x = -\fracba$.

You are watching: How many solutions does a linear equation have

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

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answered might 17 "16 at 2:55

JaredJared

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

$$a_1 x_1 + a_2 x_2 = b$$

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

$$x_1 = \fracba_1 - \left(\fraca_2a_1\right) x_2$$

Let $x_2 = \gamma$, whereby $\gamma \in \sdrta.netbbR$. Then, the solution set is a **line** parameterized as follows

$$\beginbmatrixx_1\\ x_2\endbmatrix = \beginbmatrix \fracba_1\\ 0\endbmatrix + \gamma \beginbmatrix - \fraca_2a_1\\ 1\endbmatrix$$

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answered may 17 "16 in ~ 3:11

Rodrigo de AzevedoRodrigo de Azevedo

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