# Ax+b=0 is one-variable linear equation

 P: 55 Hello Am I right in saying: ax+b=0 is one-variable linear equation ax+by+c=0 is two-variable linear equation ax^2+bx+c=0 is one-variable quadratic equation ax^2+bx+c=y is two-variable quadratic equation Every linear or quadratic equation in one or two variables can be represented in those ways. How come graph of ax+by+c=0 is point of solution set {(x,y) | ax+by+c=0}? Why not (y,x)? Since we can rearrange it as x=(y=b)/m, can we pick the y value first and then find the x value? I know graph of x=a and y=a is vertical and horizontal lines respectively because x or y will always be constant no matter what y or x is respectively. But how would we represent the solution sets? For x=a, would it be {(x,y) | x+0y=a}? Thanks.
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P: 18,019
 Quote by WannabeFeynman Hello Am I right in saying: ax+b=0 is one-variable linear equation ax+by+c=0 is two-variable linear equation ax^2+bx+c=0 is one-variable quadratic equation ax^2+bx+c=y is two-variable quadratic equation
Sure.

 Every linear or quadratic equation in one or two variables can be represented in those ways.
I would say the general two-variable quadratic equation has the form

$$ax^2 + by^2 + cxy + dx + ey + f = 0$$

 How come graph of ax+by+c=0 is point of solution set {(x,y) | ax+by+c=0}? Why not (y,x)? Since we can rearrange it as x=(y=b)/m, can we pick the y value first and then find the x value?
It doesn't matter much. You might as well choose

$$\{(y,x)~\vert~ax+ by + c = 0\}$$

It might not be the same set, but it will be an equivalent situation you're dealing with. Geometrically, this corresponds to just swapping X-axis and Y-axis.

Usually, we will of course use

$$\{(x,y)~\vert~ax + by + c = 0\}$$

but that's a convention.

 I know graph of x=a and y=a is vertical and horizontal lines respectively because x or y will always be constant no matter what y or x is respectively. But how would we represent the solution sets? For x=a, would it be {(x,y) | x+0y=a}?
Yes, that will be one possible way of writing the solution set. Another one is

$$\{(x,y)\in \mathbb{R}^2~\vert~ x =a\}$$

or

$$\{(a,y)\in \mathbb{R}^2~\vert~y\in \mathbb{R}^2\}$$
 P: 55 Thanks a lot. That cleared my doubts.
 P: 55 Ax+b=0 is one-variable linear equation Am I right in saying: graph of f(x) is graph of solution set {(x,f(x)) | f(x)=x+a} Can I right it as x=f(x)-a too?
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P: 18,019
 Quote by WannabeFeynman Am I right in saying: graph of f(x) is graph of solution set {(x,f(x)) | f(x)=x+a} Can I right it as x=f(x)-a too?
I would say the graph of a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is

$$\{(x,f(x))\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}$$

or perhaps

$$\{(x,y)\in \mathbb{R}^2~\vert~y=f(x)\}$$

In the case that ##f(x) = x+a## for all ##x##, then this becomes

$$\{(x,x+a)\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}$$

or

$$\{(x,y)\in \mathbb{R}^2~\vert~y=x+a\}$$

and you can write this of course as

$$\{(x,y)\in \mathbb{R}^2~\vert~x=y-a\}$$
 P: 55 So $${(x,f(x)) | f(x)=x+a}$$ is wrong then? Can't get { and } to show...
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P: 18,019
 Quote by WannabeFeynman So $${(x,f(x)) | f(x)=x+a}$$ is wrong then?
If your ##f## is defined by ##f(x) = x+a##, then that just says

$$\{(x,f(x))~\vert~x+a=x+a\}$$

which simplifies to

$$\{(x,f(x))~\vert~0=0\}$$

So yes, it's not really right.
 P: 55 The how come $$(x,y) | y=x+a [\tex] is correct?  P: 55 Then how come [tex] (x,y) | y=x+a$$ is correct?

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