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Ax+b=0 is onevariable linear equation 
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#1
Mar1314, 01:17 PM

P: 55

Hello
Am I right in saying: ax+b=0 is onevariable linear equation ax+by+c=0 is twovariable linear equation ax^2+bx+c=0 is onevariable quadratic equation ax^2+bx+c=y is twovariable 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. 


#2
Mar1314, 01:23 PM

Mentor
P: 18,345

[tex]ax^2 + by^2 + cxy + dx + ey + f = 0[/tex] [tex]\{(y,x)~\vert~ax+ by + c = 0\}[/tex] It might not be the same set, but it will be an equivalent situation you're dealing with. Geometrically, this corresponds to just swapping Xaxis and Yaxis. Usually, we will of course use [tex]\{(x,y)~\vert~ax + by + c = 0\}[/tex] but that's a convention. [tex]\{(x,y)\in \mathbb{R}^2~\vert~ x =a\}[/tex] or [tex]\{(a,y)\in \mathbb{R}^2~\vert~y\in \mathbb{R}^2\}[/tex] 


#3
Mar1314, 07:13 PM

P: 55

Thanks a lot. That cleared my doubts.



#4
Mar1414, 09:05 AM

P: 55

Ax+b=0 is onevariable 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? 


#5
Mar1414, 09:37 AM

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P: 18,345

[tex]\{(x,f(x))\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}[/tex] or perhaps [tex]\{(x,y)\in \mathbb{R}^2~\vert~y=f(x)\}[/tex] In the case that ##f(x) = x+a## for all ##x##, then this becomes [tex]\{(x,x+a)\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}[/tex] or [tex]\{(x,y)\in \mathbb{R}^2~\vert~y=x+a\}[/tex] and you can write this of course as [tex]\{(x,y)\in \mathbb{R}^2~\vert~x=ya\}[/tex] 


#6
Mar1414, 09:49 AM

P: 55

So
[tex] {(x,f(x))  f(x)=x+a} [/tex] is wrong then? Can't get { and } to show... 


#7
Mar1414, 09:51 AM

Mentor
P: 18,345

[tex]\{(x,f(x))~\vert~x+a=x+a\}[/tex] which simplifies to [tex]\{(x,f(x))~\vert~0=0\}[/tex] So yes, it's not really right. 


#8
Mar1414, 09:53 AM

P: 55

The how come
[tex] (x,y)  y=x+a [\tex] is correct? 


#9
Mar1414, 09:53 AM

P: 55

Then how come
[tex] (x,y)  y=x+a [/tex] is correct? 


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