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Ax+b=0 is one-variable linear equation

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WannabeFeynman
#1
Mar13-14, 01:17 PM
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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micromass
#2
Mar13-14, 01:23 PM
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Quote Quote by WannabeFeynman View Post
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

[tex]ax^2 + by^2 + cxy + dx + ey + f = 0[/tex]

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

[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 X-axis and Y-axis.

Usually, we will of course use

[tex]\{(x,y)~\vert~ax + by + c = 0\}[/tex]

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

[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]
WannabeFeynman
#3
Mar13-14, 07:13 PM
P: 55
Thanks a lot. That cleared my doubts.

WannabeFeynman
#4
Mar14-14, 09:05 AM
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?
micromass
#5
Mar14-14, 09:37 AM
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Quote Quote by WannabeFeynman View Post
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

[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=y-a\}[/tex]
WannabeFeynman
#6
Mar14-14, 09:49 AM
P: 55
So
[tex] {(x,f(x)) | f(x)=x+a} [/tex]
is wrong then?

Can't get { and } to show...
micromass
#7
Mar14-14, 09:51 AM
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Quote Quote by WannabeFeynman View Post
So
[tex] {(x,f(x)) | f(x)=x+a} [/tex]
is wrong then?
If your ##f## is defined by ##f(x) = x+a##, then that just says

[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.
WannabeFeynman
#8
Mar14-14, 09:53 AM
P: 55
The how come

[tex] (x,y) | y=x+a [\tex]

is correct?
WannabeFeynman
#9
Mar14-14, 09:53 AM
P: 55
Then how come

[tex] (x,y) | y=x+a [/tex]

is correct?


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