Ax+b=0 is one-variable linear equation

[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
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.

 

[micromass]

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\}

 

[WannabeFeynman]

Thanks a lot. That cleared my doubts.

 

[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?

 

[micromass]

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\}

 

[WannabeFeynman]

So
{(x,f(x)) | f(x)=x+a}
is wrong then?

Can't get { and } to show...

 

[micromass]

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.

 

[WannabeFeynman]

The how come

(x,y) | y=x+a [\tex]<br /> <br /> is correct?

 

[WannabeFeynman]

Then how come

(x,y) | y=x+a

is correct?