# Converting solution of linear equations to slope intercept form

I will use a very simple linear equation to make my point:

$$3x = 2$$

$x$ will always be equal to $2/3$. This equation in itself has no use in graphing. So if we want to be able to apply that equation in graphing, is it legal to convert that to this:

$$y = -3x + 2$$

All I did was moved the $3x$ to the right and made it equal to $y$. If you solve that, you see that $y$ is equal to $0$.

In fact, it's a pattern because you can use any random linear equation you want:

$$5x = 3$$

and convert it to

$$y = -5x + 3$$

and $y$ is always equal to $0$.

So if what I am doing is legal, is it safe to say that a linear equation will always give you a value of $x$ (and that $x$ value will depend on the constants and variables of the equation itself) but for that value of $x$, y will always be $0$, that is, that $x$ is essentially the point of origin for $y$. And then when you change the value of $x$ to $1$, for example, $y$ will be $-2$ and then your straight line is formed.

And if this is true (and it's probably called somwething that Im not aware of), then does that mean in the case of identity linear equations, you can do something like this:

$$4(5x + 10) = 5(4x + 8)$$

$$y = (4(5x + 10))^{-1} + 5(4x + 8)$$

So that means both $x$ and $y$ equal $0$. And identity linear equations have no application in graphing.

Looking for algebra (not calculus) explanations. thanks.

-
When you asked that in the chat, I thought that adding one variable wouldn't be viable, were it viable I guess we could repeat the same process ad infinitum: $[3x=2]\sim [3x+x_1=2]\sim [3x+x_1+...+x_n=2]$. –  Vÿska Apr 21 '13 at 5:54

You certainly can graph $3x=2$ on the x-y plane. It's graph is a vertical line where the x coordinate is $\frac{2}{3}$ and y can be anything.
The second equation, $y=-3x+2$ is different than the first. It's graph is also different. It's a line that slants down to the right, through $(\frac{2}{3},0)$ and $(0,2)$.
Your two equations are different, but do intersect at the point $(\frac{2}{3},0)$ when $y=0$.