Problem with graphing linear equations Well, I can understand how to graph basic liner equations, for example:
$$y=2x-4$$
The y-intercept would be  -4 and the slope would be 2. The coordinates could then be (0,-4)(1, -2)
However, how would I solve a linear equation like this: $$y = \frac{2x}{4}$$
What are the steps to find out the coordinates? The only relationship that I know that can possibly help me is: $$\frac{x}{4}=\frac{1}{4}x$$
 A: $$
\begin{align}
y & = \frac{2}{4} \\ \\
\iff y & = \frac{1}{2} + \;0 \\ 
 &\quad\; \vdots \qquad\vdots \\
y & = m x + b \\ \\
\therefore m & = \frac 12; \quad b = 0 \\ \\
\therefore & (0, 0) \in \;\text{line} \\
\end{align}
$$
And since $m = \; \text{slope} = \dfrac 12,\;\; (2, 1) \in \;\text{line}$, too.
Double check $m = \dfrac{ 1-0}{2 - 0} = \dfrac 12$
A: $ y=mx+n $
$m$=slope
$n=y-intercept$
$-n/m=x-intercept      $
A: Strong Hint:
The equation $y = \dfrac{2x}{4}$ is read as the linear relationship described by multiplying the $x$-value by $\dfrac{2}{4}$.

Since the numerator $2$ and the denominator $4$ are both even, we can simplify this fraction, meaning to reduce the fraction to lowest terms. To do that, we divide the numerator and the denominator by their greatest common divisor.
In symbols, the greatest common divisor of two integers $a$ and $b$ is expressed as $\gcd(a, b) = c$ such that $c$ is their greatest common divisor. By order of substitution, we consider $a = 2$ and $b = 4$.
Since $a$ and $b$ are even, their greatest common divisor is a product of $2$, namely itself: $\gcd(2, 4) = 2$
Therefore, we can simplify the fraction $\dfrac{2}{4}$ by dividing the numerator and denominator each by $2$.
Since the numerator is $2$ and the denominator is $4$, we simplify our fraction to $\dfrac{1}{2}$.
Therefore, $y = \dfrac{2x}{4} = \bigg(\dfrac{2}{4}\bigg)x = \dfrac{1}{2}x$.

We cannot simplify the fraction anymore because $\gcd(1, 2) = 1$ and so this fraction is most simplified. It is irreducible. Now, we look back at the general linear equation, which is the general linear relationship of the $x$-value to the $y$-value. This equation is in the following form: $$y = mx + c$$ for which $x$ is the independent variable and $y$ is the dependent variable. 
Since $y = \dfrac{1}{2}x$, then by order of substituion, $m = \dfrac12$ and $c = 0$. Therefore, $y = \dfrac12 x + 0$.

What is the value of the slope?
What is the value of the $y$-intercept?
What coordinates would this linear relationship generate?
