$6x -3y -18 =0$ What does $x$ and $y$ equal on a straight line graph? I need to sketch a graph of $6x-3y-18=0$.
I don't know if I'm on the right track but this is what I tried to get $x$:
$6x-3y-18=0$
$\Rightarrow 6x-3y+18=0+18$
$\Rightarrow 6x-3y=18$
$\Rightarrow 6x-3(0)=18$
$\Rightarrow x=18/6$
$\Rightarrow x=3$
Am I doing this the right way?
 A: It is very simple to plot graph of a straight line, by finding out the points of intersection of the given line with the coordinate axes as follows  


*

*$\color{red}{\text{Intersection point with the x-axis}}$:


Setting $y=0$ in the equation of the line: $6x-3y-18=0$ we get $$6x-0-18=0$$$$x=\frac{18}{6}=3$$ Hence the point of intersection with the x-axis is $(3, 0)$


*$\color{red}{\text{Intersection point with the y-axis}}$:


Setting $x=0$ in the equation of the line: $6x-3y-18=0$ we get $$0-3y-18=0$$$$y=-\frac{18}{3}=-6$$ Hence the point of intersection with the y-axis is $(0, -6)$ 
Now, specify the points of intersection $(3, 0)$ & $(0, -6)$ on the coordinate axes respectively & join them by a straight line. This will be the graph of the given line. 
A: A very simple formul everyone should know avoids to always redo these computation: is a striaght line passes though the points $(a,0)$ on the $x$-axis and $(0,b)$ on the $y$-axis, an equation of the straight line is:
$$\frac xa+\frac yb=1.$$
Furthermore , this formula generalises in $3$-space to an equation of a plane, given its intersections with the axes: $(a,0,0)$, $(0,b,0)$, $(0,0,c)$:
$$\frac xa+\frac yb+\frac zc=1.$$
Here the equation can be simplified to $\;\dfrac x3-\dfrac y6-1=0$, hence the points on the axes are $(3,0)$  and $(0,-6)$.
