$6x -3y -18 =0$ What does $x$ and $y$ equal on a straight line graph?

Question

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?

Answer 1

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

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

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 $(3, 0)$

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

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.

Answer 2

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 $(0,-6)$.