Step 1: Make a table of points for $x+3y=6$, i.e. $y=\dfrac{6-x}{3}$.
| $x$ | 0 | 6 | 3 |
| --- | --- | --- | --- |
| $y$ | 2 | 0 | 1 |
Step 2: Make a table of points for $2x-3y=12$, i.e. $y=\dfrac{2x-12}{3}$.
| $x$ | 0 | 6 | 3 |
| --- | --- | --- | --- |
| $y$ | $-4$ | 0 | $-2$ |
Step 3: Plot the points of the first equation, i.e. $(0,2),(6,0),(3,1)$, and join them to get a straight line.
Step 4: Plot the points of the second equation, i.e. $(0,-4),(6,0),(3,-2)$, and join them to get another straight line.
Step 5: The two lines are seen to intersect at the point $(6,0)$.
Step 6: Verification by algebra — adding the two equations:
$$(x+3y)+(2x-3y)=6+12 \;\Rightarrow\; 3x=18 \;\Rightarrow\; x=6$$
$$6+3y=6 \;\Rightarrow\; y=0$$
Hence the solution is $x=6,\; y=0$.