Question 1820
any straight line can be sketched if you know it in the form y = mx + c where m is the gradient and c in the y-intercept...so we need to re-arrange your equation slightly

{{{2x+3y=6}}} becomes {{{3y = -2x + 6}}} then becomes {{{y = (-2x + 6)/3}}} which is {{{y = (-2/3)x + 2}}}.

so, the gradient is -2/3 and the y-intercept is at y=2

We can also find where the line crosses the x-axis, as this is where y=0, so putting y=0 into the equation (any version) gives x=3.

You need to know what a +ve and -ve gradient looks like:

+ve is /
-ve is \

cheers
Jon