Posted by fishcakes on Friday, March 20, 2009 at 7:52pm.
make a table of values for thsi equation with at least 4 points in each table.
2x3y=4
then I isolated y to y=2/3x4/3
how do I find points where both x and y are whole numbers? How do I find the "nice" points so it is easier to graph?

math  GanonTEK, Friday, March 20, 2009 at 8:45pm
Trial and error really for x and y being whole numbers. You have to see, for example, how many 2's [the 2x bit] to take away from 4 [on the RHS] before it's divisible by 3 [from the 3y bit]
For example, from 3y = 4  2x
x = 5 will give 3y = 6 which means y = 2
The best points are usually where it cuts the axes.
Set y=0 to find where it cuts the x axis
Set x=0 to find where it cuts the y axis
you will get the points
(2, 0) and (0, 4/3) respectively.
Hope that helps

math  Reiny, Friday, March 20, 2009 at 8:47pm
y=2/3x4/3 or
y = (2x4)/3
you want the numerator to be a multiple of 3
by "trial and error", if x = 5
y = (104)/3
= 2
so one "nice" is (5,2)
once you have one point to find more just add multiples of 3 to your choice of x
e.g x = 5+12 or 17 should work
check: y = (344)/3 = 10
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