make a table of values for thsi equation with at least 4 points in each table.
then I isolated y to y=2/3x-4/3
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 = (2x-4)/3
you want the numerator to be a multiple of 3
by "trial and error", if x = 5
y = (10-4)/3
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 = (34-4)/3 = 10