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.

2x-3y=4

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=2/3x-4/3 or

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

= 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 = (34-4)/3 = 10

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