Post a New Question


posted by .

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 -

    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 -

    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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question