A bag contains balls that are red, green, or blue. One third of the balls are red, and 2/7 are blue. The number of green balls is eight fewer than the twice the number of blue balls. How many green balls are in the bag?

Since these are balls, the answers have to be whole numbers. The equations representing the different colors contain fractions and have denominators 3 and 7. So the total number of balls, represented by x, needs to be a factor of 3 and 7. So I did a trial and error thing here:

First possible common denominator = 21, which is 3x7

We test that there are 21 balls in the bag. Substituting 21 into the equations for red, blue and green we get:

red = 1/3(21) = 7

blue = 2/7(21) = 6

green = -8+2(2/7)(21) = 4

But if you add up the colors: 7+6+4 = 14 which is < 21

So we test another common denominator

Next possible common denominator = 42, which is 3x7x2

So we test that there are 42 balls in the bag. Substituting 42 into the equations for red, blue and green we get:

red = 1/3(42) = 14

blue = 2/7(42) = 12

green = -8+2(2/7)(42) = 16

If you add up the colors: 14+12+16 = 42