Bacteria of species A and species B are kept in a single test tube, where they are fed two nutrients. Each day the test tube is supplied with 19,740 units of the first nutrient and 31,990 units of the second nutrient. Each bacterium of species A requires 4 units of the first nutrient and 5 units of the second, and each bacterium of species B requires 1 unit of the first nutrient and 6 units of the second. What populations of each species can coexist in the test tube so that all the nutrients are consumed each day?

To find the populations of species A and species B that can coexist while consuming all the nutrients each day, we need to set up a system of equations based on the nutrient requirements of each bacterium species.

Let's assume the population of species A is represented by 'a' and the population of species B is represented by 'b'.

The total units of the first nutrient consumed each day would be 4a + 1b, and the total units of the second nutrient consumed each day would be 5a + 6b.

Since the test tube is supplied with 19,740 units of the first nutrient and 31,990 units of the second nutrient each day, we can set up the following equations:

4a + 1b = 19,740 (equation 1)
5a + 6b = 31,990 (equation 2)

We can now solve this system of equations to determine the populations of species A and species B.

Multiplying equation 1 by 6 and equation 2 by 1, we can eliminate the 'b' variable:

24a + 6b = 118,440 (equation 3)
5a + 6b = 31,990 (equation 2)

Next, subtract equation 2 from equation 3:

(24a + 6b) - (5a + 6b) = 118,440 - 31,990
19a = 86,450
a = 86,450 / 19
a = 4,550

Now, substitute the value of 'a' back into equation 1 to solve for 'b':

4(4,550) + 1b = 19,740
18,200 + b = 19,740
b = 19,740 - 18,200
b = 1,540

Therefore, the population of species A can be 4,550 and the population of species B can be 1,540 to consume all the nutrients in the test tube each day.