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 19,880 units of the first nutrient and 32,070 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?

Population A has 4590 and population B has 1520 bacteria

Explanation:

Let's mark bacteria:

a - bacteria from population A

b - bacteria from population B

Every bacteria a needs 4 units of the first nutrient and every bacteria b needs 1 unit of the same nutrient, so we can write it down as:

4*a + 1*b = 19880

Let's do the same for the second nutrient. Every bacteria a needs 5 units of the second nutrient and every bacteria b needs 6 units of the second nutrient:

5*a + 6*b = 32070

Now there are two equations and two unknowns. From the first equation, we can separate b:

b = 19880 - 4a

Let's place this b to the second equation:

5a + 6 * (19880 - 4a) = 32070

5a + 119280 - 24a = 32070

119280 - 32070 = 24a - 5a

19a = 87210

a = 4590

This can be brought back to: b = 19880 - 4a

b = 1520