Bacteria of species A and species B are kept in a single environment, where they are fed two nutrients. Each day the environment is supplied with 19,560 units of the first nutrient and 31,670 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 environment so that all the nutrients are consumed each day?
you want
4A+1B = 19560
5A+6B = 31670
Now just solve as usual.