SOLUTION: 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 19,820 units of the first nutrient

Let x be the number of species A  and y be the number of species B.


The most difficult part of the solution is to setup the equations.


From the condition, you have these two equation


    4x + 1y = 19820  units of the first nutrient     (1)

    5x + 6y = 32090  units of the second nutrient    (2)


These equations -- this system of equations -- is your setup.


At this point, the setup is just completed.


Now you should solve it.

You can use different methods: Substitution, Elimination, using determinants (the Cramer's rule).


For example, you may express  y = 19820-4x from the first equation and then substitute it into the second


    5x + 6*(19820-4x) = 32090.


It gives you  x =  = 4570.

Then  y = 19820 - 4*4570 = 1540.


ANSWER.  4570 species of the type A  and  1540 species of the type B.