Answer:
We require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day
Step-by-step explanation:
Let n₁ be the population of A required and n₂ be the population of B required.
Now we require 2 units of the first nutrient for species A and one unit of the first nutrient for species B. The total nutrients required by species A is 2n₁ and that by species B is 1n₂ = n₂. So, the total nutrients required by both species A and B is 2n₁ + n₂. Since this equals the quantity of the first nutrient which is 10,560, then 2n₁ + n₂ = 10,560 (1)
Now we require 5 units of the second nutrient for species A and 6 units of the second nutrient for species B. The total nutrients required by species A is 5n₁ and that by species B is 6n₂. So, the total nutrients required by both species A and B is 5n₁ + 6n₂. Since this equals the quantity of the first nutrient which is 31,510, then 5n₁ + 6n₂ = 31,510 (2).
So, we have two simultaneous equations which we would solve to find the populations of A and B which satisfy both equations.
2n₁ + n₂ = 10,560 (1)
5n₁ + 6n₂ = 31,510 (2)
From (1) n₂ = 10,560 - 2n₁ (3)
Substituting equation (3) into (2), we have
5n₁ + 6(10,560 - 2n₁) = 31,510
expanding the brackets, we have
5n₁ + 63,360 - 12n₁ = 31,510
collecting like terms, we have
5n₁ - 12n₁ = 31,510 - 63,360
simplifying, we have
- 7n₁ = -31,850
dividing both sides by -7, we have
n₁ = -31,850/-7
n₁ = 4,550
Substituting n₁ = 4,550 into (3), we have
n₂ = 10,560 - 2(4,550)
n₂ = 10,560 - 9,100
n₂ = 1,460
So, we require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day
