Question

We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day, an individual of bacteria 1 can each 3 units of food A, 2 units of food B, and 1 unit of food C. An individual of bacteria 2 can each 1 unit of food A, 4 units of food B, and 1 unit of food C. An individual of bacteria 3 can eat 2 units of food A and food B but does not eat food C. In one day, the bacteria eat a total of 58 units of food A, 70 units of food B, and 20 units of food C. How many of each bacteria are there

Answer 1

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Step-by-step explanation:

From the given information:

For food source A; we have:

3P₁ + P₂ + 2P₃ = 58 units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70 units of food B ---- (2)

For food source C; we have:

P₁ + P₂ = 20 units of food C ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

3P₁ + P₂ + 2P₃ = 58

-

2P₁ + 4P₂ + 2P₃ = 70

P₁ - 3P₂ + 0 = - 12

P₁ = -12 + 3P₂ ---- (4)

Replace, the value of P₁ in (4) into equation (3)

P₁ + P₂ = 20

-12 + 3P₂ + P₂ = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂ = 20

P₁ + 8 = 20

P₁ = 20 - 8

P₁ = 12

To find P₃; replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ = 7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.