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

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

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.

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