Answer:
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
<u> </u>
<u> P₁ - 3P₂ + 0 = - 12 </u>
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.
