To boost the growth of a crop, a farmer decided to use different combinations of three fertilizers, A, B, and C. The first combination costs $384 and consists of 6 liters of fertilizer A, 5 liters of fertilizer B, and 3 liters of fertilizer C. The second combination consists of 10

liters of A, 2 liters of B, and 6 liters of C, and it costs $516. The last combination consists of 4 liters of A, 8 liters of B, and 2 liters of C,

with a cost of $368. Let x be the price of fertilizer A, y be the price of fertilizer B, and z be the price of fertilizer C. Use matrices to

determine the cost of each type of fertilizer.

X=

y =

ZE

The answer is explained below

Step-by-step explanation:

Let x be the price of fertilizer A, y be the price of fertilizer B, and z be the price of fertilizer C

The first combination  costs $384 and consists of 6 liters of fertilizer A, 5 liters of fertilizer B, and 3 liters of fertilizer C. The first combination is given by the equation:

6X + 5Y + 3Z = 384

The second combination consists of 10 liters of A, 2 liters of B, and 6 liters of C, and it costs $516. The second combination is given by the equation:

10X + 2Y + 6Z = 516

The last combination consists of 4 liters of A, 8 liters of B, and 2 liters of C,  with a cost of $368. The last combination is given by the equation:

4X + 8Y + 2Z = 368

In Matrix form it can be represented as:

Therefore:

X = $30, Y = $24, Z = $28

The price of fertilizer A = $30 per liter, The price of fertilizer B = $24 per liter and The price of fertilizer c = $28 per liter

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068