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

it is contrapositive

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