Amixture of 5 pounds of fertilizer a, 13 pounds of fertilizer b, and 4 pounds of fertilizer c provides the optimal nutrients for a plant. commercial brand x contains equal parts of fertilizer b and fertilizer c. commercial brand y contains one part of fertilizer a and two parts of fertilizer b. commercial brand z contains two parts of fertilizer a, five parts of fertilizer b, and two parts of fertilizer c. how much of each fertilizer brand is needed to obtain the desired mixture?

2 pounds of brand X, 3 pounds of brand Y and 1 pound of Brand Z should be added to get the desired mixture.

Step-by-step explanation:

Mixture that gives the optimal nutrients for a plant is

Fertilizer A = 5 pounds

Fertilizer B = 13 pounds

Fertilizer C = 4 pounds

Commercial brand X contains equal part of B and C,

X = B + C

Commercial brand Y contains one part of fertilizer A and two parts of fertilizer B.

Y = A + 2B

Similarly, commercial Z contains two parts of fertilizer A, five parts of fertilizer B and two parts of fertilizer C.

Z = 2A + 5B + 2C

Let x, y and z pounds of brands X, Y and Z are mixed to get the desired mixture.

Therefore, the mixture we get

x(B + C) + y(A + 2B) + z(2A + 5B + 2C)

A(y + 2z) + B(x + 2y + 5z) + C(x + 2z) (a)

Now we know the desired combination will contain A = 5 pounds, B = 13 pounds and C = 4 pounds

(5A + 13B + 4C) (b)

By comparing expression (a) and (b)

y + 2z = 5 (1)

x + 2y + 5z = 13 (2)

x + 2z = 4 (3)

Subtract equation 1 from 3

x - y = -1 (4)

equation (2)×2 - equation (1)×5

2(x + 2y + 5z) - 5(y + 2z) = 26 - 25

2x + 4y + 10z - 5y - 10z = 1

2x - y = 1 (5)

Now subtract equation (4) from equation (5)

2x - y - x + y = 1 + 1

x = 2

form equation 4

2 - y = -1

y = 2 + 1

y = 3

From equation (1)

3 + 2z = 5

2z = 2

z = 1

Therefore, 2 pounds of brand X, 3 pounds of brand Y and 1 pound of Brand Z should be added to get the desired mixture.

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