Question

A mixture 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?

Answer 1

Answer:

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.