Question

type the correct answer in each box.
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 =
z =

Explanation:

Step1: Set up linear equations

From the problem, we derive:

$$\begin{cases} 6x + 5y + 3z = 384 \\ 10x + 2y + 6z = 516 \\ 4x + 8y + 2z = 368 \end{cases}$$

Simplify the third equation by dividing by 2:
$$2x + 4y + z = 184 \implies z = 184 - 2x - 4y$$

Step2: Substitute z into first equation

Substitute $z = 184 - 2x - 4y$ into $6x + 5y + 3z = 384$:

$$\begin{align*} 6x + 5y + 3(184 - 2x - 4y) &= 384 \\ 6x + 5y + 552 - 6x - 12y &= 384 \\ -7y &= 384 - 552 \\ -7y &= -168 \\ y &= 24 \end{align*}$$

Step3: Substitute z and y into second equation

Substitute $z = 184 - 2x - 4y$ and $y=24$ into $10x + 2y + 6z = 516$:

$$\begin{align*} 10x + 2(24) + 6(184 - 2x - 4(24)) &= 516 \\ 10x + 48 + 6(184 - 2x - 96) &= 516 \\ 10x + 48 + 6(88 - 2x) &= 516 \\ 10x + 48 + 528 - 12x &= 516 \\ -2x &= 516 - 576 \\ -2x &= -60 \\ x &= 30 \end{align*}$$

Step4: Calculate z using x and y

Substitute $x=30$, $y=24$ into $z = 184 - 2x - 4y$:

$$\begin{align*} z &= 184 - 2(30) - 4(24) \\ z &= 184 - 60 - 96 \\ z &= 28 \end{align*}$$

Answer:

$x = 30$
$y = 24$
$z = 28$