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Chapter 4: Problem 39

At a food co-op where co-op members are required to work 18 times per year, almonds cost \(\$ 9.46\) per pound and Brazil nuts cost \(\$ 7.31\) per pound. Find the amount of almonds and Brazil nuts needed to make \(40 \mathrm{lb}\) of a mixture that costs \(\$ 8\) per pound. Round to the nearest tenth.

Short Answer

Expert verified
12.8 lb of almonds and 27.2 lb of Brazil nuts.

Step by step solution

01

Define Variables

Let the amount of almonds needed be denoted as \( x \) pounds and the amount of Brazil nuts as \( y \) pounds.
02

Set Up Equations

We have two equations from the problem statement. The first equation represents the total weight: \[ x + y = 40 \]. The second equation represents the total cost: \[ 9.46x + 7.31y = 8 \times 40 \].
03

Simplify Cost Equation

Simplify the cost equation: \[ 9.46x + 7.31y = 320 \].
04

Solve for One Variable

From the first equation, solve for \( y \): \[ y = 40 - x \].
05

Substitute and Solve

Substitute \( y = 40 - x \) into the cost equation: \[ 9.46x + 7.31(40 - x) = 320 \]. Simplify this to: \[ 9.46x + 292.4 - 7.31x = 320 \]. Combine like terms to get: \[ 2.15x + 292.4 = 320 \].
06

Isolate x

Isolate \( x \) by subtracting 292.4 from both sides: \[ 2.15x = 27.6 \]. Then, divide by 2.15: \[ x = \frac{27.6}{2.15} \approx 12.8 \].
07

Solve for y

Substitute \( x \) back into \( y = 40 - x \): \[ y = 40 - 12.8 = 27.2 \].
08

Round to the Nearest Tenth

The amounts of almonds and Brazil nuts are already to the tenth place: \( x = 12.8 \) and \( y = 27.2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

linear equations
Linear equations are equations that form a straight line when graphed. They generally have the form: \[ ax + by = c \]. Here, the exercise uses linear equations to represent the relationship between the total weight and the total cost of almonds and Brazil nuts. In our problem, the equations are: 1. \( x + y = 40 \) - representing the total weight of the nuts mixture. 2. \( 9.46x + 7.31y = 320 \) - representing the total cost of the nuts mixture. Linear equations are useful for describing real-world situations like budgeting and mixing solutions. Understanding how to set up and solve these equations is fundamental in algebra.
cost equations
Cost equations are used to determine how costs add up in different scenarios. In this problem, we are figuring out the cost of mixing two types of nuts based on their individual costs per pound. The cost equation derived from the problem is: \( 9.46x + 7.31y = 320 \). Here, - \( 9.46 \) represents the cost per pound of almonds, - \( x \) is the amount of almonds, - \( 7.31 \) signifies the cost per pound of Brazil nuts, - \( y \) is the amount of Brazil nuts, - and \( 320 \) is the total cost for a \( 40 \) pound mixture at \( 8 \$ \) per pound. Understanding how to create and solve cost equations can help in financial planning, project budgets, and pricing strategies.
algebraic substitution
Algebraic substitution involves replacing one variable with an expression involving another variable. This helps in solving systems of equations as it simplifies the problem. In our exercise, we used substitution to solve the linear equations. From the first equation \( x + y = 40 \), we solved for \( y \): \( y = 40 - x \). This expression was then substituted into the cost equation \( 9.46x + 7.31y = 320 \) to get: \( 9.46x + 7.31(40 - x) = 320 \). Substituting in algebra reduces the number of variables and makes the system easier to solve.
solving systems of equations
Solving systems of equations involves finding values for variables that satisfy all equations in the system. In this problem, we used both the substitution and combination methods. After substituting \( y = 40 - x \) into the cost equation, we simplified: \( 9.46x + 292.4 - 7.31x = 320 \). Then, by combining like terms, we got: \( 2.15x + 292.4 = 320 \). By isolating \( x \) and solving, we found: \( x = 12.8 \). Finally, substituting back to find \( y \): \( y = 40 - 12.8 = 27.2 \). Understanding how to solve systems of equations is essential for more complex problem-solving in math and various real-world applications, such as mixing ratios, budgeting, and logistics.

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