Problem 34 - Algebra A Combined Function

Question

Macadamia nuts cost an astounding \(\$ 16.50\) per pound, but research by an independent firm says that mixed nuts sell better if macadamias are included. The standard mix costs \(\$ 9.25\) per pound. Find how many pounds of macadamias and how many pounds of the standard mix should be combined to produce 40 pounds that will cost \(\$ 10\) per pound. Find the amounts to the nearest tenth of a pound.

Step-by-Step Solution

Answer

4.1 pounds of macadamias and 35.9 pounds of standard mix.

Step 1: Define the Variables

Let \( x \) be the number of pounds of macadamia nuts, and \( y \) be the number of pounds of the standard mix. We are tasked with finding \( x \) and \( y \) such that the totals are 40 pounds, and the price is \$10 per pound.

Step 2: Set Up the Equations

We need two equations based on the problem description. The weight equation is \( x + y = 40 \). The cost equation based on the total price is \( 16.50x + 9.25y = 10 \times 40 \).

Step 3: Substitute and Simplify

Substitute the value of \( y \) from the weight equation, \( y = 40 - x \), into the cost equation: \[ 16.50x + 9.25(40 - x) = 400 \]Expand and simplify the equation.

Step 4: Solve for \( x \)

Simplify the cost equation:\[ 16.50x + 370 - 9.25x = 400 \]Combine like terms:\[ 7.25x = 30 \]Solve for \( x \):\[ x = \frac{30}{7.25} \approx 4.14 \]

Step 5: Solve for \( y \)

Substitute \( x \approx 4.14 \) back into the equation for \( y \):\[ y = 40 - 4.14 \approx 35.86 \]

Step 6: Verify the Solution

Check that the values satisfy the cost equation:\[ 16.50(4.14) + 9.25(35.86) = 68.31 + 331.19 \approx 399.50 \]The approximate calculation shows the result is close to \$400 due to rounding. Thus, the solution is validated.

Key Concepts

linear equations

Understanding linear equations is the first step in solving many algebra word problems like this one. A linear equation is an algebraic statement where each term is either a constant or the product of a constant and a single variable. They form a straight line when graphed and are the simplest form of equations to manage.
For example, in the given problem, we have two basic linear equations derived from the conditions:

• The first equation is based on weight: \( x + y = 40 \), which tells us the total number of pounds of nuts required.
• The second equation is based on cost: \( 16.50x + 9.25y = 400 \), which tells us about the pricing constraint.

By solving these linear equations, we can identify the right balance of macadamia nuts (\( x \)) and standard mix nuts (\( y \)) to meet the desired conditions.
The magic lies in using these linear equations to create a system that models the problem accurately.

system of equations

When we need to solve problems with two variables that interact with each other under different conditions, we use a system of equations. A system of equations is simply two or more equations that share the same set of variables. Solving these systems helps us find values for the variables that satisfy all the equations simultaneously.
In our problem, we have two main equations:

• The weight constraint: \( x + y = 40 \)
• The cost constraint: \( 16.50x + 9.25y = 400 \)

Together, these equations form a system that helps us determine the exact amount of each type of nut needed. By understanding each equation’s role in the system, we can use methods like substitution or elimination to find the solution.
Using a system of equations provides a structured way to approach complex problems, making it easier to identify the relationships between different elements of a real-world scenario.

substitution method

The substitution method is a common technique for solving systems of linear equations. In this method, you solve one of the equations for one variable and then substitute that expression into the other equation.
This approach initially simplifies the equations, making it easier to solve for the variables. In our nut problem, we first expressed one variable in terms of another using the first equation:

• \( y = 40 - x \)

Then, we substituted \( y \) from this equation into the second cost equation:

• \( 16.50x + 9.25(40 - x) = 400 \)

This substitution transformed our system into a single equation with only one variable, \( x \).
Solving this equation gave us the value of \( x \). After finding \( x \), we substituted back into the first equation to find \( y \).
The substitution method is helpful because it allows you to reduce the number of variables, making complex problems more straightforward to unravel.