Problem 63 - Algebra and Trigonometry Real Mathematics, Real People

Question

A grocer mixes peanuts that cost \(\$ 2.50\) per pound and walnuts that cost \(\$ 8.00\) per pound to make 100 pounds of a mixture that costs \(\$ 5.25\) per pound. How much of each kind of nut is put into the mixture?

Step-by-Step Solution

Answer

Therefore, the grocer needs to use 50 pounds of peanuts and 50 pounds of walnuts in his mixture to achieve the desired price per pound.

Step 1: Define the Variables

Let's denote the number of pounds of peanuts the grocer uses as \(x\), and the number of pounds of walnuts as \(y\).

Step 2: Write Down the Equations

From the question we can derive two equations. 1) The cost of the mixture is \(\$5.25\) per pound, hence the total cost for 100 pounds is \(\$5.25 * 100\). This must equal the total cost of the peanuts (\(\$2.50 * x\)) plus the total cost of the walnuts (\(\$8.00 * y\). Formulate it like this: \(2.50x + 8y = 5.25 * 100\)2) The total weight of the mixture is 100 pounds, so we know that \(x + y = 100\).

Step 3: Solve the System of Equations

Now we have a system of equations: \(2.50x + 8y = 525\) (equation 1)\(x + y = 100\) (equaton 2)To solve these equations, we can first isolate \(x\) in equation 2 and then substitute in equation 1: From equation 2, we get \(x = 100 - y\). Substituting this in equation 1 gives \[2.50(100 - y) + 8y = 525\] This simplifies to a equation in one variable (y), which can be solved as follows:\(250 - 2.50y + 8y = 525\)Solving for \(y\) we get \(y = 50\) pounds. Substituting \(y = 50\) into equation 2 yields \(x = 50\) pounds.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when plotted on a graph. They form the foundation for solving many algebraic problems, such as finding unknown values. In our exercise, we have two linear equations:

• Equation 1 is derived from the cost requirements: \(2.50x + 8y = 525\), which accounts for the cost components of peanuts and walnuts within the mixture.
• Equation 2 stems from the total weight constraint: \(x + y = 100\), representing the total amount of nuts mixed.

These equations involve the variables \(x\) and \(y\) where \(x\) is the pounds of peanuts and \(y\) is the pounds of walnuts. Each equation provides information about relationships between the variables, allowing us to model and solve the problem effectively.
Linear equations are a core component of many real-life applications because they allow us to express relationships between different quantities in a clear mathematical format. Often, these equations are written in the form \(ax + by = c\), which you can compare to our problem's setup.

Mathematical Modeling

Mathematical modeling involves using mathematical concepts and structures to represent real-world situations. For our problem, we model a real-life context of mixing peanuts and walnuts and determining the amounts required for a specific mixture cost.
To build a mathematical model, we identified key factors: the cost per pound for each type of nut, the total weight of the mixture, and the mixture's cost per pound. These insights led us to create a system of two linear equations. This model simplifies complex real-world conditions into easily solvable mathematical expressions.
By abstracting and framing the problem with equations, mathematical modeling allows us to explore the relationships between variables (here, the quantities of nuts) and to determine precise solutions. This approach can be applied broadly, aiding problem-solving in engineering, economics, and everyday scenarios.

Algebraic Solution

Algebraic solutions involve manipulating algebraic expressions to find the value of unknown variables. In our system of equations, we used substitution, a classic algebraic method, to find the solution.
Starting with Equation 2: \(x + y = 100\), we expressed \(x\) in terms of \(y\) (\(x = 100 - y\)). This set the stage for substitution into Equation 1: \(2.50x + 8y = 525\). With this, we transformed the problem into a single-variable equation:\[2.50(100 - y) + 8y = 525\]
Solving this equation through simplification and basic arithmetic enabled us to find \(y = 50\) and consequently \(x = 50\). This step-by-step process highlighted how algebraic methods can efficiently unlock answers, even in multi-variable scenarios.

• The power of algebra lies in its ability to simplify complex problems.
• It reduces multiple unknowns to single-variable contexts.
• Algebraic solutions are systematic and allow you to check your work easily.

Understanding such algebraic techniques equips students with tools necessary for tackling a wide range of mathematical challenges.