Problem 52 - Elementary Algebra

Question

A man bought 2 pounds of coffee and 1 pound of butter for a total of \(\$ 18.75\). A month later the prices had not changed (this makes it a fictitious problem), and he bought 3 pounds of coffee and 2 pounds of butter for \(\$ 29.50\). Find the price per pound of both the coffee and the butter.

Step-by-Step Solution

Answer

Coffee costs \( \$8 \) per pound, and butter costs \( \$2.75 \) per pound.

Step 1: Setting Up Equations

Since we know the total costs for the items purchased on the two different occasions, let's define two variables: let the price per pound of coffee be \( c \) and the price per pound of butter be \( b \). From the first purchase, we have the equation: \( 2c + 1b = 18.75 \). From the second purchase, we have the equation: \( 3c + 2b = 29.50 \).

Step 2: Solving for One Variable

We will solve for one of the variables by eliminating it. Let's start by solving for \( b \) in the first equation: \( b = 18.75 - 2c \).

Step 3: Substituting Back

Substitute the expression for \( b \) from the first equation into the second equation: \( 3c + 2(18.75 - 2c) = 29.50 \).

Step 4: Simplifying the Equation

Expand and simplify this equation: \( 3c + 37.50 - 4c = 29.50 \). Simplifying further, we get: \( -c + 37.50 = 29.50 \).

Step 5: Solving for Coffee Price

Subtract 37.50 from both sides of the equation: \( -c = 29.50 - 37.50 \). Thus, \( -c = -8 \). Hence, \( c = 8 \). So, the price per pound of coffee is \( \$8 \).

Step 6: Finding the Price of Butter

Now, substitute \( c = 8 \) back into the equation \( b = 18.75 - 2c \): \( b = 18.75 - 16 \), which simplifies to \( b = 2.75 \). So, the price per pound of butter is \( \$2.75 \).

Key Concepts

Systems of Equations

In algebra, a **system of equations** is a collection of two or more equations with the same set of unknown variables. In solving these systems, the goal is to find the values of the variables that satisfy all the equations simultaneously.
This approach is critical when dealing with problems involving multiple relationships and unknowns, like our example problem where we had to figure out the pricing formula for coffee and butter.
Here, we identified:

• The equations: \( 2c + b = 18.75 \) and \( 3c + 2b = 29.50 \) captured the cost relationships of the purchases.
• The variables: \( c \) and \( b \) represented the coffee and butter's price per pound, respectively.

These equations form the basis of our problem-solving strategy. We used them to determine the cost of individual items, knowing the total costs of combined purchases.
Utilizing systems of equations enables us to describe multiple constraints and solve for them by using various methods like substitution or elimination. The process you choose depends on which will simplify the problem into easily solvable parts.

Problem-Solving in Algebra

**Problem-solving in algebra** involves systematic strategies that break down complex problems into simpler parts. In solving the exercise, we adopted the substitution method. This method is highly effective when you can easily express one variable in terms of another.
Steps taken to tackle the problem:

• **Identifying variables and their relationships:** We started by setting variable representations, \( c \) for coffee and \( b \) for butter.
• **Formulating relevant equations:** From the purchase scenarios, we came up with two equations.
• **Simplifying the problem:** We isolated \( b \) from the first equation to substitute into the second equation, which helped us reduce it to an equation with just one variable.

These steps provide a logical flow to tackle tricky algebraic questions. Once the equations are simplified to contain only one variable, solving becomes more straightforward.

Price Per Unit Calculation

Calculating the **price per unit** of items is essential when trying to understand the value of individual components in a combined purchase.
In our example problem:

• We used the derived equations \( 2c + b = 18.75 \) and \( 3c + 2b = 29.50 \) to solve for individual prices.
• After calculating, the price per pound came out to \( \\(8 \) for coffee and \( \\)2.75 \) for butter.

Understanding prices per unit is valuable not only in academics but also in real-world scenarios like budgeting and cost analysis.
This concept teaches us how to efficiently allocate resources and assess product value, considering each item's contribution to the whole cost.