Problem 58 - Intermediate Algebra

Question

A man bought 2 pounds of coffee and 1 pound of butter for a total of $$\$9.25.$$ 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 $$\$ 15.50$. Find the price per pound of both the coffee and the butter.

Step-by-Step Solution

Answer

The price per pound of coffee is $3.00, and the price per pound of butter is $3.25.

Step 1: Define Variables

Let's define some variables to solve the problem. Let \( x \) be the price per pound of coffee and \( y \) be the price per pound of butter.

Step 2: Set Up Equations

Based on the information given, we can set up the following system of equations:1. For the first purchase: \( 2x + y = 9.25 \)2. For the second purchase: \( 3x + 2y = 15.50 \)

Step 3: Solve the First Equation for y

From the first equation, solve for \( y \):\[ y = 9.25 - 2x \]

Step 4: Substitute y in the Second Equation

Substitute \( y = 9.25 - 2x \) into the second equation:\[ 3x + 2(9.25 - 2x) = 15.50 \]

Step 5: Simplify and Solve for x

Expand the substituted equation:\[ 3x + 18.50 - 4x = 15.50 \]Combine like terms:\[ -x + 18.50 = 15.50 \]Subtract 18.50 from both sides:\[ -x = -3.00 \]Multiply throughout by -1 to solve for \( x \):\[ x = 3.00 \]

Step 6: Substitute x in the Expression for y

Substitute \( x = 3.00 \) back into the expression for \( y \):\[ y = 9.25 - 2(3.00) \]\[ y = 9.25 - 6.00 \]\[ y = 3.25 \]

Key Concepts

Linear Equations

Linear equations are a fundamental concept in algebra. These are equations in which each term is either a constant or the product of a constant and a single variable.
For example, the linear equation from our problem is structured as \(2x + y = 9.25\). Here, \(x\) and \(y\) are unknowns — often referred to as variables.

Linear equations can have one or multiple variables; however, they are linear because they form a straight line if graphed on a coordinate plane. They have no variables raised to a power greater than one.
This property makes them straightforward to solve using methods such as substitution, elimination, or graphical solutions due to their predictable pattern.

Algebraic Problem Solving

Algebraic problem solving involves finding the values of variables that satisfy given mathematical equations or problems.
It requires logical thinking and the use of systematic approaches.

In our exercise, we employ a step-by-step approach:

• First, define unknown variables. This lets us translate the word problem into mathematical equations.
• Second, set up a system of equations based on the information given.
• Third, solve the equations using algebraic techniques such as substitution or elimination.

By breaking down the problem into smaller, manageable parts, we simplify the problem-solving process. This effective method is widely used in mathematics to tackle various complex equations.

Variable Definition

Defining variables is an essential part of solving mathematical problems, especially in algebra.

By assigning variables, we transform real-world situations into an algebraic form that can be analyzed and solved.
In our case, we defined:

• \(x\) as the price per pound of coffee.
• \(y\) as the price per pound of butter.

These definitions allow us to create equations that represent the relationships and conditions given by the problem. Without clearly defined variables, it would be much harder to systematically approach and solve algebraic equations.
Effective variable definition is crucial for a clear analysis and solution of any algebraic problem.