Problem 61 - Precalculus: Mathematics for Calculus

Question

Value of Coins A man has 14 coins in his pocket, all of which are dimes and quarters. If the total value of his change is \(\$ 2.75,\) how many dimes and how many quarters does he have?

Step-by-Step Solution

Answer

The man has 5 dimes and 9 quarters.

Step 1: Define the Variables

Let \( x \) represent the number of dimes and \( y \) represent the number of quarters.

Step 2: Set Up the Equations

We have two equations based on the problem statement: \( x + y = 14 \) and \( 0.10x + 0.25y = 2.75 \), where the first equation represents the total number of coins and the second equation represents the total value of the coins in dollars.

Step 3: Solve for One Variable

From the equation \( x + y = 14 \), we solve for \( y \): \( y = 14 - x \).

Step 4: Substitute and Solve

Substitute \( y = 14 - x \) into the second equation: \( 0.10x + 0.25(14 - x) = 2.75 \). This simplifies to \( 0.10x + 3.5 - 0.25x = 2.75 \). Simplifying further gives \( -0.15x + 3.5 = 2.75 \).

Step 5: Isolate the Variable

Subtract 3.5 from both sides: \( -0.15x = -0.75 \).

Step 6: Solve for \( x \)

Divide both sides by -0.15 to solve for \( x \): \( x = 5 \).

Step 7: Find \( y \)

Using \( y = 14 - x \), substitute \( x = 5 \) to find \( y = 14 - 5 = 9 \).

Step 8: Verify the Solution

Check if the values satisfy both original equations. For the total count, \( 5 + 9 = 14 \), and for the total value, \( 0.10(5) + 0.25(9) = 0.50 + 2.25 = 2.75 \). Both conditions are satisfied.

Key Concepts

Linear Equations

Linear equations are fundamental mathematical expressions characterized by variables raised to the power of one. They are basically straight-line equations when graphed on a coordinate plane. In coin problems like the one in this exercise, linear equations help us represent relationships with variables.
For example, we define variables for unknown quantities, such as the number of dimes (\( x \)) and quarters (\( y\)). Each variable is involved in establishing the conditions given by the problem. In linear equations, we use coefficients and constants to represent the value of these unknowns in a standard form which can be something like \( ax + by = c \).
In our exercise:

• We first create an equation representing the total number of coins: \( x + y = 14 \)
• Then, we set up another equation for total value: \( 0.10x + 0.25y = 2.75 \)

These equations are solved simultaneously to find the values of \( x \) and \( y \), unwrapping the relationship between the number of each type of coin. Understanding linear equations enables us to solve various problems by expressing complex relations plainly.

Problem-Solving

Problem-solving is the art of identifying a challenge and developing a method to address it successfully. It involves phases such as understanding the problem, devising a plan, carrying out the plan, and evaluating the solution.
In our exercise on coins, the problem is clearly defined: find how many dimes and quarters there are. The information provided includes the total number of coins and their total value. The first step in problem-solving is to comprehend these existing conditions and translate them into mathematical terms using variables and equations.
Next is planning. We formulate a strategy, which involves setting up equations from the given data and choosing appropriate methods to solve them. Here, substitution was the chosen method, which allowed us to isolate and determine the value of \( x \) and then \( y \). Each solution step builds upon the previous one.

• Solving for one variable and substituting into another equation simplifies the problem.
• Checking the solution ensures that both ‘number of coins’ and ‘total value’ conditions are being met.

This structured approach organizes thoughts and calculations, making it easier to arrive at the correct solution.

Coin Value Problems

Coin value problems are intriguing exercises commonly encountered in elementary algebra. They require understanding both numerical and value-based distinctly as total quantities and their respective values create two different equations. These problems exemplify the application of linear systems to everyday situations.
In coin value problems, such as the one involving dimes and quarters, specific points must be addressed:

• Define the coins and their respective values, giving clear purpose to variables \( x \) and \( y \) representing the number of coins.
• Set up two equations: one for the total coin count and another for the total monetary value. The latter combines the value contributions of each type of coin. In this instance, a dime is valued at 10 cents, written as \( 0.10 \), and a quarter at 25 cents, \( 0.25 \).

Coin problems aid in honing one's skills in forming and solving linear systems. They not only boost arithmetic proficiency but also crucial analytical thinking, relevant to tackling real-world monetary issues. Grasping how these problems work can significantly enhance one's ability to address similar challenges efficiently.