Problem 248 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve. Lucinda had a pocketful of dimes and quarters with a value of \(\$ \$ 6.20\). The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have?

Step-by-Step Solution

Answer

42 dimes, 8 quarters.

Step 1: Define Variables

Let the number of dimes be represented by \(d\) and the number of quarters be represented by \(q\).

Step 2: Formulate Equations

Based on the problem, we can write two equations: 1. The total value equation: \(0.10d + 0.25q = 6.20\) 2. The number relationship: \(d = 3q + 18\).

Step 3: Substitute the Second Equation into the First

Substitute \(d\) from the second equation into the first equation: \(0.10(3q + 18) + 0.25q = 6.20\).

Step 4: Simplify the Equation

Expand and simplify the equation: \[ 0.10 \times 3q + 0.10 \times 18 + 0.25q = 6.20 \] \[ 0.30q + 1.80 + 0.25q = 6.20 \] Combine like terms: \[ 0.55q + 1.80 = 6.20 \].

Step 5: Solve for the Number of Quarters

Isolate \(q\): \[ 0.55q = 6.20 - 1.80 \] \[ 0.55q = 4.40 \] \[ q = \frac{4.40}{0.55} \] \[ q = 8 \].

Step 6: Solve for the Number of Dimes

Use the value of \(q\) to find \(d\): \[ d = 3q + 18 \] \[ d = 3(8) + 18 \] \[ d = 24 + 18 \] \[ d = 42 \].

Step 7: Final Answer

Lucinda has 42 dimes and 8 quarters.

Key Concepts

Variable Elimination

Variable elimination is a key method used in solving systems of equations. It allows us to eliminate one variable so we can solve for the other. In this exercise, we started by defining our variables: the number of dimes as \(d\) and the number of quarters as \(q\). From the given information, we set up two equations: the total value equation (\(0.10d + 0.25q = 6.20\)) and the number relationship equation (\(d = 3q + 18\)).

By substituting the second equation into the first, we eliminate \(d\) and are left with an equation in terms of \(q\). This reduction in the number of variables makes it easier to solve for \(q\). After finding the value of \(q\), we can then substitute it back into the original relationship to find \(d\).

This method is efficient because it reduces a problem involving two variables down to a problem involving only one. Once the value of \(q\) is found, the value of \(d\) is straightforward to determine.

Equation Formulation

Formulating the correct equations from a word problem is an essential step. It involves translating the problem's text into mathematical expressions. In our coin problem, we were given two pieces of information:

• The total value of the dimes and quarters is $6.20.
• The number of dimes is eighteen more than three times the number of quarters.

From these statements, we formulated two equations. The total value equation, \(0.10d + 0.25q = 6.20\), converts the total monetary value into an equation using the value of each type of coin. The second equation, \(d = 3q + 18\), encapsulates the relationship between the number of dimes and quarters.

Consistently translating the problem's information into equations allows us to use mathematical methods to find a solution. Always carefully read the problem and determine what each variable represents and how they relate to each other.

Coin Problem

Coin problems are a common type of algebraic word problem where you're asked to figure out the quantities of different coins based on their values and relationships. Usually, you will be given:

• The total value of the coins.
• A relationship between the quantities of the different coins.

In this problem, we had to find the number of dimes and quarters. By defining the variables for each type of coin and setting up equations based on their values and given relationships, we could use algebra to solve for the quantities.

Key steps in solving a coin problem:

• Define your variables.
• Formulate equations based on the problem's conditions.
• Use substitution or elimination to solve the system of equations.
• Verify the solution by plugging the values back into the original conditions.

Practicing coin problems will help you become more comfortable with setting up and solving systems of equations, a skill useful beyond just coin-related questions.