Problem 59 - Algebra and Trigonometry

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 S2. \(75,\) how many dimes and how many quarters does he have?

Step-by-Step Solution

Answer

He has 5 dimes and 9 quarters.

Step 1: Set up the Variables

Let \( d \) be the number of dimes and \( q \) be the number of quarters. We know that the total number of coins is 14. Therefore, we have the equation \( d + q = 14 \).

Step 2: Establish the Value Equation

Dimes are worth 10 cents each and quarters are worth 25 cents each. The total value of the coins is $2.75, which is 275 cents. So the equation for the value is \( 10d + 25q = 275 \).

Step 3: Simplify the Value Equation

To make calculations easier, simplify the value equation by dividing everything by 5, giving \( 2d + 5q = 55 \).

Step 4: Solve the System of Equations

With the equations \( d + q = 14 \) and \( 2d + 5q = 55 \), use substitution or elimination to solve for \( d \) and \( q \). Let's use substitution: from \( d + q = 14 \), express \( d \) as \( d = 14 - q \).

Step 5: Substitute and Solve for One Variable

Substitute \( d = 14 - q \) into the simplified value equation: \( 2(14 - q) + 5q = 55 \), which simplifies to \( 28 - 2q + 5q = 55 \). Simplify further to \( 3q = 27 \), giving \( q = 9 \).

Step 6: Solve for the Other Variable

Substitute \( q = 9 \) back into the equation \( d + q = 14 \): \( d + 9 = 14 \). Solving for \( d \) gives \( d = 5 \).

Step 7: Verify the Solution

Check that 5 dimes and 9 quarters give a total of 14 coins and a value of 275 cents. Calculate: \( 5 \, \text{dimes} = 50 \, \text{cents} \) and \( 9 \, \text{quarters} = 225 \, \text{cents} \), together \( 50 + 225 = 275 \) cents. This confirms our solution.

Key Concepts

Coin Problems

Coin problems are a popular type of word problem in algebra that involve dealing with different denominations of coins, their counts, and their total values.
Typically, you are given a total number of coins, a total monetary value, and asked to find how many of each type of coin there are.
These problems are excellent for practicing setting up and solving systems of equations.When faced with a coin problem:

• Assign variables to each type of coin (e.g., let \( d \) represent the number of dimes and \( q \) the number of quarters).
• Set up one equation to represent the total number of coins.
• Create a second equation for the total value in cents or dollars.

Using these strategies, you can solve for the number of each type of coin.

Substitution Method

The substitution method is a fundamental technique for solving systems of linear equations.
It involves solving one of the equations for one variable and then substituting that expression into the other equation.
This allows you to reduce the number of variables and solve for one at a time.Here's how the substitution method works:

• First, solve one of the equations for one variable. For example, from the equation \( d + q = 14 \), you can express \( d \) as \( d = 14 - q \).
• Next, substitute this expression into the second equation. Replace the \( d \) in \( 2d + 5q = 55 \) with \( 14 - q \).
• Simplify the equation you've substituted into, and solve for the remaining variable.
• Finally, substitute back to find the other variable.

Substitution is particularly useful when one variable is easily isolated.

Elimination Method

The elimination method, also known as the addition method, is another effective strategy for solving systems of linear equations.
This method involves adding or subtracting the equations to eliminate one of the variables, allowing you to solve for the other. To use the elimination method, follow these steps:

• Arrange the equations so that corresponding variables and their coefficients are aligned.
• If necessary, multiply one or both equations by constants to obtain coefficients that will cancel when added or subtracted.
• Add or subtract the equations to eliminate one of the variables.
• Solve the resulting equation for the remaining variable.
• Substitute the value obtained back into one of the originals to find the second variable.

This method is often easier when both equations are already in a similar form.

Linear Equations

Linear equations are algebraic equations in which each term is either a constant or the product of a constant and a single variable.
These equations are called 'linear' because their graph is a straight line.
In the problem of coin counting, both equations formed are linear:

• The equation for the total number of coins: \( d + q = 14 \).
• The equation for the total value: \( 10d + 25q = 275 \), which was simplified to \( 2d + 5q = 55 \).

Solving systems of linear equations requires finding the values of the variables that satisfy both equations simultaneously.
These intersections represent solutions to the problem, meaning the particular counts of each type of coin.
Understanding how to manipulate and solve linear equations is essential to tackling these kinds of algebraic challenges.