Problem 21 - Elementary and Intermediate Algebra

Question

A collection of 42 coins consists of dimes and nickels. The total value is \(\$ 3.00 .\) How many dimes and how many nickels are there?

Step-by-Step Solution

Answer

There are 18 dimes and 24 nickels.

Step 1 - Define Variables

Let the number of dimes be denoted by variable: x, and the number of nickels be denoted by y.

Step 2 - Set Up Equations

We have two pieces of information: the total number of coins is 42, and the total value is $3.00. This gives us two equations: 1) x + y = 42 2) 0.10x + 0.05y = 3.00.

Step 3 - Simplify the Value Equation

To simplify the value equation, multiply all terms by 100 to eliminate the decimals: 10x + 5y = 300.

Step 4 - Solve the System of Equations

First, solve the first equation for y: y = 42 - x. Next, substitute y in the second equation: 10x + 5(42 - x) = 300.

Step 5 - Combine Like Terms

Distribute and simplify the equation: 10x + 210 - 5x = 300 5x + 210 = 300.

Step 6 - Isolate Variable x

Subtract 210 from both sides: 5x = 90 Divide by 5: x = 18.

Step 7 - Find y Using x

Substitute x = 18 back into the first equation: y = 42 - 18 y = 24.

Key Concepts

linear equations

Linear equations are equations of the first order. This means the variables are not raised to any power higher than one. They appear in the form: ax + b = 0, where a and b are constants, and x is the variable.
In our coin problem, we formed two linear equations based on the known quantities: the number of coins and their total value:

• The first equation, representing the total number of coins, is: x + y = 42
• The second equation, representing the total value of the coins, is: 0.10x + 0.05y = 3.00

Understanding how to set up and manipulate linear equations is crucial for solving many algebra problems.

systems of equations

A system of equations is a set of two or more equations with the same variables. The solution to the system is the set of variable values that simultaneously satisfy all included equations. In the coin problem, we have two equations:

• x + y = 42
• 0.10x + 0.05y = 3.00

To solve the system of equations, we must find values of x and y that make both equations true. This is typically done using methods such as substitution or elimination.

substitution method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. Here's how it works in our coin problem:
1. Solve the first equation for y: We get y = 42 - x.
2. Substitute this expression into the second equation: We replace y with 42 - x in 0.10x + 0.05(42 - x) = 3.00.
This method simplifies a system of equations by reducing it to one equation with a single variable, making it easier to solve.

coin problems

Coin problems are a type of word problem in algebra that involve determining the number of coins of different denominations based on given totals. Here’s the step-by-step approach we used in our problem:

• Define variables for each coin type.
• Form equations based on the total count and value.
• Use methods like substitution to solve the equations.

Practice with coin problems reinforces the understanding and application of linear equations and systems of equations.

elementary algebra

Elementary algebra is the foundation of all higher-level math. It deals with the basics of manipulating equations and expressions involving variables and constants. Key concepts in elementary algebra include:

• Variable definitions and operations.
• Setting up and solving equations.
• Understanding and applying properties of operations.

In our coin problem:

• Defined x and y for dimes and nickels.
• Set up two equations from the problem statement.
• Solved the system to find how many dimes and nickels are there.

If you grasp these fundamental concepts, you'll have a strong base for tackling more complex math problems in the future.