Problem 53 - Algebra 1: Concepts and Skills

Question

You have 12 coins worth \(\$ 1.95 .\) If you only have dimes and quarters, how many of each do you have?

Step-by-Step Solution

Answer

9 dimes and 3 quarters

Step 1: Define the problem using system of equations

From the problem we extract two equations. The total number of coins gives a linear equation: x + y = 12. The total value of the coins gives us another linear equation: 0.1x + 0.25y = 1.95

Step 2: Rearranging of one equation to solve for one variable

Let's rearrange the first equation to solve for y. Doing this leads to: y = 12 - x

Step 3: Substituting into second equation

Substitute y in the second equation with the right side of the rearranged first equation. Thus, the second equation becomes: 0.1x + 0.25(12 - x) = 1.95

Step 4: Solving for 'x'

Solve the equation from step 3 to find x. That results in: x = 9. Therefore the number of dimes is 9.

Step 5: Finding 'y'

Substitute x = 9 into equation from step 2. Following this process, you get: y= 12 - 9 = 3. Therefore, the number of quarters is 3.

Step 6: Check your solution

Verify that these numbers make sense in the context of the original problem and make the original equations true. Indeed, 9 dimes plus 3 quarters does equal 12 coins, and the total value of all these coins is indeed $1.95, validating the results.

Key Concepts

Linear Equations

A linear equation is an equation that forms a straight line when it is graphed. In simple terms, it is an equation where each variable is raised to the power of one. This type of equation is usually represented in the form \(ax + by = c\). In the context of our coin problem, we have two linear equations:

• \(x + y = 12\)
• \(0.1x + 0.25y = 1.95\)

The first equation tells us how many coins we have in total. The second equation tells us what their total value is. Linear equations are easy to manipulate, which makes them ideal for solving problems involving multiple variables, like the number of dimes and quarters we have in this problem.

Coin Problems

Coin problems are a classic type of word problem in algebra where you determine the quantity or value of coins. This type of problem usually provides information about the number of coins and their worth in the context of a system of equations.
A common approach to solving coin problems involves:

• Identifying the different types of coins involved—in our case, dimes and quarters.
• Understanding the values of these coins: dimes are worth \(0.10\) dollars, and quarters are worth \(0.25\) dollars.
• Setting up equations to represent the total number and total value of the coins.

In our specific problem, we used these ideas to create a system of equations that helps us find the counts of each type of coin that make up the total number of coins and their total value.

Substitution Method

The substitution method is a technique for solving a system of equations. This approach involves solving one of the equations for one variable and then substituting that expression into the other equation. Here’s a simple breakdown of how it worked in the coin problem:

• First, we rearranged the first equation \(x + y = 12\) and solved it for \(y\): \(y = 12 - x\).
• Then, we substituted this expression for \(y\) in the second equation, transforming \(0.1x + 0.25y = 1.95\) into \(0.1x + 0.25(12 - x) = 1.95\).

This substitution changes the problem to a single-variable equation, which can be easily solved to find \(x\). Once \(x\) is determined, this value is substituted back to find \(y\).
The substitution method is especially useful in systems where one of the equations is easily rearranged, simplifying the process significantly.