Problem 12 - Algebra 2

Question

Mykia had 25 quarters and dimes. The total value of all the coins was \(\$ 4.00 .\) How many quarters and dimes did Mykia have?

Step-by-Step Solution

Answer

Mykia had 10 quarters and 15 dimes.

Step 1: Define Variables

Let the number of quarters be \( q \) and the number of dimes be \( d \). We know that there are a total of 25 coins, so we can write the equation \( q + d = 25 \).

Step 2: Translate Total Value Condition

The value of the quarters is \( 0.25q \) dollars and the value of the dimes is \( 0.10d \) dollars. The total value of all the coins is \( 4.00 \) dollars, so we can write the equation \( 0.25q + 0.10d = 4.00 \).

Step 3: Solve Equation System

We now have a system of two equations: \( q + d = 25 \) and \( 0.25q + 0.10d = 4.00 \). First, let's solve the first equation for \( q \): \( q = 25 - d \). Then, substitute this expression into the second equation: \( 0.25(25 - d) + 0.10d = 4.00 \).

Step 4: Simplify and Solve for d

Expanding the substituted equation gives \( 6.25 - 0.25d + 0.10d = 4.00 \). Simplify this to get \( 6.25 - 0.15d = 4.00 \). Subtract 6.25 from both sides to get \( -0.15d = -2.25 \), then divide both sides by -0.15, resulting in \( d = 15 \).

Step 5: Find the Number of Quarters

Using the equation \( q = 25 - d \), substitute \( d = 15 \): \( q = 25 - 15 = 10 \). Thus, there are 10 quarters.

Key Concepts

Variables

In mathematics, variables act as placeholders for numbers that we do not yet know. They allow us to create equations that represent real-world situations. Consider the problem where Mykia has a mix of quarters and dimes. We define two variables:

• Let \( q \) represent the number of quarters.
• Let \( d \) represent the number of dimes.

Using these variables, we translate the word problem into equations so that we can solve for the unknowns. Variables simplify the complex information, helping us systematically find answers.

Linear Equations

Linear equations are mathematical statements that relate two variables using constant coefficients. They form a straight line when graphed. In the context of Mykia's coins:

• The first equation \( q + d = 25 \) shows the total count of coins.
• The second equation \( 0.25q + 0.10d = 4.00 \) represents the total value of coins.

Each equation aligns with a condition given in the problem. With linear equations, we describe a relationship between variables that can be solved with algebraic methods, such as finding the exact number of dimes and quarters.

Problem-Solving Techniques

Problem-solving in algebra often begins by carefully reading the problem and translating the words into mathematical statements. Here are some common techniques to tackle such problems:

• Identify what you are solving for and assign variables accordingly.
• Set up equations based on conditions given.
• Use known mathematical methods like substitution or elimination to solve the equations.

By systematically approaching each problem in this way, you can break down complicated problems into manageable parts, making them easier to solve.

Substitution Method

The substitution method is a powerful algebraic tool for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. For example, with the coin problem:

• Solve \( q + d = 25 \) for \( q \), giving \( q = 25 - d \).
• Substitute \( q = 25 - d \) into the second equation \( 0.25q + 0.10d = 4.00 \).
• Solve for \( d \) and then back-substitute to find \( q \).

This method reduces the number of unknowns and simplifies calculations, leading systematically to the solution. It's especially useful when equations can be easily manipulated to isolate one variable.