Problem 247 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve. Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value \(\$ 13.10\). The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?

Step-by-Step Solution

Answer

Peter has 11 dimes and 48 quarters.

Step 1 - Define the Variables

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

Step 2 - Set Up the Equations

Translate the problem into equations:1. The value equation: \( 0.10d + 0.25q = 13.10 \)2. The relationship between quarters and dimes: \( q = 3d + 15 \)

Step 3 - Substitute the Second Equation into the First

Substitute \( q = 3d + 15 \) into \( 0.10d + 0.25q = 13.10 \) to get the equation in one variable.\( 0.10d + 0.25(3d + 15) = 13.10 \)

Step 4 - Simplify and Solve for \( d \)

Distribute and combine like terms:\( 0.10d + 0.75d + 3.75 = 13.10 \)Combine like terms:\( 0.85d + 3.75 = 13.10 \)Subtract 3.75 from both sides:\( 0.85d = 9.35 \)Divide both sides by 0.85:\( d = 11 \)

Step 5 - Solve for \( q \)

Substitute \( d = 11 \) back into \( q = 3d + 15 \) to find \( q \):\( q = 3(11) + 15 \)\( q = 33 + 15 \)\( q = 48 \)

Step 6 - Verify the Solution

Check that the values satisfy both original equations.1. Value equation: \( 0.10(11) + 0.25(48) = 1.10 + 12.00 = 13.10 \)2. Relationship: \( 48 = 3(11) + 15 = 33 + 15 = 48 \)Both equations are satisfied.

Key Concepts

Linear Equations

Linear equations form the foundation of solving many mathematical problems. A linear equation is an equation that involves variables raised only to the first power. They typically look like this:

• ax + by = c

where 'a,' 'b,' and 'c' are constants, and 'x' and 'y' are variables. In the exercise example, the linear equation specifying the value of dimes and quarters is:

• 0.10d + 0.25q = 13.10

Every term here is either a constant or a variable raised to the first power. This simple form makes them straightforward to manage mathematically. Understanding how to setup and solve linear equations is crucial for tackling more complex math problems.

Variable Substitution

Variable substitution is a powerful technique to solve systems of equations. It involves expressing one variable in terms of another and substituting this expression into another equation. In our exercise, we start with two equations:

• 0.10d + 0.25q = 13.10
• q = 3d + 15

The second equation expresses 'q' in terms of 'd,' which can be substituted back into the first. This reduces the system to a single equation with one variable: oh.10d + 0.25(3d + 15) = 13.10Thus, solving the system becomes easier since we focus on just one variable. Once we find 'd,' plugging its value back into one of our original equations gives us the value of 'q.' This process is crucial for breaking down complex systems into manageable parts.

Word Problems

Word problems involve translating real-life situations into mathematical language. This skill is valuable and practical. In the exercise, Peter's collection of coins is a real-life scenario, transformed into mathematical equations. Here are the steps we followed:

• Define Variables: Let 'd' be the number of dimes and 'q' be the number of quarters.
• Set Up Equations: Identify the relationships and translate them into equations based on the problem's context.

Translating the problem means noting key information, like the value of coins and their numbers. This helps in forming correct equations, ensuring we solve the problem accurately and effectively.

Solving Equations

Solving equations is the final step in finding the solution. This involves manipulating equations to isolate the variable and determine its value. Here, after substituting the second equation into the first, we had:

• 0.10d + 0.25(3d + 15) = 13.10

Let's break it down:

• Expand: 0.10d + 0.75d + 3.75 = 13.10
• Combine like terms: 0.85d + 3.75 = 13.10
• Isolate 'd': 0.85d = 9.35
• Solve: d = 11

Once we know 'd,' finding 'q' becomes easy:

• q = 3(11) + 15
• q = 48

Lastly, always verify the solution by plugging the values back into the original equations to ensure they hold true. This guarantees accuracy and completeness in your solution.