Problem 47 - Algebra for College Students

Question

Solve each equation. Suppose that Javier has a handful of coins, consisting of pennies, nickels, and dimes, worth \(\$ 2.63\). The number of nickels is 1 less than twice the number of pennies, and the number of dimes is 3 more than the number of nickels. How many coins of each kind does he have?

Step-by-Step Solution

Answer

8 pennies, 15 nickels, and 18 dimes.

Step 1: Define the Variables

Let \( p \) represent the number of pennies, \( n \) represent the number of nickels, and \( d \) represent the number of dimes.

Step 2: Establish Equations Based on Conditions

From the problem, we know:1. The total value is \(263\) cents (since \(\$2.63 = 263\) cents).2. \( n = 2p - 1 \) (since the number of nickels is 1 less than twice the number of pennies).3. \( d = n + 3 \) (since the number of dimes is 3 more than the number of nickels).

Step 3: Value Equation for Total Coins

The value equation considering the coins' worth is:\[1p + 5n + 10d = 263\]This accounts for pennies being \(1\) cent each, nickels \(5\) cents, and dimes \(10\) cents.

Step 4: Substitute Relationships into Value Equation

Substitute \( n = 2p - 1 \) and \( d = (2p - 1) + 3 = 2p + 2 \) into the value equation:\[1p + 5(2p - 1) + 10(2p + 2) = 263\]

Step 5: Simplify and Solve for \( p \)

Distribute and simplify the equation:\[p + 10p - 5 + 20p + 20 = 263\]Combine like terms:\[31p + 15 = 263\]Subtract 15 from both sides:\[31p = 248\]Divide both sides by 31:\[p = 8\]

Step 6: Calculate \( n \) and \( d \)

Using \( p = 8 \):\[n = 2(8) - 1 = 15\]\[d = 15 + 3 = 18\]

Step 7: Verify the Solution

Check the total value: \\(8(1) + 15(5) + 18(10) = 8 + 75 + 180 = 263\) cents. The solution satisfies the given conditions.

Key Concepts

Variable Substitution

Variable substitution is an essential concept in solving systems of equations and is particularly useful when dealing with complex problems involving relationships between variables. In the process of solving Javier's coin problem, we used variable substitution to transform the problem into a solvable equation.Here's how it works. You start by identifying relationships between different variables. In our example, we used the relationships given in the problem:

• The number of nickels ( \(\( n \)\) ) is 1 less than twice the number of pennies ( \(\( p \)\) ). This gives us the equation \(\( n = 2p - 1 \)\) .
• The number of dimes ( \(\( d \)\) ) is 3 more than the number of nickels, creating the equation \(\( d = n + 3 \)\) .

To solve, substitute these expressions into the main equation where needed. For instance, we replaced \(\( n \)\) and \(\( d \)\) in the total value equation to eventually solve for \(\( p \)\) . This transformation turns a complex problem into a series of simpler steps that are easier to manage and solve.

Coins Problem

Coins problems are classic examples used to teach systems of equations. These problems involve finding the number of coins in different denominations given particular constraints or totals. The goal is to find how many of each type of coin is present using logical reasoning and algebraic equations.In Javier's coins problem, we have three types of coins - pennies, nickels, and dimes. The problem provides specific conditions:

• The total value of the coins is \(\( \\)2.63 \)\( , which equates to \)\( 263 \)\( cents.
• The number of nickels is 1 less than twice the number of pennies ( \)\( n = 2p - 1 \)\( ).
• The number of dimes is 3 more than the number of nickels ( \)\( d = n + 3 \)$ ).

Using these conditions, the problem requires setting up equations to model each relationship and solve them simultaneously to find the solution. This method teaches the foundational skills needed to tackle similar real-world situations or more complex mathematical challenges.

Linear Equations

Linear equations are equations of the first degree, meaning they involve only the first powers of the variables. They are represented by lines when graphed. In the context of Javier's coins problem, linear equations help us express the relationships between different types of coins and their values.We formulated several linear equations based on the problem's conditions:

• The equation representing total value: \(\( 1p + 5n + 10d = 263 \)\) . This accounts for the values of pennies, nickels, and dimes in terms of cents.
• The relationship between nickels and pennies provided the linear equation: \(\( n = 2p - 1 \)\) .
• Likewise, the relation between dimes and nickels was modeled as \(\( d = n + 3 \)\) .

Solving these linear equations often involves substituting and simplifying to find the values of unknown variables. In this problem, we substituted \(\( n \)\) and \(\( d \)\) into the value equation, reduced it to a single variable equation, and solved it by isolating the variable. Understanding how to build and solve linear equations is crucial as they are used extensively in various mathematical disciplines and real-life applications.