Problem 14 - Intermediate Algebra

Question

In 10 minutes, the top three finishers in the 2010 Nathan's Hot Dog Eating Contest consumed a total of 136 hot dogs. The winner, Joey Chestnut, ate 9 more hot dogs than the runner-up, Tim Janus. Pat Bertoletti finished a distant third, 8 hot dogs behind Janus. How many hot dogs did each person eat?

Step-by-Step Solution

Answer

Joey ate 54, Tim ate 45, and Pat ate 37 hot dogs.

Step 1: Define the Variables

Let \( J \) be the number of hot dogs Joey Chestnut ate, \( T \) be the number of hot dogs Tim Janus ate, and \( P \) be the number of hot dogs Pat Bertoletti ate.

Step 2: Set Up the Equations

According to the problem, Joey ate 9 more hot dogs than Tim, so we have: \( J = T + 9 \). Tim ate 8 more hot dogs than Pat, so \( T = P + 8 \). The total number of hot dogs eaten by all three is 136, giving us the equation: \( J + T + P = 136 \).

Step 3: Substitute and Solve for One Variable

Substitute the expressions for \( J \) and \( T \) in terms of \( P \) into the total equation: \( (T + 9) + T + P = 136 \). Simplify to get \( 2T + P + 9 = 136 \).

Step 4: Simplify and Solve for P

Substitute \( T = P + 8 \) into the equation: \( 2(P + 8) + P + 9 = 136 \).Simplify to \( 2P + 16 + P + 9 = 136 \), which simplifies to \( 3P + 25 = 136 \). Solve for \( P \): \( 3P = 111 \), so \( P = 37 \).

Step 5: Solve for T

Using \( T = P + 8 \), substitute \( P = 37 \): \( T = 37 + 8 = 45 \). Thus, Tim Janus ate 45 hot dogs.

Step 6: Solve for J

Using \( J = T + 9 \), substitute \( T = 45 \): \( J = 45 + 9 = 54 \). Thus, Joey Chestnut ate 54 hot dogs.

Key Concepts

Variables

In solving any mathematical problem, especially systems of equations, understanding variables is crucial. Variables are symbols or letters used to represent unknown values. In our scenario, the variables are:

• \( J \): represents the number of hot dogs Joey Chestnut ate.
• \( T \): stands for the number of hot dogs Tim Janus consumed.
• \( P \): is the number of hot dogs Pat Bertoletti ate.

Each variable holds a specific numeric value that satisfies the given conditions of the problem. The use of variables allows us to formulate equations that describe relationships between the quantities we are dealing with. This makes it easier to manipulate the expressions algebraically to find answers.
By assigning a variable to each unknown, we can then proceed to express the relationships mathematically, paving the way for systematic solving.

Substitution Method

The substitution method is a key strategy used to solve systems of equations. This method involves expressing one variable in terms of another and then substituting this expression into the other equations. Let's break down how it was applied in the problem:

• First, we expressed \( J \) in terms of \( T \): \( J = T + 9 \).
• Next, \( T \) was expressed in terms of \( P \): \( T = P + 8 \).

These expressions were then plugged into the total equation \( J + T + P = 136 \). This substitution simplifies the equation to include only one variable, \( P \), making it much easier to solve.
By dealing with just one variable instead of three, the complex problem becomes more manageable. With this simplified equation, we can directly solve for \( P \) and subsequently find \( T \) and \( J \) using the earlier substitutions. This step-by-step substitution is what renders large systems of equations approachable.

Algebraic Expressions

At the core of solving equations using variables is the manipulation of algebraic expressions. An algebraic expression is a combination of numbers, variables, and arithmetic operations. In systems of equations like this one, the algebraic expressions serve as the groundwork for forming the equations:

• The expression \( J = T + 9 \) demonstrates Joey's total as a result of adding more hot dogs to Tim's count.
• The equation \( T = P + 8 \) aligns Tim's quantity as a sum exceeding Pat's.
• The total equation \( J + T + P = 136 \) encapsulates the overall scenario to ensure all conditions are met.

Notice how each equation is an algebraic expression that defines a specific part of the problem. By simplifying and substituting within algebraic expressions, we reduce complexity and solve for unknowns efficiently. In essence, understanding and manipulating these expressions is the key to unlocking solutions in algebraic problems.