Problem 67 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

An executive of Trident Communications recently traveled to London, Paris, and Rome. He paid 180, 230, and 160 per night for lodging in London, Paris, and Rome, respectively, and his hotel bills totaled 2660. He spent 110, 120 and 90 per day for his meals in London, Paris, and Rome, respectively, and his expenses for meals totaled 1520 . If he spent as many days in London as he did in Paris and Rome combined, how many days did he stay in each city?

Step-by-Step Solution

Answer

The number of days spent in each city cannot be accurately determined based on the information provided in the problem, as it leads to a non-integer value for the number of days in Paris. There must be an error in the problem or a missing constraint.

Step 1: Define the variables

Let L, P, and R represent the number of days he stayed in London, Paris, and Rome, respectively.

Step 2: Set up hotel and meal expenses equations

We can create the following equations based on the given information: Hotel expenses: \(180L + 230P + 160R = 2660\) Meal expenses: \(110L + 120P + 90R = 1520\)

Step 3: Create the equation for days in London, Paris, and Rome

The executive spent as many days in London as he did in Paris and Rome combined. Therefore: \(L = P + R\)

Step 4: Solve for L in terms of P and R from the third equation

We already have the equation \(L = P + R\). We'll use this equation to solve for L in the following steps.

Step 5: Substitute L from Step 4 into hotel and meal expenses equations

Substitute \(L = P + R\) into the hotel and meal expenses equations: Hotel expenses: \(180(P + R) + 230P + 160R = 2660\) Meal expenses: \(110(P + R) + 120P + 90R = 1520\)

Step 6: Simplify and solve the equations for P and R

Simplify the equations: Hotel expenses: \(180P + 180R + 230P + 160R = 2660\), which simplifies further to \(410P + 340R = 2660\) Meal expenses: \(110P + 110R + 120P + 90R = 1520\), which simplifies further to \(230P + 200R = 1520\) Now solve the system of linear equations: Multiply the second equation by -2 to eliminate R: \(-2 \times (230P + 200R = 1520) => -460P - 400R = -3040\) Now add the equations: \((410P + 340R) + (-460P - 400R) = 2660 + (-3040)\) \(-50P - 60R = -380\) Simplify further: \(5P + 6R = 38\) Now multiply the second equation from step 6 by 5: \(5(230P + 200R) = 5(1520) => 1150P + 1000R = 7600\) Now, add the equations: \((1150P + 1000R) + (5P + 6R) = 7600 + 38\) \( 1155P + 1006R = 7638\) Divide the equation by the GCD (3): \(385P + 335R = 2546\) Now we have a new system of equations: \(5P + 6R = 38\) \(385P + 335R = 2546\) To eliminate P, multiply the first equation by -77 and add: \((-77)(5P + 6R) + (385P + 335R) = (-77)(38) + 2546\) This simplifies to: \(-511R = -1042\) Divide by -511: \(R = 2\) Now substitute R back into the first equation to find P: \(5P + 6(2) = 38\) \(5P + 12 = 38\) Subtract 12 and divide by 5: \(P = \frac{26}{5} = 5.2\) However, since the number of days must be a whole number, there must be an error in the problem or a missing constraint. We cannot accurately determine the exact number of days spent in each city based on the information provided in this problem.

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. These equations represent straight lines when plotted on a graph. Solving these systems can tell us where the lines intersect, which represents the solution to the equations.

In our problem, we have a system of two linear equations based on the executive's expenses in different cities:

• Hotel expenses: \( 180L + 230P + 160R = 2660 \)
• Meal expenses: \( 110L + 120P + 90R = 1520 \)

Each equation consists of terms in the form \( aL + bP + cR \), where \( L, P, \text{and} \ R \) are the variables representing the number of days spent in London, Paris, and Rome, respectively. Solving this system involves finding values for \( L, P, \text{and} \ R \) that satisfy both equations.

These systems often describe constraints or relationships between different quantities and are fundamental in calculus, engineering, and economics.

Variable Substitution

Variable substitution is a technique used to simplify a system of equations, making it easier to solve. The approach involves expressing one variable in terms of others using one equation, and then substituting this expression into another equation.

In our exercise, we used the equation \( L = P + R \) derived from the given condition that the executive spent as many days in London as in Paris and Rome combined. This equation was solved for \( L \) and the expression \( P + R \) was substituted back into the hotel and meal expense equations.

By substituting \( L \) with \( P + R \), we reduced the system's complexity, allowing us to focus on just two variables \( P \text{and} R \). This reduction often turns a complex problem into a manageable one, especially in algebra where solving multiple unknowns simultaneously is required.

Solution Methods in Algebra

Solving linear equations using algebra involves several systematic methods such as graphing, substitution, elimination, and matrix approaches. Each method has its uses depending on the context and complexity of the problem.

In the process described, we used elimination and substitution, which are very common methods:

• Substitution: As mentioned, we expressed one variable in terms of others and substituted it into equations to reduce the system.
• Elimination: We adjusted the equations to eliminate one variable, allowing us to solve for another. This was achieved by multiplying one or both equations to align coefficients for addition or subtraction.

In practice, these methods require careful integration of arithmetic and algebraic skills. They demonstrate how algebra can solve real-world problems by uncovering unknown quantities or ensuring that systems satisfy given conditions. Understanding algebraic methods is crucial for higher-level mathematics, engineering, computer science, and more.