Math Problem Statement

A new restaurant is to contain two-seat tables and four-seat tables. Fire codes limit the restaurant's maximum occupancy to 68 customers. If the owners have hired enough servers to handle 20 tables of customers, how many of each kind of table should they purchase?

Solution

The problem asks to determine how many two-seat and four-seat tables a restaurant should purchase, given the following conditions:

  • The total number of tables is 20.
  • The total seating capacity is 68.
  • Each two-seat table seats 2 people, and each four-seat table seats 4 people.

Let's define the variables:

  • Let xx be the number of two-seat tables.
  • Let yy be the number of four-seat tables.

Now, form the system of linear equations:

  1. Since the total number of tables is 20:
    x+y=20x + y = 20

  2. Since the total number of seats is 68:
    2x+4y=682x + 4y = 68

These are the two equations that represent the situation. The correct system of linear equations is:

x + y = 20 \\ 2x + 4y = 68 \end{cases}$$ Next steps: 1. You can solve this system using either substitution or elimination methods. 2. Substituting one equation into the other would help in finding the values of $$x$$ and $$y$$. Would you like me to solve this system step by step for you? Here are 5 related questions: 1. How would the system of equations change if the seating capacity increased? 2. What if the restaurant could handle 25 tables, but the seating capacity remains 68? 3. How would you set up the system if the two-seat tables were replaced by three-seat tables? 4. Could you solve this system using matrix methods? 5. How would you interpret the solution graphically? **Tip:** When working with word problems, clearly defining variables at the start simplifies creating accurate equations.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Systems of Equations
Algebra

Formulas

x + y = 20
2x + 4y = 68

Theorems

System of Linear Equations

Suitable Grade Level

Grades 7-10

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