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Chapter 8: Problem 28

A local buffet charges \(\$ 7.50\) per person for the basie buffet and \(\$ 9.25\) for the deluxe buffet (which includes crab legs.) If 27 diners went out to eat and the total bill was \(\$ 227.00\) before taxes, how many chose the basic buffet and how many chose the deluxe buffet?

Short Answer

Expert verified
13 people chose the basic buffet, 14 chose the deluxe buffet.

Step by step solution

01

Define Variables

Let \( x \) represent the number of people who chose the basic buffet, and \( y \) represent the number of people who chose the deluxe buffet.
02

Set Up Equations

We have two equations based on the problem statement. The first equation represents the total number of diners, which is 27: \( x + y = 27 \). The second equation represents the total bill: \( 7.50x + 9.25y = 227 \).
03

Solve for One Variable

From the first equation, express \( y \) in terms of \( x \): \( y = 27 - x \).
04

Substitute and Simplify

Substitute \( y = 27 - x \) into the second equation: \( 7.50x + 9.25(27 - x) = 227 \). Simplify the equation: \( 7.50x + 249.75 - 9.25x = 227 \).
05

Solve the Simplified Equation

Combine like terms in the equation: \( -1.75x + 249.75 = 227 \). To isolate \( x \), subtract 249.75 from both sides: \( -1.75x = -22.75 \). Solve for \( x \): \( x = \frac{-22.75}{-1.75} = 13 \).
06

Find the Second Variable

Substitute \( x = 13 \) back into the equation \( y = 27 - x \): \( y = 27 - 13 = 14 \).
07

Verify the Solution

Check that these values satisfy the second equation: \( 7.50(13) + 9.25(14) = 97.50 + 129.50 = 227 \). Both conditions of the problem are satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Systems of Equations
When solving real-world problems, like figuring out the number of people choosing different types of buffets, we often encounter situations where there are multiple unknowns. These unknowns can usually be expressed as variables. A system of equations allows us to solve for these variables.
  • An equation system is a set of two or more equations that share the same variables. In our buffet problem, we have two equations that involve the variables \( x \) (for the basic buffet) and \( y \) (for the deluxe buffet).
  • The first equation represents the total number of diners. Since 27 people ate, we get\( x + y = 27 \).
  • The second equation represents the total money spent, providing us with another equation: \( 7.50x + 9.25y = 227 \).
Our goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously. The solution to a system of equations is a set of values for the variables that makes all the equations true. Systems of equations are essential tools in algebra that help us model and solve many practical problems.
Problem-Solving
Solving a problem involving a system of equations requires clearly defined steps. This approach not only helps in organizing thoughts but also ensures accuracy. Let's break down the steps required to tackle such a problem.
  • Understand the problem: Read the problem statement carefully, identifying what is known and what needs to be found.
  • Define the variables: Represent the unknowns with variables. In this case, we used \( x \) for the basic buffet group and \( y \) for the deluxe group.
  • Set up the equations: Based on the problem details, construct equations that include the defined variables. The equations should represent all relationships described in the problem.
  • Solve the equations: Use algebraic methods, like substitution or elimination, to find the values of the variables.
  • Verify the solution: Substitute your answers back into the original equations to check that all conditions are satisfied. If the equations hold true, you've found the correct solution.
This organized methodical approach to problem-solving helps prevent errors and ensures a successful and efficient resolution to mathematical problems.
Variable Substitution
One of the most powerful methods to solve a system of equations is variable substitution. This technique is particularly handy when one equation is easy to solve for one variable. Here鈥檚 how substitution works in our buffet example.
  • Express one variable in terms of the other: From the first equation, \( x + y = 27 \), we can solve for \( y \) to get \( y = 27 - x \). This expression will be used to substitute into the other equation.
  • Substitute into the second equation: Replace \( y \) in the second equation with \( 27 - x \), transforming it into a single-variable equation: \( 7.50x + 9.25(27 - x) = 227 \).
  • Simplify and solve: Simplify the equation to isolate \( x \). In this instance, simplifying gives \( -1.75x + 249.75 = 227 \), which solves to \( x = 13 \).
  • Substitute back to find the second variable: With \( x = 13 \), find \( y \) using the expression \( y = 27 - x \). Therefore, \( y = 14 \).
This method simplifies complex problems into smaller manageable chunks, making it an efficient approach for tackling systems of equations. Understanding substitution provides a foundation for more advanced algebraic techniques.

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Most popular questions from this chapter

In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{x^{6}+5 x^{5}+16 x^{4}+80 x^{3}-2 x^{2}+6 x-43}{x^{3}+5 x^{2}+16 x+80} $$

Consider the following scenario. In the small village of Pedimaxus in the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, \(90 \%\) of those who subscribe to the Pedimaxus Tribune want to keep getting it, but \(10 \%\) want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, \(80 \%\) want to continue with it and \(20 \%\) want switch to the Tribune. We can express this situation using matrices. Specifically, let \(X\) be the 'state matrix' given by $$ X=\left[\begin{array}{l} T \\ P \end{array}\right] $$ where \(T\) is the number of people who get the Tribune and \(P\) is the number of people who get the Picayune in a given week. Let \(Q\) be the 'transition matrix' given by $$ Q=\left[\begin{array}{ll} 0.90 & 0.20 \\ 0.10 & 0.80 \end{array}\right] $$ such that \(Q X\) will be the state matrix for the next week. Explain why the transition matrix does what we want it to do.

Let \(z=a+b i\) and \(w=c+d i\) be arbitrary complex numbers. Associate \(z\) and \(w\) with the matrices $$ Z=\left[\begin{array}{rr} a & b \\ -b & a \end{array}\right] \text { and } W=\left[\begin{array}{rr} c & d \\ -d & c \end{array}\right] $$ Show that complex number addition, subtraction and multiplication are mirrored by the associated matrix arithmetic. That is, show that \(Z+W, Z-W\) and \(Z W\) produce matrices which can be associated with the complex numbers \(z+w, z-w\) and \(z w,\) respectively.

In Exercises \(32-36,\) consider the following definitions. A square matrix is said to be an upper triangular matrix if all of its entries below the main diagonal are zero and it is said to be a lower triangular matrix if all of its entries above the main diagonal are zero. For example,$$ E=\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 4 & -9 \\ 0 & 0 & -5 \end{array}\right] $$ from Exercises 8 - 21 above is an upper triangular matrix whereas $$ F=\left[\begin{array}{ll} 1 & 0 \\ 3 & 0 \end{array}\right] $$ is a lower triangular matrix. (Zeros are allowed on the main diagonal.) Discuss the following questions with your classmates. Give an example of a matrix which is neither upper triangular nor lower triangular.

Solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. $$ \left\\{\begin{aligned} x+2 y^{2} &=2 \\ x^{2}+4 y^{2} &=4 \end{aligned}\right. $$
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