Problem 7 - Algebra 2

Question

Jambalaya is a Cajun dish made from chicken, sausage, and rice. Simone is making a large pot of of jambalaya for a party. Chicken costs \(\$ 6\) per pound, sausage costs \(\$ 3\) per pound, and rice costs \(\$ 1\) per pound. She spends \(\$ 42\) on 13.5 pounds of food. She buys twice as much rice as sausage. Write a system of three equations that represents how much food Simone purchased.

Step-by-Step Solution

Answer

The system is: 1) \(6c + 3s + r = 42\), 2) \(c + s + r = 13.5\), 3) \(r = 2s\).

Step 1: Identify Variables

Let \( c \) be the pounds of chicken, \( s \) be the pounds of sausage, and \( r \) be the pounds of rice Simone purchased.

Step 2: Set Up Cost Equation

Simone spends \( \$42 \) on the food. The cost equation is based on the cost per pound: \( 6c + 3s + 1r = 42 \).

Step 3: Set Up Total Weight Equation

Simone purchased a total of 13.5 pounds of food. The equation for total weight is \( c + s + r = 13.5 \).

Step 4: Set Up Rice to Sausage Ratio Equation

Simone buys twice as much rice as sausage. Therefore, the equation reflecting this relationship is \( r = 2s \).

Step 5: Combine and Present the System of Equations

The complete system of equations is:\[ \begin{align*} 6c + 3s + r &= 42 \ c + s + r &= 13.5 \ r &= 2s \end{align*} \]

Key Concepts

Linear Equations

Linear equations are a fundamental part of algebra and mathematics. They are equations that graph as straight lines when plotted on a graph. For these mathematical expressions, the highest power of the variable is always 1, which is why they are called 'linear'. Linear equations are frequently used to model real-world situations, like calculating costs or quantities, just as Simone did for her jambalaya ingredient purchases.

In this exercise, we have three linear equations:

• The cost equation: \( 6c + 3s + r = 42 \)
• The total weight equation: \( c + s + r = 13.5 \)
• The relationship between rice and sausage: \( r = 2s \)

Each equation involves the summation of terms multiplied by constants or coefficients. These coefficients tell us how much each variable contributes to the total sum. For example, 6 in \( 6c \) represents that every pound of chicken costs $6. This is a crucial element in setting up and solving any system of linear equations.

Algebraic Expressions

An algebraic expression consists of numbers, variables, and operation symbols that together represent a value. They do not contain an equal sign, unlike equations. However, they serve as building blocks for forming equations. In the scenario given, Simone’s decision to purchase certain amounts of chicken, sausage, and rice can each be represented by algebraic expressions.

For instance:

• \( 6c \) represents the total cost for the chicken.
• \( 3s \) represents the total cost for the sausage.
• \( r \) simply accounts for the total cost of the rice.

Together, these expressions and the numerical operations on them help form our cost equation \( 6c + 3s + r \). The beauty of using algebraic expressions lies in their ability to convey complex relationships in a simple, symbolic form. This allows us to analyze and solve problems more efficiently.

Variables

Variables play an integral role in both algebraic expressions and linear equations. They serve as placeholders for unknown values that we aim to discover or analyze. In our exercise, they let us represent quantities of chicken, sausage, and rice without having explicit values yet.

In our system:

• \( c \) is the variable for the pounds of chicken.
• \( s \) represents the pounds of sausage.
• \( r \) represents the pounds of rice.

Variables provide the flexibility needed to create equations that model real-world scenarios. As we manipulate and solve these equations, we eventually uncover the specific values each variable stands for. By establishing their relationships in the system of equations, we can solve for each variable, determining exactly how much of each ingredient Simone used. This makes variables crucial in mathematical modeling and problem-solving.