Problem 8 - 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. How much chicken, sausage, and rice will she use in her dish?

Step-by-Step Solution

Answer

Simone will use 4.5 pounds of chicken, 3 pounds of sausage, and 6 pounds of rice.

Step 1: Define Variables

Let \( c \) represent pounds of chicken, \( s \) represent pounds of sausage, and \( r \) represent pounds of rice. We know \( r = 2s \) because Simone buys twice as much rice as sausage.

Step 2: Set Up the Equations

From the problem, we have two main pieces of information: the total weight of the food and the total cost. These give us the equations: \( c + s + r = 13.5 \) for the total weight and \( 6c + 3s + 1r = 42 \) for the total cost.

Step 3: Substitute and Simplify the Equations

Substitute \( r = 2s \) into the two equations. The weight equation becomes \( c + s + 2s = 13.5 \) or \( c + 3s = 13.5 \). The cost equation becomes \( 6c + 3s + 2s = 42 \) or \( 6c + 5s = 42 \).

Step 4: Solve the System of Equations

From \( c + 3s = 13.5 \), we can express \( c \) as \( c = 13.5 - 3s \). Substitute this into the second equation: \( 6(13.5 - 3s) + 5s = 42 \). Simplify to \( 81 - 18s + 5s = 42 \), which simplifies further to \( -13s = -39 \), so \( s = 3 \).

Step 5: Calculate Amount of Rice and Chicken

With \( s = 3 \), use \( r = 2s \) to find \( r = 6 \). Substitute \( s = 3 \) into \( c = 13.5 - 3s \) to find \( c = 13.5 - 9 = 4.5 \).

Step 6: Verify the Solution

Check the total pounds: \( 4.5 + 3 + 6 = 13.5 \). Check the total cost: \( 6(4.5) + 3(3) + 1(6) = 27 + 9 + 6 = 42 \). Both checks confirm the solution is correct.

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations, particularly useful when dealing with linear equations. This method involves solving one of the equations for one variable and then substituting that expression into the other equation. In the jambalaya problem, we knew that Simone buys twice as much rice as sausage, so we expressed rice in terms of sausage, specifically as \( r = 2s \). By substituting \( r \) in the equations, we simplified the system to involve just two variables, chicken \( c \) and sausage \( s \). This makes the equations easier to solve because we can focus on finding values for just \( c \) and \( s \), which we eventually use to find \( r \).

The substitution method not only reduces the number of variables but also often simplifies the arithmetic involved, which is why it's a favored technique when equations are easy to manipulate like these linear ones.

Linear Equations

Linear equations are fundamental in algebra and appear as the backbone of many real-world problems, just like in this jambalaya scenario. These are equations of the first order, meaning they involve only the first powers of the variables. In the context of Simone's dish, our linear equations were based on the constraints of weight and cost. We had two linear equations: \( c + s + r = 13.5 \) representing the total weight, and \( 6c + 3s + 1r = 42 \) representing the total cost.

Linear equations often appear in forms \( ax + by + cz = d \) and can be solved using methods like substitution, elimination, or graphical methods. Understanding linear equations helps break down complex word problems into manageable pieces that can be solved step by step. These equations model relationships where changes between variables maintain consistent rates, which is why they fit so well for budgeting problems like our jambalaya ingredients.

Problem Solving

Problem solving in mathematics requires a strategic approach to dissect, understand, and ultimately solve the problem. In the case of the jambalaya problem, the key steps involved translating a real-life situation into mathematical expressions and equations. By defining variables for the unknowns — chicken \( c \), sausage \( s \), and rice \( r \) — we took the first step in framing the problem mathematically. This step is crucial as it sets up a clear path for finding solutions.

Once the variables were set, establishing the right equations was essential; they were derived from the problem statement's constraints: total weight and total cost. Using mathematical methods like substitution, we could simplify and solve these equations for unknown values. The art of problem-solving is also about verifying solutions, ensuring that they meet all given conditions. Checking both the weight and cost verified our solution, assuring that our logical path from start to finish was consistent and correct. Effective problem-solving is not just about the right answer but about building confidence through a thorough understanding of the process.