Problem 33 - College Algebra with Modeling and Visualization

Question

Three students buy lunch in the cafeteria. One student buys 2 hamburgers, 2 orders of fries, and 1 soda for \(\$ 9 .\) Another student buys 1 hamburger, 1 order of fries, and 1 soda for \(\$ 5 .\) The third student buys 1 hamburger and 1 order of fries for \(\$ 5 .\) If possible, find the cost of each item. Interpret the results.

Step-by-Step Solution

Answer

The setup results in no solutions; possibly due to error or misinterpretation.

Step 1: Assign Variables

Assign variables to each item. Let the price of a hamburger be \( h \), the price of an order of fries be \( f \), and the price of a soda be \( s \).

Step 2: Write Equations

Based on the purchases:- The first student buys 2 hamburgers, 2 fries, and 1 soda for \( \\(9 \). This gives the equation: \[ 2h + 2f + s = 9 \]- The second student buys 1 hamburger, 1 fries, and 1 soda for \( \\)5 \). This gives the equation: \[ h + f + s = 5 \]- The third student buys 1 hamburger and 1 fries for \( \$5 \). This gives the equation: \[ h + f = 5 \]

Step 3: Simplify Using Equations

Notice that the third equation \( h + f = 5 \) can replace \( h + f \) in the second equation. Hence, substitute \( h + f = 5 \) in the second equation: \[ (h + f) + s = 5 \] This simplifies to \[ 5 + s = 5 \] leading to \( s = 0 \).

Step 4: Substitute Soda Price into Other Equations

With \( s = 0 \), substitute into the first equation: \[ 2h + 2f + 0 = 9 \] Simplifies to: \[ 2h + 2f = 9 \]

Step 5: Solve for Hamburgers and Fries

With two equations: 1. \( h + f = 5 \)2. \( 2h + 2f = 9 \)Substitute \( h + f = 5 \) into \( 2h + 2f = 9 \): \[ 2(h + f) = 9 \] \[ 2 \times 5 = 9 \] This leads to a contradiction, as \( 10 eq 9 \). Hence, there is no consistent solution for \( h \) and \( f \) with \( s = 0 \).

Step 6: Analyze Contradiction and Rectify

The previous steps revealed a contradiction. Re-evaluate the third student's equation: - Correctly solve starting from \( 2h + 2f = 9 \) and \( h + f = 5 \) \[ h + f = 5 \] leads to: \( h = 5 - f \). Substitute into the first equation: \[ 2(5-f) + 2f = 9 \] \[ 10 - 2f + 2f = 9 \] \[ 10 = 9 \] is wrong, thus reassess starting from the second equation.

Step 7: Re-solve for Error

The contradiction indicates an assumption error. Reassess prior steps, as substituting specific values did not yield solutions, confirming misalignment in setup rather than approach.

Key Concepts

Algebraic Substitution

Algebraic substitution is a crucial method in solving systems of equations. In the context of this exercise, the idea is to identify expressions that can replace variables in other equations. The third equation, \( h + f = 5 \), provides a straightforward relationship. Instead of handling three variables at once, substitution allows us to address two easier equations.
By expressing \( h + f \) and noting it equals 5, we can substitute this directly into the second equation, \( h + f + s = 5 \). This simplifies to solving for one less variable. Sadly, even when resolving the adjusted equations, a contradiction arises, indicating an error is present, which means a revisiting of the assumptions or equations is necessary.
The core principle of substitution is to simplify complex problems by breaking them into smaller parts. This strategy is incredibly valuable in algebra and assists in minimizing errors when solving equations by reducing the amount of simultaneous information to process.

Equations in Three Variables

Working with equations in three variables can be quite challenging, especially when trying to isolate individual elements. In this problem, the items' costs act as our variables: hamburgers (\( h \)), fries (\( f \)), and sodas (\( s \)).
The relationships presented translate to three separate equations built from the transactions described. These are:

• First student: \( 2h + 2f + s = 9 \)
• Second student: \( h + f + s = 5 \)
• Third student: \( h + f = 5 \)

Initially, these equations seem plausible and manageable. However, solving them reveals inconsistencies leading to a wrong value for soda. This contradiction suggests a need to revise the mathematical setup. Typically, such contradictions mean the system doesn't have a unique or valid solution, possibly due to misplaced values or assumptions. Re-evaluating each step is essential to verify each equation's authenticity and alignment.

Problem Solving in Algebra

Problem-solving in algebra involves strategically approaching equations to find meaningful solutions. When handling exercises like these, one must be careful in understanding what each part of the problem implies. For example, assigning distinct variables to each item of purchase helps in organizing thoughts. From there, writing equations based on the scenario allows for the problem's numerical aspects to be analyzed.
In this exercise, attention is required to not only substitute and simplify but to also interpret results. The contradiction noted in the final steps indicated something deeper – possibly an error in the initial equation setup, or a misinterpretation of the problem's context. Getting these elements right is central to finding logical conclusions in algebra.
In problem-solving, noticing contradictions isn't always a setback; it's an opportunity to refine understanding. By reassessing or trying different approaches, one can often find the correct pathway to a valid solution. This flexibility is key in mastering algebraic concepts and excelling in solving systems of equations.