4. Gordon Purchased 5 False Teeth And 10 Fakespiders And Spent A Total Of $35. The Price Of the False is a fascinating problem that combines elements of basic arithmetic, algebra, and problem-solving skills. This scenario prompts us to analyze the costs associated with two different items—false teeth and fakespiders—based on the total amount spent and the quantity purchased. Understanding such problems can enhance critical thinking and mathematical reasoning, which are essential in everyday financial decisions and academic pursuits.
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Understanding the Problem
Before diving into calculations, it’s essential to clearly understand the details provided:
- Gordon bought 5 false teeth.
- He also bought 10 fakespiders.
- The total amount spent was $35.
- The goal is to determine the price of the false teeth and, potentially, the price of the fakespiders.
This problem is a typical example of solving a system of linear equations where the total cost is split between two items with unknown individual prices.
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Breaking Down the Key Components
Items Purchased and Their Quantities
- False Teeth: 5 units
- Fakespiders: 10 units
Total Expenditure
- Total amount spent: $35
Unknown Variables
- Let x be the price of one false tooth.
- Let y be the price of one fakespider.
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Setting Up the Mathematical Equations
Given the information, we can set up the following equation based on the total expenditure:
\[ 5x + 10y = 35 \]
This equation means that five false teeth at price x each, plus ten fakespiders at price y each, sum to $35.
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Solving the System: Determining the Price of False Teeth
Simplify the Equation
Divide the entire equation by 5 to make calculations easier:
\[ x + 2y = 7 \]
Now, this simplified equation expresses the relationship between x and y.
Expressing One Variable in Terms of the Other
Rearranged, the equation becomes:
\[ x = 7 - 2y \]
This indicates that the price of a false tooth depends on the price of a fakespider.
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Exploring Possible Solutions
Since prices are generally non-negative and likely to be reasonable, we can consider various values of y and compute corresponding x.
Assumptions and Constraints
- Prices should be positive numbers (greater than zero).
- The prices should be realistic, e.g., less than or equal to a certain amount per item.
Calculating Prices for Different Values of y
| y (price of one fakespider) | x (price of one false tooth) | Explanation |
|------------------------------|------------------------------|--------------|
| 0 | 7 | If fakespiders are free, false teeth cost $7 each. |
| 1 | 5 | Fakespiders at $1 each, false teeth at $5 each. |
| 2 | 3 | Fakespiders at $2 each, false teeth at $3 each. |
| 3 | 1 | Fakespiders at $3 each, false teeth at $1 each. |
| 3.5 | 0 | Fakespiders at $3.5 each, false teeth free. |
Validity of Solutions
- When y = 0, x = $7 — plausible.
- When y = 1, x = $5 — plausible.
- When y = 2, x = $3 — plausible.
- When y = 3, x = $1 — plausible.
- When y > 3.5, x becomes negative, which isn't realistic for prices.
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Finding a Reasonable Cost for the False Teeth
Based on typical pricing, and the plausible solutions above, the most reasonable options are:
- False teeth cost around $5 or $3 per unit.
- Fakespiders cost around $1 to $2 each.
Suppose we assume the fakespiders cost $1 each; then, the false teeth cost:
\[ x = 7 - 2(1) = 7 - 2 = \$5 \]
This scenario looks reasonable, with false teeth at $5 each and fakespiders at $1 each.
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Final Calculation and Verification
Total Cost Calculation
Using the assumed prices:
- 5 false teeth at $5 each = 5 × $5 = $25
- 10 fakespiders at $1 each = 10 × $1 = $10
Total = $25 + $10 = $35, which matches the total expenditure.
Conclusion:
- Price of one false tooth = $5
- Price of one fakespider = $1
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Additional Considerations
Variability of Prices
Depending on the context, prices could vary. For example, if fakespiders are priced higher, the false teeth would be cheaper, and vice versa. The key is that the sum must always total $35 given the quantities.
Practical Application
Understanding how to solve such problems helps in real-life scenarios like budgeting, shopping, and evaluating deals. It also reinforces foundational algebra skills.
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Summary
In this scenario, by setting up a simple algebraic equation, simplifying it, and exploring various solutions, we determined that:
- The price of each false tooth is approximately $5.
- The price of each fakespider is approximately $1.
This analysis not only answers the original question but also demonstrates how to approach similar problems involving multiple items and total costs.
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Final Thoughts
Solving problems like "Gordon purchased 5 false teeth and 10 fakespiders for $35" illustrates the importance of critical thinking and mathematical reasoning. Whether for academic purposes or everyday life, mastering these problem-solving techniques allows for better financial decision-making and analytical skills. Remember, always check your solutions to ensure they make sense within the given context, and consider multiple scenarios to understand the full range of possibilities.
