Part A: Algebraic Solution
Let's denote the cost of each pen as \( p \). David bought 3 notebooks at a cost of $4 each. Therefore, the total cost for the notebooks is:
\[ 3 \times 4 = 12 \text{ dollars} \]
He also bought 2 pens, which costs \( 2p \). The total amount David spent is given as $16. Thus, we can write the equation:
\[ 12 + 2p = 16 \]
Now, we can solve for \( p \):
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Subtract 12 from both sides:
\[ 2p = 16 - 12 \] \[ 2p = 4 \]
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Divide both sides by 2:
\[ p = \frac{4}{2} \] \[ p = 2 \]
Thus, the cost of each pen is $2.
Part B: Arithmetic Solution
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Calculate the total spent on notebooks:
- David buys 3 notebooks for $4 each: \[ 3 \times 4 = 12 \text{ dollars} \]
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Determine how much money he has left for pens:
- His total expenditure is $16, and he has already spent $12 on notebooks: \[ 16 - 12 = 4 \text{ dollars} \]
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Calculate the cost of each pen:
- He bought 2 pens with the remaining $4: \[ \text{Cost per pen} = \frac{4 \text{ dollars}}{2} = 2 \text{ dollars} \]
Thus, through this arithmetic solution, we also find that the cost of each pen is $2.
Part C: Compare Solutions
Operations Used:
- Algebraic Solution: The method involves forming an equation based on a known total, substituting known values, and performing operations to isolate a variable.
- Arithmetic Solution: This method breaks down the problem into sequential calculations involving direct subtraction and division without the formal setup of an equation.
Efficiency:
- The algebraic solution might seem more abstract but can be quicker for more complex problems. It succinctly represents relationships through equations, allowing for clear and scalable problem-solving.
- The arithmetic solution, while straightforward and intuitive, may require more steps to reach the assessment of each component and can become cumbersome with larger numbers or more variables.
Conclusion: Both solutions are valid, but I think the algebraic solution can be viewed as more efficient for more complex scenarios. It provides a systematic approach that can be scaled up, while the arithmetic method might be easier for quick, simple calculations.