David is buying notebooks and pens for his class. He buys 3 notebooks, and each notebook costs $4. Then, he buys 2 pens. David spends a total of $16. He wants to know the amount he spent on each pen.

Question

Part A: Algebraic Solution
Write an algebraic equation to represent the total amount David spent, and solve it to find the cost of each pen.
Part B: Arithmetic Solution
Use an arithmetic method (break the problem into smaller steps using only numbers and operations) to find the cost of each pen. Show your work and explain the sequence of operations.
Part c: Compare Solutions
Compare the two methods: How are the operations used in the algebraic solution different from the operations used in arithmetic solution? Which approach do you think is more efficient, and why?

Answer 1

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 \( p \):

1. Subtract 12 from both sides:

   \[ 2p = 16 - 12 \] \[ 2p = 4 \]

2. Divide both sides by 2:

   \[ p = \frac{4}{2} \] \[ p = 2 \]

Thus, the cost of each pen is $2.

Part B: Arithmetic Solution

1. Calculate the total spent on notebooks:

   • David buys 3 notebooks for $4 each: \[ 3 \times 4 = 12 \text{ dollars} \]

2. Determine how much money he has left for pens:

   • His total expenditure is $16, and he has already spent $12 on notebooks: \[ 16 - 12 = 4 \text{ dollars} \]

3. 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.