At a certain photoprocessing shop, the first standard-size print of a negative costs $4, and each additional print of the...

1. Translate the problem requirements: Clarify the pricing structure (first print vs additional prints) and what we're calculating (total cost for multiple negatives with multiple prints each)
2. Analyze the cost structure for a single negative: Break down how costs accumulate when ordering multiple prints of one negative
3. Scale up to multiple negatives: Apply the single-negative cost pattern across all x negatives
4. Verify the expression matches answer choices: Ensure our derived formula aligns with the given options

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening at this photoprocessing shop in plain English:

• The first print of any negative costs \(\$4\)
• Each additional print of that same negative costs \(\$1\)
• We have x different negatives
• We want y prints of each negative
• We need to find the total cost for everything

So if I have 3 different negatives and I want 5 prints of each, I'm getting 15 total prints, but the pricing isn't simply \(15 \times \text{some price}\) because the first print of each negative is more expensive.

Process Skill: TRANSLATE - Converting the pricing structure into mathematical understanding

2. Analyze the cost structure for a single negative

Let's think about what happens when I order y prints of just one negative:

• The first print costs \(\$4\)
• The remaining prints cost \(\$1\) each
• How many remaining prints are there? If I want y total prints, then after the first print, I need \((\mathrm{y} - 1)\) more prints
• So the cost for one negative with y prints is: \(\$4 + \$1 \times (\mathrm{y} - 1)\)

Let me check this with a concrete example:
If I want 5 prints of one negative:

• First print: \(\$4\)
• Remaining 4 prints: \(4 \times \$1 = \$4\)
• Total: \(\$4 + \$4 = \$8\)

Using our formula: \(\$4 + \$1 \times (5 - 1) = \$4 + \$4 = \$8\) ✓

3. Scale up to multiple negatives

Now I need to apply this same cost structure to all x negatives:

• Each of the x negatives costs \(\$4 + \$1 \times (\mathrm{y} - 1)\)
• Total cost = \(\mathrm{x} \times [\$4 + \$1 \times (\mathrm{y} - 1)]\)
• Total cost = \(\mathrm{x} \times [\$4 + (\mathrm{y} - 1)]\)
• Total cost = \(\$4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\)

Let me verify with numbers:
If \(\mathrm{x} = 2\) negatives and \(\mathrm{y} = 3\) prints each:

• Negative 1: \(\$4 + \$1 \times 2 = \$6\)
• Negative 2: \(\$4 + \$1 \times 2 = \$6\)
• Total: \(\$12\)

Using our formula: \(\$4(2) + 2(3 - 1) = \$8 + \$4 = \$12\) ✓

4. Verify the expression matches answer choices

Our derived formula is: \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\)

Looking at the answer choices:

1. \(4\mathrm{xy} + (\mathrm{x} - 1)\mathrm{y}\)
2. \(4\mathrm{xy} + (\mathrm{y} - 1)\mathrm{x}\)
3. \(4\mathrm{x} + \mathrm{xy}\)
4. \(4\mathrm{x} + \mathrm{xy} - 1\)
5. \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\)

Our expression exactly matches choice E: \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\)

Let me double-check by expanding choice E:
\(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1) = 4\mathrm{x} + \mathrm{xy} - \mathrm{x} = 3\mathrm{x} + \mathrm{xy}\)

And checking our logic: x negatives each costing \(\$4\) gives us \(4\mathrm{x}\), plus x negatives each with \((\mathrm{y}-1)\) additional prints at \(\$1\) each gives us \(\mathrm{x}(\mathrm{y}-1)\).

4. Final Answer

The total cost in dollars of y standard-size prints of each of x different negatives is \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\).

The answer is E.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the pricing structure

Students often misread the problem and think that ALL prints cost \(\$4\) each, or that only the FIRST print overall costs \(\$4\) while all others cost \(\$1\). The key detail is that the FIRST print of EACH NEGATIVE costs \(\$4\), while additional prints of that SAME negative cost \(\$1\). This means if you have 3 different negatives, you'll pay \(\$4\) three times (once for each negative's first print).

2. Confusing what 'y prints of each of x negatives' means

Some students interpret this as y total prints distributed among x negatives, rather than y prints PER negative. The problem specifically states 'y prints of EACH of x negatives,' meaning if \(\mathrm{x}=3\) and \(\mathrm{y}=5\), you're getting 15 total prints (5 prints each of 3 different negatives), not 5 prints total shared among 3 negatives.

3. Setting up the wrong cost equation structure

Students might try to calculate total prints first \((\mathrm{xy})\) and then apply some average price, missing that the pricing is tiered per negative. The correct approach requires calculating the cost per negative first \([\$4 + (\mathrm{y}-1)\times\$1]\), then multiplying by x negatives.

Errors while executing the approach

1. Incorrect algebraic distribution

When expanding expressions like \(\mathrm{x}[\$4 + (\mathrm{y}-1)]\), students often make errors such as getting \(4\mathrm{x} + \mathrm{x}(\mathrm{y}+1)\) instead of \(4\mathrm{x} + \mathrm{x}(\mathrm{y}-1)\), or incorrectly distributing to get \(4\mathrm{x} + \mathrm{xy} - 1\) instead of \(4\mathrm{x} + \mathrm{xy} - \mathrm{x}\).

2. Confusing which variable represents additional prints

Students might use \((\mathrm{x}-1)\) instead of \((\mathrm{y}-1)\) for additional prints, thinking about additional negatives rather than additional prints per negative. Since we want y prints of each negative, we need \((\mathrm{y}-1)\) additional prints after the first print.

Errors while selecting the answer

1. Choosing algebraically equivalent but incorrectly structured expressions

Students might expand their correct expression \(4\mathrm{x} + \mathrm{x}(\mathrm{y}-1)\) to get \(4\mathrm{x} + \mathrm{xy} - \mathrm{x} = 3\mathrm{x} + \mathrm{xy}\), then mistakenly select choice C \((4\mathrm{x} + \mathrm{xy})\) thinking it's close enough, when they should have kept the original form to match choice E exactly.

2. Selecting expressions that give correct values for simple test cases

When checking their work with small numbers, students might find that multiple answer choices give the same result for their test case (like \(\mathrm{x}=1, \mathrm{y}=2\)) and then guess among the 'working' options rather than testing with different values or checking their logical setup more carefully.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient values
Let's use \(\mathrm{x} = 2\) negatives and \(\mathrm{y} = 3\) prints of each negative. These are small, manageable numbers that will make our calculations clear.

Step 2: Calculate cost for one negative
For each negative with 3 prints:
• First print: \(\$4\)
• Additional prints: \((3 - 1) = 2\) prints at \(\$1\) each = \(\$2\)
• Total per negative: \(\$4 + \$2 = \$6\)

Step 3: Calculate total cost for all negatives
Total cost for 2 negatives: \(2 \times \$6 = \$12\)

Step 4: Test answer choices
Using \(\mathrm{x} = 2\) and \(\mathrm{y} = 3\):

1. \(4\mathrm{xy} + (\mathrm{x} - 1)\mathrm{y} = 4(2)(3) + (2-1)(3) = 24 + 3 = 27\) ❌
2. \(4\mathrm{xy} + (\mathrm{y} - 1)\mathrm{x} = 4(2)(3) + (3-1)(2) = 24 + 4 = 28\) ❌
3. \(4\mathrm{x} + \mathrm{xy} = 4(2) + (2)(3) = 8 + 6 = 14\) ❌
4. \(4\mathrm{x} + \mathrm{xy} - 1 = 4(2) + (2)(3) - 1 = 8 + 6 - 1 = 13\) ❌
5. \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1) = 4(2) + 2(3-1) = 8 + 2(2) = 8 + 4 = 12\) ✓

Answer: E