If x, y, and k are positive and x is less than y, then (x + k)/(y + k) is

1. Translate the problem requirements: We need to compare two fractions: \(\mathrm{x/y}\) and \(\mathrm{(x+k)/(y+k)}\), where x, y, and k are all positive and \(\mathrm{x < y}\). The question asks how adding the same positive value k to both numerator and denominator changes the fraction.
2. Set up the comparison framework: Establish what it means for \(\mathrm{(x+k)/(y+k)}\) to be greater than, less than, or equal to \(\mathrm{x/y}\) by setting up the inequality \(\mathrm{(x+k)/(y+k)}\) compared to \(\mathrm{x/y}\).
3. Apply cross-multiplication to eliminate fractions: Cross-multiply to get \(\mathrm{(x+k)·y}\) compared to \(\mathrm{x·(y+k)}\), which simplifies the comparison without dealing with fraction arithmetic.
4. Simplify and analyze the relationship: Expand the products and use the given constraint that \(\mathrm{x < y}\) to determine which expression is larger, leading directly to the answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're actually comparing here. We have two fractions that we need to compare:

• The original fraction: \(\mathrm{x/y}\)
• The new fraction after adding k to both parts: \(\mathrm{(x+k)/(y+k)}\)

We know that x, y, and k are all positive numbers, and crucially, x is less than y. This means our original fraction \(\mathrm{x/y}\) is less than 1 (since the numerator is smaller than the denominator).

The question is asking: when we add the same positive number k to both the top and bottom of a fraction, what happens to the value of that fraction?

Process Skill: TRANSLATE - Converting the abstract relationship into a clear comparison question

2. Set up the comparison framework

To determine the relationship between these fractions, let's think about what it means for one fraction to be greater than another.

We want to know whether \(\mathrm{(x+k)/(y+k)}\) is greater than, less than, or equal to \(\mathrm{x/y}\).

In mathematical terms, we're asking whether:
\(\mathrm{(x+k)/(y+k) \, ? \, x/y}\)

where the ? represents the relationship we need to determine (>, <, or =).

To compare fractions effectively, we'll use cross-multiplication, which is a reliable method that eliminates the fractions and makes the comparison clearer.

3. Apply cross-multiplication to eliminate fractions

Let's use cross-multiplication to compare \(\mathrm{(x+k)/(y+k)}\) and \(\mathrm{x/y}\).

When we cross-multiply, we're comparing:
\(\mathrm{(x+k) \times y}\) versus \(\mathrm{x \times (y+k)}\)

Since all our variables (x, y, k) are positive, we don't need to worry about flipping any inequality signs during this process.

So our comparison becomes:
\(\mathrm{(x+k) \times y \, ? \, x \times (y+k)}\)

Let's expand both sides:

• Left side: \(\mathrm{(x+k) \times y = xy + ky}\)
• Right side: \(\mathrm{x \times (y+k) = xy + xk}\)

Now we're comparing: \(\mathrm{xy + ky}\) versus \(\mathrm{xy + xk}\)

Process Skill: MANIPULATE - Using cross-multiplication to simplify the fraction comparison

4. Simplify and analyze the relationship

Since both sides have xy, we can subtract it from both sides to focus on the difference:

\(\mathrm{xy + ky}\) versus \(\mathrm{xy + xk}\)

Subtracting xy from both sides:
\(\mathrm{ky}\) versus \(\mathrm{xk}\)

Since k is positive, we can divide both sides by k:
\(\mathrm{y}\) versus \(\mathrm{x}\)

Now we use our given constraint: we know that \(\mathrm{x < y}\), which means \(\mathrm{y > x}\).

Working backwards through our logic:

• Since \(\mathrm{y > x}\), we have \(\mathrm{ky > xk}\)
• This means \(\mathrm{xy + ky > xy + xk}\)
• Therefore \(\mathrm{(x+k) \times y > x \times (y+k)}\)
• Since we cross-multiplied from \(\mathrm{(x+k)/(y+k)}\) versus \(\mathrm{x/y}\), this tells us:

\(\mathrm{(x+k)/(y+k) > x/y}\)

Process Skill: APPLY CONSTRAINTS - Using the given condition x < y to reach the final conclusion

Final Answer

We have proven that \(\mathrm{(x+k)/(y+k)}\) is greater than \(\mathrm{x/y}\).

This makes intuitive sense: when we add the same positive amount to both the numerator and denominator of a fraction that's less than 1, we're adding a larger proportion to the smaller numerator than to the larger denominator, which increases the overall value of the fraction.

The answer is B. greater than x/y

Common Faltering Points

Errors while devising the approach

1. Overlooking the constraint that x < y

Students often focus on the fact that all variables are positive but miss the crucial constraint that x is less than y. This constraint is essential because it determines that \(\mathrm{x/y < 1}\), which is key to understanding how adding k affects the fraction. Without recognizing this constraint, students might think the relationship could vary depending on the values.

2. Attempting to substitute specific numbers without understanding the general relationship

Some students immediately try to plug in specific values (like \(\mathrm{x=1, y=2, k=1}\)) without first establishing the algebraic relationship. While this can work, it doesn't provide the comprehensive understanding needed for similar problems and may lead to incomplete analysis if they don't test enough cases.

3. Confusion about what relationship needs to be determined

Students might misunderstand what they're comparing - they may think they need to compare \(\mathrm{(x+k)/(y+k)}\) to 1, or compare it to some other expression, rather than clearly identifying that they need to compare \(\mathrm{(x+k)/(y+k)}\) with \(\mathrm{x/y}\).

Errors while executing the approach

1. Sign errors during cross-multiplication

When cross-multiplying \(\mathrm{(x+k)/(y+k)}\) versus \(\mathrm{x/y}\), students might incorrectly flip inequality signs or make algebraic errors in the expansion, leading to \(\mathrm{(x+k)×y}\) being compared incorrectly to \(\mathrm{x×(y+k)}\).

2. Incorrect algebraic simplification

After expanding to \(\mathrm{xy + ky}\) versus \(\mathrm{xy + xk}\), students might make errors when canceling xy from both sides, or they might incorrectly divide by k, forgetting that since k is positive, this operation preserves the inequality direction.

3. Misapplying the constraint x < y

Even after correctly simplifying to \(\mathrm{ky}\) versus \(\mathrm{xk}\), students might incorrectly conclude which is larger, either by confusing x and y or by incorrectly reasoning about how the constraint \(\mathrm{x < y}\) affects the final comparison.

Errors while selecting the answer

1. Selecting the opposite relationship

Students might correctly determine that \(\mathrm{ky > xk}\) but then incorrectly trace back through their work, concluding that \(\mathrm{(x+k)/(y+k) < x/y}\) instead of the correct \(\mathrm{(x+k)/(y+k) > x/y}\), thus choosing option D instead of B.

2. Choosing the variable answer option E

Students who are unsure of their algebraic work might default to option E, thinking that the relationship depends on the specific value of k, when in fact the relationship is always the same regardless of k's value (as long as k is positive).

Alternate Solutions

Smart Numbers Approach

Step 1: Choose smart numbers that satisfy the constraints

We need x, y, and k to be positive with \(\mathrm{x < y}\). Let's choose values that make calculations simple:

• \(\mathrm{x = 1}\)
• \(\mathrm{y = 2}\) (satisfies \(\mathrm{x < y}\))
• \(\mathrm{k = 1}\) (positive value)

Step 2: Calculate the original fraction x/y

\(\mathrm{x/y = 1/2 = 0.5}\)

Step 3: Calculate the new fraction (x+k)/(y+k)

\(\mathrm{(x+k)/(y+k) = (1+1)/(2+1) = 2/3 ≈ 0.667}\)

Step 4: Compare the fractions

Since \(\mathrm{2/3 > 1/2}\), we have \(\mathrm{(x+k)/(y+k) > x/y}\)

Step 5: Verify with different smart numbers

Let's try \(\mathrm{x = 2, y = 5, k = 3}\):

• Original: \(\mathrm{x/y = 2/5 = 0.4}\)
• New: \(\mathrm{(x+k)/(y+k) = (2+3)/(5+3) = 5/8 = 0.625}\)
• Again, \(\mathrm{5/8 > 2/5}\), confirming \(\mathrm{(x+k)/(y+k) > x/y}\)

Step 6: Understanding why this always works

When we add the same positive number k to both numerator and denominator of a proper fraction (where \(\mathrm{x < y}\)), we're adding a larger proportion to the smaller numerator than to the larger denominator, which increases the overall fraction value.

Answer: B - greater than x/y