Loading...
If \(\mathrm{x}\), \(\mathrm{y}\), and \(\mathrm{k}\) are positive and \(\mathrm{x}\) is less than \(\mathrm{y}\), then \(\frac{\mathrm{x} + \mathrm{k}}{\mathrm{y} + \mathrm{k}}\) is
Let's start by understanding what we're actually comparing here. We have two fractions that we need to compare:
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
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.
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:
Now we're comparing: \(\mathrm{xy + ky}\) versus \(\mathrm{xy + xk}\)
Process Skill: MANIPULATE - Using cross-multiplication to simplify the fraction comparison
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:
\(\mathrm{(x+k)/(y+k) > x/y}\)
Process Skill: APPLY CONSTRAINTS - Using the given condition x < y to reach the final conclusion
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
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.
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.
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 \(\mathrm{x/y}\).
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 \(\mathrm{x×(y+k)}\).
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.
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.
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.
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).
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:
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 \(\mathrm{x = 2, y = 5, k = 3}\):
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