What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x...

1. Translate the problem requirements: We need values of x where there exists a corresponding y such that both \(\mathrm{xy > 0}\) (x and y have the same sign) AND \(\mathrm{xy = x + y}\) (the product equals the sum)
2. Solve for y in terms of x: Use the equation \(\mathrm{xy = x + y}\) to express y as a function of x, determining when this relationship is valid
3. Apply the sign constraint: Use the condition \(\mathrm{xy > 0}\) to determine which values of x allow for valid y values that satisfy both conditions
4. Verify the solution set: Test boundary values and confirm the range of x that works

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We need to find values of x where we can find a corresponding y such that BOTH of these things happen:

• \(\mathrm{xy > 0}\): This means x and y have the same sign (both positive or both negative)
• \(\mathrm{xy = x + y}\): The product of x and y equals their sum

Think of it this way: imagine you have two numbers x and y. When you multiply them together, you get the same result as when you add them together. And on top of that, both numbers must have the same sign.

Process Skill: TRANSLATE - Converting the mathematical conditions into clear requirements we can work with

2. Solve for y in terms of x

Now let's use the equation \(\mathrm{xy = x + y}\) to find y in terms of x. This will help us understand when such a y can exist.

Starting with: \(\mathrm{xy = x + y}\)
Let's rearrange this to get all terms with y on one side:
\(\mathrm{xy - y = x}\)
Factor out y: \(\mathrm{y(x - 1) = x}\)
So: \(\mathrm{y = \frac{x}{x - 1}}\)

But wait! This formula only works when \(\mathrm{x \neq 1}\), because we can't divide by zero when \(\mathrm{x - 1 = 0}\).

Let's check what happens when \(\mathrm{x = 1}\): If \(\mathrm{x = 1}\), our original equation \(\mathrm{xy = x + y}\) becomes \(\mathrm{1 \cdot y = 1 + y}\), which gives us \(\mathrm{y = 1 + y}\). This is impossible since it would mean \(\mathrm{0 = 1}\).

So for any valid solution, we need \(\mathrm{x \neq 1}\), and \(\mathrm{y = \frac{x}{x - 1}}\).

Process Skill: MANIPULATE - Algebraic rearrangement to express y in terms of x

3. Apply the sign constraint

Now we use the condition \(\mathrm{xy > 0}\), which means x and y must have the same sign.

We know \(\mathrm{y = \frac{x}{x - 1}}\), so \(\mathrm{xy = x \cdot \frac{x}{x - 1} = \frac{x^2}{x - 1}}\)

For \(\mathrm{xy > 0}\), we need \(\mathrm{\frac{x^2}{x - 1} > 0}\).

Since \(\mathrm{x^2}\) is always positive (except when \(\mathrm{x = 0}\)), the sign of this expression depends entirely on the sign of \(\mathrm{(x - 1)}\):

• If \(\mathrm{x - 1 > 0}\) (meaning \(\mathrm{x > 1}\)), then \(\mathrm{\frac{x^2}{x - 1} > 0}\) ✓
• If \(\mathrm{x - 1 < 0}\) (meaning \(\mathrm{x < 1}\)), then \(\mathrm{\frac{x^2}{x - 1} < 0}\) ✗

But wait, we also need to check what happens when \(\mathrm{x = 0}\):
If \(\mathrm{x = 0}\), then \(\mathrm{y = \frac{0}{0-1} = 0}\), and \(\mathrm{xy = 0 \cdot 0 = 0}\), which doesn't satisfy \(\mathrm{xy > 0}\).

Therefore, we need \(\mathrm{x > 1}\) for both conditions to be satisfied.

Process Skill: APPLY CONSTRAINTS - Using the sign requirement to determine valid x values

4. Verify the solution set

Let's test our conclusion with a concrete example:

Try \(\mathrm{x = 2}\): Then \(\mathrm{y = \frac{2}{2-1} = \frac{2}{1} = 2}\)
Check: \(\mathrm{xy = 2 \cdot 2 = 4}\), and \(\mathrm{x + y = 2 + 2 = 4}\) ✓
Also: \(\mathrm{xy = 4 > 0}\) ✓

Try \(\mathrm{x = 0.5}\): Then \(\mathrm{y = \frac{0.5}{0.5-1} = \frac{0.5}{-0.5} = -1}\)
Check: \(\mathrm{xy = 0.5 \cdot (-1) = -0.5}\), and \(\mathrm{x + y = 0.5 + (-1) = -0.5}\) ✓
But: \(\mathrm{xy = -0.5 < 0}\) ✗

This confirms that \(\mathrm{x > 1}\) is indeed the correct condition.

5. Final Answer

The values of x that satisfy both conditions are \(\mathrm{x > 1}\).

Looking at the answer choices, this corresponds to choice D. x > 1.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint xy > 0

Students often misunderstand what \(\mathrm{xy > 0}\) means. They might think it simply means "x and y are both positive" instead of understanding that it means "x and y have the same sign" (both positive OR both negative). This leads them to only consider positive values and miss part of the analysis.

2. Treating the conditions as separate instead of simultaneous

Some students approach this by first finding all x where \(\mathrm{xy > 0}\) could be satisfied, then separately finding where \(\mathrm{xy = x + y}\) could be satisfied, rather than recognizing that we need to find x values where BOTH conditions can be satisfied by the SAME y value simultaneously.

3. Forgetting to check for domain restrictions

Students may dive straight into algebraic manipulation without considering whether certain values of x might make the problem undefined or impossible to solve, particularly when \(\mathrm{x = 1}\) makes the denominator zero.

Errors while executing the approach

1. Algebraic manipulation errors when solving for y

When rearranging \(\mathrm{xy = x + y}\) to get \(\mathrm{y = \frac{x}{x-1}}\), students commonly make errors such as incorrectly factoring out y or making sign errors. They might get \(\mathrm{y = \frac{x}{1-x}}\) instead of \(\mathrm{y = \frac{x}{x-1}}\).

2. Incorrect sign analysis for the inequality

When analyzing \(\mathrm{\frac{x^2}{x-1} > 0}\), students often forget that \(\mathrm{x^2}\) is always non-negative, and incorrectly conclude that both the numerator and denominator need to be analyzed for sign changes. They may create incorrect sign charts or miss the fact that \(\mathrm{x = 0}\) makes the entire expression equal to 0.

3. Missing the special case x = 0

Even after correctly finding that \(\mathrm{x > 1}\) works, students often forget to explicitly check what happens when \(\mathrm{x = 0}\), since \(\mathrm{x^2 = 0}\) makes the analysis different from other cases where \(\mathrm{x^2 > 0}\).

Errors while selecting the answer

1. Including the boundary point x = 1

After determining that we need \(\mathrm{x - 1 > 0}\), some students incorrectly conclude that \(\mathrm{x \geq 1}\) satisfies the conditions, forgetting that \(\mathrm{x = 1}\) was specifically excluded because it makes the original equation impossible to solve.

2. Confusing which inequality direction to use

Students might correctly identify that the boundary is at \(\mathrm{x = 1}\) but then select \(\mathrm{x < 1}\) instead of \(\mathrm{x > 1}\), especially if they made a sign error earlier and didn't catch it through verification.