If x neq -y, then is (x-y)/(x + y) > 1? x > 0 \(\mathrm{y}...

Understanding the Question

We need to determine whether the fraction \(\frac{\mathrm{x-y}}{\mathrm{x+y}}\) is greater than 1, given that \(\mathrm{x} \neq -\mathrm{y}\).

What We Need to Determine

The question asks: Is \(\frac{\mathrm{x-y}}{\mathrm{x+y}} > 1\)?

This is a yes/no question. We need to determine if we can definitively answer "yes" or "no" to whether this inequality holds true.

Key Insight

For a fraction to be greater than 1, its numerator must exceed its denominator AND both must have the same sign. But here's the crucial catch: when the denominator \(\mathrm{(x+y)}\) is negative, the entire fraction becomes negative and cannot possibly be greater than 1. This makes the sign of \(\mathrm{(x+y)}\) our "master switch."

The given constraint \(\mathrm{x} \neq -\mathrm{y}\) ensures that \(\mathrm{x + y} \neq 0\), so division by zero is not a concern.

Analyzing Statement 1

Statement 1: \(\mathrm{x} > 0\)

This tells us that x is positive, but we know nothing about y.

Let's test different scenarios to see what happens:


Scenario 1: Small negative y (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\))
  

  
\(\mathrm{x + y} = 3 + (-1) = 2\) (positive denominator)
  
\(\mathrm{x - y} = 3 - (-1) = 4\) (positive numerator)
  
The fraction: \(\frac{4}{2} = 2\), which IS > 1 ✓
  

Scenario 2: Large negative y (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\))
  

  
\(\mathrm{x + y} = 1 + (-3) = -2\) (negative denominator)
  
\(\mathrm{x - y} = 1 - (-3) = 4\) (positive numerator)
  
The fraction: \(\frac{4}{-2} = -2\), which is NOT > 1 ✗
  



Since we get different answers ("yes" in Scenario 1, "no" in Scenario 2), Statement 1 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: \(\mathrm{y} < 0\)

This tells us that y is negative, but we know nothing about x.

Testing different scenarios:


Scenario 1: Large positive x (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\))
  

  
\(\mathrm{x + y} = 3 + (-1) = 2\) (positive denominator)
  
\(\mathrm{x - y} = 3 - (-1) = 4\) (positive numerator)
  
The fraction: \(\frac{4}{2} = 2\), which IS > 1 ✓
  

Scenario 2: Small positive x (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\))
  

  
\(\mathrm{x + y} = 1 + (-3) = -2\) (negative denominator)
  
\(\mathrm{x - y} = 1 - (-3) = 4\) (positive numerator)
  
The fraction: \(\frac{4}{-2} = -2\), which is NOT > 1 ✗
  



Again, we get different answers to our yes/no question, so Statement 2 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices B and D (already eliminated).

Combining Statements

Let's see what happens when we use BOTH statements together.

From both statements, we know:

\(\mathrm{x} > 0\) (x is positive)
\(\mathrm{y} < 0\) (y is negative)


The Critical Factor: Relative Magnitudes

Even with both pieces of information, the key question remains: Is x larger than \(|\mathrm{y}|\) or smaller?

This determines the sign of our denominator \(\mathrm{(x+y)}\):

When \(\mathrm{x} > |\mathrm{y}|\): The positive x "wins," so \(\mathrm{(x+y)} > 0\)
When \(\mathrm{x} < |\mathrm{y}|\): The negative y "wins," so \(\mathrm{(x+y)} < 0\)


Let's verify with concrete examples:

Case 1: \(\mathrm{x} > |\mathrm{y}|\) (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\))

\(\mathrm{x + y} = 3 + (-1) = 2 > 0\)
\(\mathrm{x - y} = 3 - (-1) = 4 > 0\)
Fraction: \(\frac{4}{2} = 2 > 1\) ✓
Answer to question: YES


Case 2: \(\mathrm{x} < |\mathrm{y}|\) (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\))

\(\mathrm{x + y} = 1 + (-3) = -2 < 0\)
\(\mathrm{x - y} = 1 - (-3) = 4 > 0\)
Fraction: \(\frac{4}{-2} = -2 < 1\) ✗
Answer to question: NO


Since we can construct examples that give different answers to our yes/no question even with both statements, they are NOT sufficient together.

[STOP - Not Sufficient!] This eliminates choice C.

The Answer: E

Even with both statements combined, we cannot definitively answer whether \(\frac{\mathrm{x-y}}{\mathrm{x+y}} > 1\) because the answer depends on the relative magnitudes of x and \(|\mathrm{y}|\), which we cannot determine.

Answer Choice E: "The statements together are not sufficient."