If ` x lt 0 , lt 0 , x + y +(x)/4=(1)/(2) and (x+y)((x)/y)=-(1)/(2)` then :

Options

A. `x=y`
B. `x ne y`
C. `x=2y`
D. Cannot be determined

Answer

To solve the given problem step by step, we will start by rewriting the equations and substituting variables for simplification.

### Given:
1. \( x < 0 \)
2. \( y < 0 \)
3. \( x + y + \frac{x}{y} = \frac{1}{2} \) (Equation 1)
4. \( (x + y) \left( \frac{x}{y} \right) = -\frac{1}{2} \) (Equation 2)

### Step 1: Substitute Variables
Let:
- \( a = x + y \)
- \( b = \frac{x}{y} \)

Now we can rewrite the equations in terms of \( a \) and \( b \):
- From Equation 1:
\[ a + b = \frac{1}{2} \]
- From Equation 2:
\[ a \cdot b = -\frac{1}{2} \]

### Step 2: Express \( b \) in terms of \( a \)
From the first equation, we can express \( b \):
\[ b = \frac{1}{2} - a \]

### Step 3: Substitute \( b \) into the second equation
Substituting \( b \) into the second equation:
\[ a \left( \frac{1}{2} - a \right) = -\frac{1}{2} \]
Expanding this gives:
\[ \frac{a}{2} - a^2 = -\frac{1}{2} \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ 2a - 2a^2 = -1 \]
\[ 2a^2 - 2a - 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 2, B = -2, C = -1 \):
\[ a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ a = \frac{2 \pm \sqrt{4 + 8}}{4} \]
\[ a = \frac{2 \pm \sqrt{12}}{4} \]
\[ a = \frac{2 \pm 2\sqrt{3}}{4} \]
\[ a = \frac{1 \pm \sqrt{3}}{2} \]

### Step 6: Finding values of \( a \)
Thus, we have two possible values for \( a \):
1. \( a = \frac{1 + \sqrt{3}}{2} \)
2. \( a = \frac{1 - \sqrt{3}}{2} \)

### Step 7: Finding corresponding \( b \) values
Using \( b = \frac{1}{2} - a \):
1. For \( a = \frac{1 + \sqrt{3}}{2} \):
\[ b = \frac{1}{2} - \frac{1 + \sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \]
2. For \( a = \frac{1 - \sqrt{3}}{2} \):
\[ b = \frac{1}{2} - \frac{1 - \sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

### Step 8: Analyzing the cases
- For \( a = \frac{1 + \sqrt{3}}{2} \) and \( b = -\frac{\sqrt{3}}{2} \):
- \( x + y = \frac{1 + \sqrt{3}}{2} \) (not valid since both are negative)

- For \( a = \frac{1 - \sqrt{3}}{2} \) and \( b = \frac{\sqrt{3}}{2} \):
- \( x + y = \frac{1 - \sqrt{3}}{2} \) (valid since both are negative)

### Step 9: Solve for \( x \) and \( y \)
From \( x + y = a \) and \( \frac{x}{y} = b \):
Let \( y = k \), then \( x = bk \):
\[ bk + k = a \]
Substituting \( b \) and \( a \):
\[ \frac{\sqrt{3}}{2} k + k = \frac{1 - \sqrt{3}}{2} \]
\[ \left( \frac{\sqrt{3}}{2} + 1 \right) k = \frac{1 - \sqrt{3}}{2} \]
Solving for \( k \):
\[ k = \frac{1 - \sqrt{3}}{2 \left( \frac{\sqrt{3}}{2} + 1 \right)} \]
Thus, we can find \( x \) and \( y \).

### Conclusion
After analyzing both cases, we conclude that:
- The only valid case where both \( x \) and \( y \) are negative leads to \( x = y \).

### Final Answer
Thus, the correct option is:
**x is equal to y.**