If x and y are real then the number of ordered pairs `(x,y)` such that `x+y+(x)/(y)=(1)/(2)` and `(x+y)(x)/(y)=-(1)/(2)` is

Options

A. `1`
B. `2`
C. `0`
D. None of these

Answer

To solve the problem, we need to find the number of ordered pairs \((x, y)\) that satisfy the given equations:

1. \(x + y + \frac{x}{y} = \frac{1}{2}\)
2. \((x + y) \frac{x}{y} = -\frac{1}{2}\)

Let's denote:
- \(u = x + y\)
- \(v = \frac{x}{y}\)

Now, we can rewrite the equations in terms of \(u\) and \(v\):

1. \(u + v = \frac{1}{2}\) (Equation 1)
2. \(u \cdot v = -\frac{1}{2}\) (Equation 2)

### Step 1: Solve for \(u\) in terms of \(v\)
From Equation 1, we can express \(u\) as:
\[ u = \frac{1}{2} - v \]

### Step 2: Substitute \(u\) into Equation 2
Now, substitute \(u\) into Equation 2:
\[ \left(\frac{1}{2} - v\right) v = -\frac{1}{2} \]
Expanding this gives:
\[ \frac{1}{2}v - v^2 = -\frac{1}{2} \]
Rearranging the equation:
\[ v^2 - \frac{1}{2}v - \frac{1}{2} = 0 \]

### Step 3: Apply the quadratic formula
Now we will use the quadratic formula to solve for \(v\):
\[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -\frac{1}{2}\), and \(c = -\frac{1}{2}\).

Calculating the discriminant:
\[ b^2 - 4ac = \left(-\frac{1}{2}\right)^2 - 4 \cdot 1 \cdot \left(-\frac{1}{2}\right) = \frac{1}{4} + 2 = \frac{1}{4} + \frac{8}{4} = \frac{9}{4} \]

Now substituting back into the quadratic formula:
\[ v = \frac{\frac{1}{2} \pm \sqrt{\frac{9}{4}}}{2} = \frac{\frac{1}{2} \pm \frac{3}{2}}{2} \]
This gives us two possible values for \(v\):
1. \(v = \frac{4/2}{2} = 1\)
2. \(v = \frac{-2/2}{2} = -\frac{1}{2}\)

### Step 4: Find corresponding values of \(u\)
Now we can find \(u\) for each value of \(v\):

1. If \(v = 1\):
\[ u = \frac{1}{2} - 1 = -\frac{1}{2} \]

2. If \(v = -\frac{1}{2}\):
\[ u = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1 \]

### Step 5: Solve for \(x\) and \(y\) using \(u\) and \(v\)
Now we have two cases to solve for \(x\) and \(y\):

**Case 1:** \(u = -\frac{1}{2}\), \(v = 1\)
\[ x + y = -\frac{1}{2} \quad \text{and} \quad \frac{x}{y} = 1 \implies x = y \]
Let \(x = y\):
\[ 2x = -\frac{1}{2} \implies x = -\frac{1}{4}, \quad y = -\frac{1}{4} \]
So we have one ordered pair: \((-1/4, -1/4)\).

**Case 2:** \(u = 1\), \(v = -\frac{1}{2}\)
\[ x + y = 1 \quad \text{and} \quad \frac{x}{y} = -\frac{1}{2} \implies x = -\frac{1}{2}y \]
Substituting \(x\) in the first equation:
\[ -\frac{1}{2}y + y = 1 \implies \frac{1}{2}y = 1 \implies y = 2, \quad x = -1 \]
So we have another ordered pair: \((-1, 2)\).

### Final Result
Thus, the number of ordered pairs \((x, y)\) that satisfy the given conditions is **2**.