For which of the following system of equations, x = 6, y = -4 is the solution?
(I) `(1)/(2x)-(1)/(y)=-1` and `(1)/(x)+(1)/(2y)=8`
(II) `(2)/(x)+(2)/(3y)=1/6` and `(3)/(x)+(2)/(y)=0`

Options

A. Only (I)
B. Only (II)
C. Both (I) and (II)
D. Neither (I) nor (II)

Answer

To determine which system of equations has the solution \( x = 6 \) and \( y = -4 \), we will check both systems of equations one by one.

### System of Equations I:
1. **Equations:**
\[ \frac{1}{2x} - \frac{1}{y} = -1 \]
\[ \frac{1}{x} + \frac{1}{2y} = 8 \]

2. **Substituting \( x = 6 \) and \( y = -4 \):**
- For the first equation:
\[ \frac{1}{2(6)} - \frac{1}{-4} = -1 \]
Simplifying:
\[ \frac{1}{12} + \frac{1}{4} = -1 \]
Finding a common denominator (12):
\[ \frac{1}{12} + \frac{3}{12} = -1 \]
\[ \frac{4}{12} = -1 \quad \text{(This is false)} \]

- For the second equation:
\[ \frac{1}{6} + \frac{1}{2(-4)} = 8 \]
Simplifying:
\[ \frac{1}{6} - \frac{1}{8} = 8 \]
Finding a common denominator (24):
\[ \frac{4}{24} - \frac{3}{24} = 8 \]
\[ \frac{1}{24} = 8 \quad \text{(This is also false)} \]

### Conclusion for System I:
Neither equation is satisfied, so this system does not have \( x = 6 \) and \( y = -4 \) as a solution.

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### System of Equations II:
1. **Equations:**
\[ \frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \]
\[ \frac{3}{x} + \frac{2}{y} = 0 \]

2. **Substituting \( x = 6 \) and \( y = -4 \):**
- For the first equation:
\[ \frac{2}{6} + \frac{2}{3(-4)} = \frac{1}{6} \]
Simplifying:
\[ \frac{1}{3} - \frac{1}{6} = \frac{1}{6} \]
Finding a common denominator (6):
\[ \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \]
\[ \frac{1}{6} = \frac{1}{6} \quad \text{(This is true)} \]

- For the second equation:
\[ \frac{3}{6} + \frac{2}{-4} = 0 \]
Simplifying:
\[ \frac{1}{2} - \frac{1}{2} = 0 \]
\[ 0 = 0 \quad \text{(This is also true)} \]

### Conclusion for System II:
Both equations are satisfied, so this system has \( x = 6 \) and \( y = -4 \) as a solution.

### Final Answer:
The correct option is **II**.

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