Solve for ` x and y` :
` (1)/(2x) - (1)/(y) = - 1, (1)/(x) + (1)/(2y)= 8 ( x ne 0 , y ne 0 )`

Options

A. `(6,4)`
B. `(1/6 , 1/4)`
C. `(3,2)`
D. `(1/3 , 1/2)`

Explanation

To solve the equations

\[ \frac{1}{2x} - \frac{1}{y} = -1 \]
\[ \frac{1}{x} + \frac{1}{2y} = 8 \]

we will use substitution to simplify the problem. Let's define:

\[ u = \frac{1}{x} \quad \text{and} \quad v = \frac{1}{y} \]

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

1. From the first equation:
\[ \frac{u}{2} - v = -1 \]
Rearranging gives:
\[ \frac{u}{2} = v - 1 \quad \Rightarrow \quad u = 2(v - 1) \quad \Rightarrow \quad u = 2v - 2 \quad \text{(Equation 1)} \]

2. From the second equation:
\[ u + \frac{v}{2} = 8 \]
Rearranging gives:
\[ u = 8 - \frac{v}{2} \quad \text{(Equation 2)} \]

Now we have two equations in terms of \(u\) and \(v\):

- Equation 1: \(u = 2v - 2\)
- Equation 2: \(u = 8 - \frac{v}{2}\)

Next, we can set these two expressions for \(u\) equal to each other:

\[ 2v - 2 = 8 - \frac{v}{2} \]

To eliminate the fraction, multiply the entire equation by 2:

\[ 4v - 4 = 16 - v \]

Now, rearranging gives:

\[ 4v + v = 16 + 4 \]
\[ 5v = 20 \]
\[ v = 4 \]

Now that we have \(v\), we can substitute it back into Equation 1 to find \(u\):

\[ u = 2(4) - 2 = 8 - 2 = 6 \]

Now we have:

\[ u = 6 \quad \text{and} \quad v = 4 \]

Recalling our definitions of \(u\) and \(v\):

\[ u = \frac{1}{x} \quad \Rightarrow \quad x = \frac{1}{u} = \frac{1}{6} \]
\[ v = \frac{1}{y} \quad \Rightarrow \quad y = \frac{1}{v} = \frac{1}{4} \]

Thus, the solution is:

\[ x = \frac{1}{6}, \quad y = \frac{1}{4} \]