Solve for `x` and `y`
`(5)/(x+y)+(1)/(x-y)=2`
`(15)/(x+y)-(5)/(x-y)=-2`

Answer

To solve the equations

\[ \frac{5}{x+y} + \frac{1}{x-y} = 2 \]
\[ \frac{15}{x+y} - \frac{5}{x-y} = -2 \]

we will use substitution. Let's define:

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

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

1. The first equation becomes:
\[ 5u + v = 2 \tag{1} \]

2. The second equation becomes:
\[ 15u - 5v = -2 \tag{2} \]

### Step 1: Solve for \(u\) and \(v\)

From equation (1), we can express \(v\) in terms of \(u\):

\[ v = 2 - 5u \tag{3} \]

Now, substitute equation (3) into equation (2):

\[ 15u - 5(2 - 5u) = -2 \]

### Step 2: Simplify the equation

Expanding the equation gives:

\[ 15u - 10 + 25u = -2 \]

Combining like terms results in:

\[ 40u - 10 = -2 \]

### Step 3: Solve for \(u\)

Adding 10 to both sides:

\[ 40u = 8 \]

Dividing both sides by 40:

\[ u = \frac{8}{40} = \frac{1}{5} \]

### Step 4: Substitute \(u\) back to find \(v\)

Now substitute \(u\) back into equation (3):

\[ v = 2 - 5\left(\frac{1}{5}\right) = 2 - 1 = 1 \]

### Step 5: Find \(x+y\) and \(x-y\)

Recall that:

\[ u = \frac{1}{x+y} \implies x+y = \frac{1}{u} = \frac{1}{\frac{1}{5}} = 5 \]

\[ v = \frac{1}{x-y} \implies x-y = \frac{1}{v} = \frac{1}{1} = 1 \]

### Step 6: Solve for \(x\) and \(y\)

Now we have a system of equations:

1. \(x + y = 5\) \tag{4}
2. \(x - y = 1\) \tag{5}

Adding equations (4) and (5):

\[ (x+y) + (x-y) = 5 + 1 \]
\[ 2x = 6 \implies x = 3 \]

Now substitute \(x\) back into equation (4):

\[ 3 + y = 5 \implies y = 5 - 3 = 2 \]

### Final Solution

Thus, the solution is:

\[ x = 3, \quad y = 2 \]