We are given the equation $\frac{1}{x+iy} + \frac{1}{1+3i} = 1$, where $x$ and $y$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$). The problem is to find the values of $x$ and $y$ that satisfy this equation.

1. Problem Description

We are given the equation $\frac{1}{x+iy} + \frac{1}{1+3i} = 1$, where $x$ and $y$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$). The problem is to find the values of $x$ and $y$ that satisfy this equation.

2. Solution Steps

First, isolate the term with $x$ and $y$:

$$\frac{1}{x+iy} = 1 - \frac{1}{1+3i}$$

Next, simplify the right side of the equation:

$$\frac{1}{x+iy} = \frac{(1+3i) - 1}{1+3i} = \frac{3i}{1+3i}$$

Now, take the reciprocal of both sides:

$$x+iy = \frac{1+3i}{3i}$$

To simplify the right side, multiply the numerator and denominator by the conjugate of the denominator, which is $-3i$:

$$x+iy = \frac{(1+3i)(-3i)}{(3i)(-3i)} = \frac{-3i - 9i^2}{-9i^2}$$

Since $i^2 = -1$, we have:

$$x+iy = \frac{-3i + 9}{9} = \frac{9-3i}{9} = \frac{9}{9} - \frac{3i}{9} = 1 - \frac{1}{3}i$$

Equating the real and imaginary parts, we get:

$$x = 1$$

$$y = -\frac{1}{3}$$

3. Final Answer

$x = 1$ and $y = -\frac{1}{3}$