The problem asks us to find the values of $x$ and $y$ given the following two equations: $\log_2(3x+y) = 1$ $\log_2(\frac{x}{y}) = -2$

1. Problem Description

The problem asks us to find the values of $x$ and $y$ given the following two equations:

$\log_2(3x+y) = 1$

$\log_2(\frac{x}{y}) = -2$

2. Solution Steps

First, we can rewrite the logarithmic equations in exponential form.

For the first equation, $\log_2(3x+y) = 1$, we have:

$3x + y = 2^1$

$3x + y = 2$

For the second equation, $\log_2(\frac{x}{y}) = -2$, we have:

$\frac{x}{y} = 2^{-2}$

$\frac{x}{y} = \frac{1}{2^2}$

$\frac{x}{y} = \frac{1}{4}$

$x = \frac{1}{4}y$

Now we have a system of two linear equations with two variables:

$3x + y = 2$

$x = \frac{1}{4}y$

We can substitute the second equation into the first equation:

$3(\frac{1}{4}y) + y = 2$

$\frac{3}{4}y + y = 2$

$\frac{3}{4}y + \frac{4}{4}y = 2$

$\frac{7}{4}y = 2$

$y = 2 \cdot \frac{4}{7}$

$y = \frac{8}{7}$

Now we can substitute the value of $y$ back into the equation $x = \frac{1}{4}y$:

$x = \frac{1}{4} \cdot \frac{8}{7}$

$x = \frac{8}{28}$

$x = \frac{2}{7}$

3. Final Answer

$x = \frac{2}{7}$

$y = \frac{8}{7}$