The problem is:

Solve for $x$ and $y$: $3 \log_2(x) + \log_3(y) = -5$ $- \log_2(x) - \log_3(y) = 1$

To solve this system of equations, let's denote:

1. $a = \log_2(x)$
2. $b = \log_3(y)$

Then the equations become:

1. $3a + b = -5$
2. $-a - b = 1$

We can solve this system of linear equations for $a$ and $b$.

1. Step 1: Add the two equations to eliminate $b$. $(3a + b) + (-a - b) = -5 + 1$ Simplifying, we get: $2a = -4 \Rightarrow a = -2$

2. Step 2: Substitute $a = -2$ into one of the equations to solve for $b$. Using $-a - b = 1$: $-(-2) - b = 1$ $2 - b = 1 \Rightarrow b = 1$

Now we have: $a = -2 \quad \text{and} \quad b = 1$

3. Step 3: Convert $a$ and $b$ back to $x$ and $y$.
   • Since $a = \log_2(x) = -2$, we have: $x = 2^{-2} = \frac{1}{4}$
   • Since $b = \log_3(y) = 1$, we have: $y = 3^1 = 3$

Thus, the solution is: $x = \frac{1}{4} \quad \text{and} \quad y = 3$

Would you like further details on any part of this solution, or do you have any questions?

Here are 5 related questions for further exploration:

1. How would you solve this system if the logarithmic bases were different?
2. What if $\log_2(x)$ and $\log_3(y)$ had additional coefficients? How would that affect the solution? 3