We are given the system of equations:

$\frac{1}{x} + \frac{1}{y} = \frac{1}{12} \quad \text{(1)}$ $\frac{1}{y} + \frac{1}{2x} = \frac{1}{15} \quad \text{(2)}$

We need to solve for $x$ and $y$.

Step 1: Express $\frac{1}{x}$ from Equation (1)

From Equation (1), we isolate $\frac{1}{x}$:

$\frac{1}{x} = \frac{1}{12} - \frac{1}{y} \quad \text{(3)}$

Step 2: Substitute Equation (3) into Equation (2)

Now substitute $\frac{1}{x}$ from Equation (3) into Equation (2):

$\frac{1}{y} + \frac{1}{2} \left( \frac{1}{12} - \frac{1}{y} \right) = \frac{1}{15}$

Simplify this equation:

$\frac{1}{y} + \frac{1}{24} - \frac{1}{2y} = \frac{1}{15}$

Step 3: Simplify the equation

To simplify, first get a common denominator for the terms involving $y$:

$\frac{1}{y} - \frac{1}{2y} = \frac{2 - 1}{2y} = \frac{1}{2y}$

Now, substitute back:

$\frac{1}{2y} + \frac{1}{24} = \frac{1}{15}$

Step 4: Solve for $y$

Subtract $\frac{1}{24}$ from both sides:

$\frac{1}{2y} = \frac{1}{15} - \frac{1}{24}$

Find a common denominator to subtract:

$\frac{1}{15} - \frac{1}{24} = \frac{8}{120} - \frac{5}{120} = \frac{3}{120} = \frac{1}{40}$

Thus:

$\frac{1}{2y} = \frac{1}{40}$

Multiply both sides by 2:

$\frac{1}{y} = \frac{1}{20}$

So, $y = 20$.

Step 5: Solve for $x$

Now that we know $y = 20$, substitute this back into Equation (1):

$\frac{1}{x} + \frac{1}{20} = \frac{1}{12}$

Subtract $\frac{1}{20}$ from both sides:

$\frac{1}{x} = \frac{1}{12} - \frac{1}{20}$

Again, find a common denominator:

$\frac{1}{12} - \frac{1}{20} = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30}$

Thus:

$\frac{1}{x} = \frac{1}{30}$

So, $x = 30$.

Final Answer:

The solution is: $x = 30, \quad y = 20$

Would you like more details or further clarification?

Related Questions:

1. How do you solve a system of equations using substitution?
2. What are other methods for solving systems of equations (e.g., elimination)?
3. How do you handle equations with fractions effectively?
4. How would this problem change if different denominators were involved?
5. How can you verify solutions to a system of equations?

Tip:

When dealing with fractions in equations, it often helps to clear them early by multiplying through by the least common denominator.