We have the system of linear equations:

\[
5x - 12y = 1 \tag{1}
\]
\[
13x + 9y = 16 \tag{2}
\]

### Step 1: Multiply the equations to eliminate one variable.

We'll eliminate \( y \). To do this, multiply equation (1) by 3 and equation (2) by 4 so that the coefficients of \( y \) in both equations become opposites:

\[
(3) \times (5x - 12y) = 3 \times 1 \quad \Rightarrow \quad 15x - 36y = 3 \tag{3}
\]
\[
(4) \times (13x + 9y) = 4 \times 16 \quad \Rightarrow \quad 52x + 36y = 64 \tag{4}
\]

### Step 2: Add the two new equations.

Adding equation (3) and equation (4) will eliminate \( y \):

\[
(15x - 36y) + (52x + 36y) = 3 + 64
\]
\[
15x + 52x = 67x = 67
\]

### Step 3: Solve for \( x \).

\[
x = \frac{67}{67} = 1
\]

### Step 4: Substitute \( x = 1 \) into one of the original equations.

Substitute into equation (1):

\[
5(1) - 12y = 1
\]
\[
5 - 12y = 1
\]
\[
-12y = 1 - 5 = -4
\]
\[
y = \frac{-4}{-12} = \frac{1}{3}
\]

### Final Answer:
\[
x = 1, \quad y = \frac{1}{3}
\]

Would you like any further details or have any questions?

### Related Questions:
1. How can you solve this system using substitution instead of elimination?
2. What are the graphical interpretations of the solutions for these equations?
3. How can you solve this system using matrices (Gaussian elimination)?
4. How does changing the coefficients affect the solution to the system?
5. How would the solution change if we altered one of the constants on the right-hand side?

### Tip:
Always check your solution by substituting both values of \(x\) and \(y\) back into the original equations to verify correctness.

Learn how to solve the system of linear equations 5x - 12y = 1 and 13x + 9y = 16 using the elimination method. Follow a step-by-step breakdown of eliminating variables, solving for x and y, and verifying the solution. Suitable for high school students (Grades 9-12).

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