The equation \( 13x + 11y = 1 \) is a linear Diophantine equation, which means we are looking for integer solutions for \( x \) and \( y \).
To find a particular solution, we can use the Extended Euclidean Algorithm to express \( 1 \) as a linear combination of \( 13 \) and \( 11 \).
Step 1: Apply Euclidean Algorithm
First, we can find the greatest common divisor (gcd) of \( 13 \) and \( 11 \):
\[ 13 = 1 \cdot 11 + 2 \] \[ 11 = 5 \cdot 2 + 1 \] \[ 2 = 2 \cdot 1 + 0 \]
Thus, \( \gcd(13, 11) = 1 \).
Step 2: Back Substitute to find coefficients
Now, we backtrack to express \( 1 \) as a combination of \( 13 \) and \( 11 \):
- From the second step above: \[ 1 = 11 - 5 \cdot 2 \]
- Substitute \( 2 \) from the first step: \[ 2 = 13 - 1 \cdot 11 \] Putting this into the previous equation gives: \[ 1 = 11 - 5(13 - 1 \cdot 11) = 11 - 5 \cdot 13 + 5 \cdot 11 = 6 \cdot 11 - 5 \cdot 13 \]
Thus, one particular solution to \( 13x + 11y = 1 \) is \( x = -5 \) and \( y = 6 \).
Step 3: General solution
The general solution to the equation \( 13x + 11y = 1 \) can be expressed as: \[ x = -5 + 11n \] \[ y = 6 - 13n \] where \( n \) is any integer.
Example Solutions
- For \( n = 0 \): \( (x, y) = (-5, 6) \)
- For \( n = 1 \): \( (x, y) = (6, -7) \)
- For \( n = -1 \): \( (x, y) = (-16, 19) \)
You can choose any integer \( n \) to generate different integer solutions to the equation.