Thus the general solution is given by |
x = 260 + (221 / 13) · t |
and |
y = -104 - (91 / 13) · t | where t  Z. |
Hence |
x = 260 + 17 · t |
and |
y = -104 - 7 · t | where t Z. |
1. We know from the previous section that gcd(221, 91) = 13, and that x =
260 and y = -104 is one solution to the linear Diophantine equation
91x + 221y = 676.
Let x0 = 260, y0 = -104, a = 91, b = 221 and d = 13, and apply Theorem 3.10.1.
Thus the general solution is given by |
x = 260 + (221 / 13) · t |
and |
y = -104 - (91 / 13) · t | where t Z. |
Hence |
x = 260 + 17 · t |
and |
y = -104 - 7 · t | where t Z. |
To check if there are any solutions in which both x and y are positive we must solve the two inequalities:
| 260 + 17 · t > 0 | and | -104 - 7 · t > 0 |
| 17 · t > -260 | -7 · t > 104 | |
| t > -15.3 | t < -14.9 |
Hence, the value t = -15 will be the only value of t which will give a solution in
which both x and y are positive;
so x = 260 + 17(-15) and y = -104 - 7(-15).
The only solution in which both x and y are positive is: x = 5 and y = 1.