Thus the general solution is given by |
x = -6 + (15 / 3) · t |
and |
y = 9 - (21 / 3) · t | where t  Z. |
Hence |
x = -6 + 5 · t |
and |
y = 9 - 7 · t | where t Z. |
1. We know from the previous section that gcd(21, 15) = 3, and that x = -6
and y = 9 is one solution to the linear Diophantine equation
21x + 15y = 9.
Let x0 = -6, y0 = 9, a = 21, b = 15 and d = 3, and apply Theorem 3.10.1.
Thus the general solution is given by |
x = -6 + (15 / 3) · t |
and |
y = 9 - (21 / 3) · t | where t Z. |
Hence |
x = -6 + 5 · t |
and |
y = 9 - 7 · t | where t Z. |
To check whether there are any solutions in which both x and y are positive we must solve the inequalities:
| -6 + 5 · t > 0 | and | 9 - 7 · t > 0 |
| 5 · t > 6 | -7 · t > -9 | |
| t > 1.2 | t < 1.3 |
There is no value of t which satisfies both these inequalities and hence there are no solutions in which both x and y are positive.
2. We know from the previous section that gcd(158, 57) = 1, and that m =
-440 000 and n = 1 220 000 is one solution to the linear Diophantine equation
158m + 57n = 20 000.
Let m0 = -440 000, n0 = 1 220 000, a = 158, b = 57 and d = 1, and apply Theorem 3.10.1.
Thus the general solution is given by |
m = -440 000 + (57 / 1) · t |
and |
n = 1 220 000 - (158 / 1) · t | where t Z. |
Hence |
m = -440 000 + 57 · t |
and |
n = 1 220 000 - 158 · t | where t Z. |
To check if there are any solutions in which both m and n are positive we must solve the two inequalities:
| -440 000 + 57 · t > 0 | and | 1 220 000 - 158 · t > 0 |
| 57 · t > 440 000 | -158 · t > -1 220 000 | |
| t > 7719.3 | t < 7721.5 |
Hence, the values t = 7720 and t = 7721 will be the two values of t which will give
solutions in which both m and n are positive; so
m = -440 000 + 57(7720) and n = 1 220 000 - 158(7720) or
m = -440 000 + 57(7721) and n = 1 220 000 - 158(7721).
The two solutions in which both m and n are positive are: m = 40 and n = 240 or m = 97 and n = 82.