Final answer:
The system of congruences x ≃ 2 (mod 6) and x ≃ 3 (mod 9) has no solutions, as shown by the fact that no integer can simultaneously satisfy both conditions due to their coefficients' common factor of 3, which doesn't align with the difference of the constants (1) not being a multiple of 3.
Step-by-step explanation:
To determine if the system of congruences has solutions, we should look at the moduli 6 and 9 to see if any number satisfies both congruences simultaneously. If a solution exists, it should satisfy both x ≃ 2 (mod 6) and x ≃ 3 (mod 9).
A number that is congruent to 2 modulo 6 will have the form 6k + 2, where k is an integer. Similarly, a number congruent to 3 modulo 9 will have the form 9m + 3, where m is an integer. For these two forms to represent the same number, they must be equal:
6k + 2 = 9m + 3
Rearranging the equation to isolate the constants on one side, we get:
6k - 9m = 1
Looking at the coefficients of k and m, we have multiples of 6 and 9, respectively. Since these coefficients have a common factor of 3, any linear combination of these will also be a multiple of 3. However, the right side of this equation, which is 1, is not a multiple of 3. This contradiction shows that it is impossible for any integer x to satisfy both congruences at the same time, meaning that there are no solutions to the system of congruences.