To find the general solution, we can use the Chinese Remainder Theorem (CRT):

Let p and q be two coprime integers, and let a and b be any two integers. Then there exists exactly one integer x such that:

x a(mod p)
x b(mod q).
To apply the CRT to our problem, we can first find the solutions modulo 6 (i.e., when y = 0):

x -2(mod 6), which gives us x = 4 or x = 10 (since 6 is a multiple of both 3 and 5). Next, we can solve for y using the second congruence:

y (-32/6 + k*10)(mod 17)
This simplifies to:

y -5(k*10 + 5)(mod 17), where k is an integer. To find all possible values of y, we can use the CRT again with p = 17 and q = 10 (since they are coprime):

Let r be any integer such that:

r -5(k*10 + 5)(mod 17)
Then there exists exactly one integer y such that:

y r(mod 10) and y (-32/6 + k*10)(mod 17). To find all possible values of (x, y), we can use the CRT again with p = 6 and q = 10:

Let s be any integer such that:

s x(mod 6) and s (-32/6 + k*10)(mod 17)
Then there exists exactly one pair (x, y) satisfying both congruences. In general, the solutions to our original linear Diophantine equation are given by:

x = 4k + 10l and y = -5(k*10 + 5) + r, where k, l, and r are integers such that:

-32/6 + k*10 r(mod 17)
and x s(mod 6), where s is any integer satisfying the first CRT. For example, if we choose k = -1, l = 0, and r = 5 (which satisfies the second CRT), then:

x = 4(-1) + 10*0 = -4
y = -5(k*10 + 5) + r = (-5)(-10 + 5) + 5 = 25
So one solution is (x, y) = (-4, 25). To find all possible solutions, we can vary k and l to get different values of x, while keeping the same value for y.