You may be mixing up the adorable gato pescador
with el “pez gato”
Okay, let’s look at those equations. The second one obviously has no solutions since 3 divides the left hand side and not the right. Don’t need to know much for that one. As for 3x+2y=1, that is the same as x=\frac{1-2y}3 so that in any solution, it must be the case that 3|1-2y (the vertical line means divides), and the only way that can happen is that y=3k+1 for an integer k. How do I know that? Divide y by 3 and call the remainder the quotient k and the remainder r. When r=0 or r=2, it is easy to see that 1-2y is not divisible by 3 and when r=1 it is. Then y=3k+1, we see that 1-2y=-3-6k and then x=\frac{1-2k}3=-1-2k and that gives all the solutions. Had the second equation been 123x+457y=6, the same methods would have worked, provided you knew the prime divisors of the coefficients, although more tedious.
Without looking at any other answers:
Set up
Number of dogs, cats, mice, which are all integers:
(1) 1 <= d < 100
(2) 1 <= c < 100
(3) 1 <= m < 100
Limit on number:
(4) 100 = d + c + m
Limit on cost:
(5) 100 = 15d + c + \frac{1}{4}m
Solution
Eq(5) multiplied by 4:
(6) 400 = 60d + 4c +m
Eq(6) minus Eq(4):
(7) 300 = 59d + 3c
Eq(7) solved for c:
(8) c = 100 - \frac{59}{3}d
Eq(3) solved for m:
(9) m = 100 - d - c
Eq(9) with Eq(8) substituted in:
(10) m = 100 - d - (100 - \frac{59}{3}d) = \frac{56}{3}d
From Eq(8), Eq(10), d is an integer:
(11) d = 3n for some integer n.
Eq(10) with Eq(11) substituted in:
(12) m = 56n
Rel(3) with Eq(12) substituted in:
(13) 1 <= 56n < 100
By inspection, only one integer satisfies Rel(13):
(14) n = 1
Eq(12) with Eq(14) substituted in:
(15) m = 56(1) = 56
Eq(11) with Eq(14) substituted in:
(16) d = 3(1) = 3
Eq(8) with Eq(16) substituted in:
(17) c = 100 - \frac{59}{3}(3) = 41
Therefore, 3 dogs, 41 cats, and 56 mice.
Working modulo 3, if 1-2y is divisible by 3, then y=3k+2 [not 1]; I guess we can call that algebra in the field \mathbb{Z}/3\mathbb{Z}. We get the set of solutions (x,y) = (-2k-1,3k+2).
If the numbers are big— let’s take 123x+457y=6, the same methods work— in finding the modular multiplicative inverse we might want to use Euclid’s algorithm, which gives 123\cdot(-26)+457\cdot7=1. So, now, multiplying through by 6, we get (x,y)=(457k-156,-123k+42).