Find all integer solutions to $15x + 12y + 30z = 24$ This problem is from Burton pg. 44. The hint given is this:

Put $y = 3s-5t$ and $z = -s + 2t$

First Question
Why does this hint work? How did it come to be? It doesn't seem to stem from the euclidean algorithm or from $\gcd (12, 30)$.
Second Question
Even so, how does one proceed? Using the given substitution doesn't appear to give integer solutions, at least according to what I have done (which is probably, most definitely not the right approach). I went from $3x + 2s - 8t = 8$ to try and substitute something for $x,s$ which led to $x = 1 +2f \: \:, \: y= -1 + 3f \: \:$ ($f$ is any integer). But this leads to $t$ having a fractional value. 
Any help is greatly appreciated.
 A: The hint reduces the equation to
$$
5x+2s=8
$$
which can be solved using the Euclidean algorithm, or noticing that $x=0$ and $s=4$ is a particular solution. Then we get that $x=2u$ and $s=4-5u$ is the general solution.
There are other reductions. I think the following is just as simple and maybe a bit more intuitive.

Factoring $3$ out of the equation gives
$$
5x+4y+10z=8
$$
If we set $w=x+2z$, all we really need to solve is
$$
5w+4y=8
$$
One solution is $w=0$ and $y=2$, which means the general solution is
$$
w=4t\qquad\text{and}\qquad y=2-5t
$$
Therefore, we get the general solution to the original equation to be
$$
x=4t-2z\qquad\text{and}\qquad y=2-5t
$$
A: The hint looks arbitrary to me. That's not how I would attack this problem.
After $$ 5x + 4y + 10z = 8$$ I would take advantage of "$5$" and "$10$" like so: reducing modulo $5$, it must be true that $ 4y \equiv 8 \mod 5$, or $ y \equiv 2 \mod 5$. Write $y = 2 + 5k$, substitute in the original equation, divide by $5$, and that gives you $x$ in terms of $k$ and $z$. Then $k$ and $z$ are free parameters (they can be any integers), and you have $x$ and $y$ expressed in terms of these parameters (and $z$, of course, is itself).
A: Let $ux+vy+wz=c$, with $u,v,w,c\in \mathbb{Z}$ a general such equation, to be solved in integers. Suppose that $uvw\not =0$ to simplify.
1) Put $v=dv_1$, $w=dw_1$, with $v_1$ and $w_1$ coprime.
2) Find $v_2,w_2\in \mathbb{Z}$ such that $w_2v_1-v_2w_1=1$ (they exists by Bezout).
3) Put $u_1y+w_1z=s$, $v_2y+w_2z=t$. Then we can solve this system, obtaining $y=a_1s+b_1t$, $z=a_2s+b_2t$, with $a_1,a_2,b_1,b_2\in \mathbb{Z}$. For all $s,t\in \mathbb{Z}$, we have $y,z\in \mathbb{Z}$. 
4) Replace: of course, we get that $ux+ds=c$, and it is easy to finish (and to see if such solutions exists).        
