Suppose you need to solve
$$a_1x_1 + a_2x_2 + a_3x_3 = c\qquad (1)$$
in integers.
I claim this is equivalent to solving
$$\gcd(a_1,a_2)y + a_3x_3 = c\qquad (2)$$
in integers.
To see this, note that any solution to (1) produces a solution to (2): letting $g=\gcd(a_1,a_2)$, we can write $a_1 = gk_1$, $a_2=gk_2$, so then we have:
$$c = a_1x_1 + a_2x_2 + a_3x_3 = g(k_1x_1) + g(k_2x_2) + a_3x_3 = g(k_1x_1+k_2x_2) + a_3x_3,$$
solving (2). Conversely, suppose you have a solution to (2). Since we can find $r$ and $s$ such that $g=ra_1+sa_2$, we have
$$c = gy+a_3x_3 = (ra_1+sa_2)y +a_3x_3 = a_1(ry) + a_2(sy) + a_3x_3,$$
yielding a solution to (1).
This should tell you how to solve the general case
$$a_1x_1+\cdots+a_nx_n = c$$
in terms of $\gcd(a_1,\ldots,a_n)$, which can in turn be computed recursively.
The diophantine equation $ax+by = r$ can be solved if and only if $\gcd(a,b)\mid r$. In that case, the Euclidean algorithm provides an effective way of finding all solutions.
The key to this is that the collection of all possible values of $ax+by$ is precisely the set of all multiples of $\gcd(a,b)$.
Now consider $ax+by+cz = t$. This is equivalent to solving $\gcd(a,b)w + cz = t$, since any solution to the latter yields a solution of the former (by suitable choice of $x$ and $y$ so that $ax+by = \gcd(a,b)w$) and conversely, any solution to $ax+by+cz=t$ yields a solution to $\gcd(a,b)w+cz = t$. Thus, the diophantine equation $ax+by+cz=t$ is equivalent to the diophantine equation $\gcd(a,b)w+cz=t$, which can be solved if and only if $\gcd(\gcd(a,b),c) = \gcd(a,b,c)$ divides $t$.
Similar considerations yield that the diophantine equation
$$a_1x_1+a_2x_2+\cdots+a_kx_k = d$$
with $k\gt 0$ has a solution if and only if $\gcd(a_1,a_2,\ldots,a_k)$ divides $d$. When it does, the Euclidean algorithm provides an effective way of finding the solutions.
Gerry Myerson has already addressed the other part of your question.
Best Answer
You solve $18 u + 14 v = 2 = \gcd(18,14).$ Solve $2 w + 63 z = 1.$ Combine to get $18 x + 14 y + 63 z = 1.$ Then multiply all by $5.$