Greatest Common Divisor I have calculated $d = 39$ but don't know how to find $u$ and $v$.

(source: gyazo.com)
Btw I know this is not really a cryptography question, but there isn't a tag for GCD.
 A: $\newcommand{\GCD}{\operatorname{GCD}}$
Hint:
Use extended Euclid's division Algorithm as vonbrand suggested.
$ax+by=\GCD(a,b)$

$2184=975 \times 2+234$
$975=234 \times 3+39$
$234=39 \times 6+0$
$\GCD (2184,975)=39$
Now use the method of back-substitution:
$\ 975-(234  \times 3)=39 $
$\ 975-((2184-(975 \times 2))  \times 3)=39 $
A: You can use a matrix technique. First, write
$$\left[ \begin{array}{cc|c}
 1 & 0 & 2184 \\ 0 & 1 & 975 \end{array}\right]$$
Then perform repeated row operations, e.g.:
$$\left[ \begin{array}{cc|c}
 1 & 0 & 2184 \\ 0 & 1 & 975 \end{array}\right] \stackrel{R_1-2R_2}{\longrightarrow} \left[ \begin{array}{cc|c}
 1 & -2 & 234 \\ 0 & 1 & 975 \end{array}\right] \stackrel{R_2-4R_1}{\longrightarrow} \left[ \begin{array}{cc|c}
 1 & -2 & 234 \\ -4 & 9 & 39 \end{array}\right]$$
$$\left[ \begin{array}{cc|c}
 1 & -2 & 234 \\ -4 & 9 & 39 \end{array}\right] \stackrel{R_2-4R_1}{\longrightarrow} \left[ \begin{array}{cc|c}
 1 & -2 & 234 \\ -4 & 9 & 39 \end{array}\right]$$
Since $39$ divides $234$ we stop and we get $\gcd(2184,975) = 39$ and, $(-4) \cdot 2184 + 9 \cdot 975 = 39$.
A: Cancelling the gcd: $\rm\ 56u + 25v = 1\:\Rightarrow\: mod\ 25\!:\ u \equiv \dfrac{1}{56}\equiv \dfrac{-24}6 \ \equiv -4,\ $ so $\rm\ u = -4\! +\! 25n,\: $ so $\rm\: v = (1\!-\!56u)/25\, =\, (1-56(-4\!+\!25n))/25\, =\, 9\!-\!56n,\: $ i.e. $\rm\: (u,v)\, =\, (-4,9)+(25,-26)n$.
Remark $\ $ Generally it is more efficient to employ the extended Euclidean algorithm. 
A: By using The Extended Euclidean Algorithm
$gcd(2184, 975) = x$
$2184 = 2*975 + 234$
$975 = 4*234 + 39$
$234 = 6*39$
Thus $gcd(2184, 975) = 39$
$d = 2184u + 975v$. Solve for $u$ and $v$.
$39 = 975 - 4*234$
$39 = 975 - 4(2184 - 2*975)$
$39 = 9*975 - 4*2184$
$u = -4$ and $v = 9$
Two more examples here.
