<s>[INST] Bezout's Identity: Finding A Pair of Integers 
There is only one integer $x$, between 100 and 200 such that the integer pair $(x, y)$ satisfies the equation $42x + 55y = 1$. What's the value of $x$ in this integer pair?

We know that
$$\begin{align} x &= x_0 + 5t \\ y &= y_0 - 4t \end{align}$$
But we need to know what $x_0, y_0$ are. By applying the GCD algorithm we can get the answer to be $x_0 = 17$ and $y_0 = 13$. So we need to find $100 \leq 17 + 5t \leq 200$. But treating this parametrically yields too many solutions. How do I discover the one solution?
 [/INST] First, you find a solution to $42x + 55y = 1$
\begin{array}{c|ccc|l|l}
      & 55 &  1 &   0 & &55 = 55(1)+42(0) \\
   -1 & 42 &  0 &   1 & (55,1,0)+(-1)(42, 0, 1)=(13,1,-1) & 42 = 55(0)+42(1) \\
   \hline
   -3 & 13 &  1 &  -1 & (42, 0, 1) + (-3)(13,1,-1)=(3,-3,4)& 13 = 55(1)+42(-1) \\
   -4 &  3 & -3 &   4 & (13,1,-1)+(-4)(3,-3,4)=(1, 13, -17)& 3=55(-3)+42(4) \\
      & 1  & 13 & -17 & & 1=55(13)+42(-17)
\end{array}
So $42(-17)+55(13)=1$. We get $(x_0,y_0)=(-17, 13)$.
Next you characterize the general solutions. 
$$(x,y)=(-17+55t, 13-42t)$$
Next, you search for the value of t that will put $x$ between $100$ and $200$.
\begin{equation}
   100 < -17 + 55t < 200 \\
   117 < 55t < 217 \\
   \dfrac{117}{55} < t < \dfrac{217}{55} \\
   2 \frac{7}{55} < t < 3 \frac{52}{55} \\
   t = 3 \\
\end{equation}
Finally, you compute $x$, $x = -17+55(3) = 148$
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