Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
x + 6 = y\left( {x - 1} \right)\\
 \Leftrightarrow y\left( {x - 1} \right) - \left( {x + 6} \right) = 0\\
 \Leftrightarrow y\left( {x - 1} \right) - \left( {x - 1} \right) = 7\\
 \Leftrightarrow \left( {x - 1} \right)\left( {y - 1} \right) = 7\\
7 = 1.7 = \left( { - 1} \right).\left( { - 7} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 = 1\\
y - 1 = 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 =  - 1\\
y - 1 =  - 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 = 7\\
y - 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 =  - 7\\
y - 1 =  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
y = 8
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 0\\
y =  - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 8\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 6\\
y = 0
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) \in \left\{ {\left( { - 6;0} \right);\left( {2;8} \right);\left( {0; - 6} \right);\left( {8;2} \right)} \right\}\\
b,\\
2xy - x + 4y = 20\\
 \Leftrightarrow x\left( {2y - 1} \right) + 2.\left( {2y - 1} \right) = 20 - 2\\
 \Leftrightarrow \left( {x + 2} \right)\left( {2y - 1} \right) = 18
\end{array}\) 

\(y\) là số nguyên nên \(2y - 1\) là số lẻ.

Do đó, ta có:

\(\begin{array}{l}
\left( {x + 2} \right)\left( {2y - 1} \right) = 18\\
18 = 1.18 = 2.9 = 3.6 = \left( { - 1} \right).\left( { - 18} \right) = \left( { - 2} \right).\left( { - 9} \right) = \left( { - 3} \right).\left( { - 6} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2y - 1 = 1\\
x + 2 = 18
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 1 = 9\\
x + 2 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 1 = 3\\
x + 2 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 1 =  - 1\\
x + 2 =  - 18
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 1 =  - 3\\
x + 2 =  - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 1 =  - 9\\
x + 2 =  - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 1\\
x = 16
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 5\\
x = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 2\\
x = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 0\\
x =  - 20
\end{array} \right.\\
\left\{ \begin{array}{l}
y =  - 1\\
x =  - 8
\end{array} \right.\\
\left\{ \begin{array}{l}
y =  - 4\\
x =  - 4
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) \in \left\{ {\left( { - 4; - 4} \right);\left( { - 8; - 1} \right);\left( { - 20;0} \right);\left( {4;2} \right);\left( {0;5} \right);\left( {16;1} \right)} \right\}
\end{array}\)