diff --git a/src/algebra/linear-diophantine-equation.md b/src/algebra/linear-diophantine-equation.md
index 047067c59..3f4ede86c 100644
--- a/src/algebra/linear-diophantine-equation.md
+++ b/src/algebra/linear-diophantine-equation.md
@@ -27,10 +27,10 @@ A degenerate case that need to be taken care of is when $a = b = 0$. It is easy
When $a \neq 0$ and $b \neq 0$, the equation $ax+by=c$ can be equivalently treated as either of the following:
-\begin{gather}
-ax \equiv c \pmod b,\newline
-by \equiv c \pmod a.
-\end{gather}
+\begin{align}
+ax &\equiv c \pmod b \\
+by &\equiv c \pmod a
+\end{align}
Without loss of generality, assume that $b \neq 0$ and consider the first equation. When $a$ and $b$ are co-prime, the solution to it is given as
@@ -51,6 +51,12 @@ y = \frac{c-ax}{b}.
## Algorithmic solution
+**Bézout's lemma** (also called Bézout's identity) is a useful result that can be used to understand the following solution.
+
+> Let $g = \gcd(a,b)$. Then there exist integers $x,y$ such that $ax + by = g$.
+>
+> Moreover, $g$ is the least such positive integer that can be written as $ax + by$; all integers of the form $ax + by$ are multiples of $g$.
+
To find one solution of the Diophantine equation with 2 unknowns, you can use the [Extended Euclidean algorithm](extended-euclid-algorithm.md). First, assume that $a$ and $b$ are non-negative. When we apply Extended Euclidean algorithm for $a$ and $b$, we can find their greatest common divisor $g$ and 2 numbers $x_g$ and $y_g$ such that:
$$a x_g + b y_g = g$$
@@ -119,7 +125,7 @@ $$y = y_0 - k \cdot \frac{a}{g}$$
are solutions of the given Diophantine equation.
-Moreover, this is the set of all possible solutions of the given Diophantine equation.
+Since the equation is linear, all solutions lie on the same line, and by the definition of $g$ this is the set of all possible solutions of the given Diophantine equation.
## Finding the number of solutions and the solutions in a given interval
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