<s>[INST] Find all factored pairs of (a,b) such that... 
Determine all possible ordered pairs (a, b) such that

*

*$a − b = 1$

*$2a^2 + ab − 3b^2 = 22$

I've gone as far as factoring the left side of the second equation:

*

*$(a-b)(2a + 3b) = 22$

*$(1)(2a + 3b) = 22$

*$2a + 3b = 22$
I understand that a different (and arguably easier) way of solving this question is to directly substitute the first equation into the second, but I'd like to learn/use this approach.
 [/INST] $$
\begin{cases}
a − b = 1 \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2(1 + b)^2 + (1 + b)b − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2b^2+4b+2 + b+b^2 − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
4b+2 + b = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
5b+2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
5b = 20
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
b = \frac{20}{5}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
b = 4
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + 4 \\
b = 4
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 5 \\
b = 4
\end{cases}
$$
</s>