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.
 A: You can't do much more with the second equation. Now you have to solve the system formed by the first equation and the reduced form of the second by the method that you prefer to get the solution. Doing it by Gaussian reduction (for example) you get that $5b=20$ so $b=4$ and $a=5$.
A: $$
\begin{cases}
a − b = 1 \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2(1 + b)^2 + (1 + b)b − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2b^2+4b+2 + b+b^2 − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
4b+2 + b = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
5b+2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
5b = 20
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
b = \frac{20}{5}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
b = 4
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + 4 \\
b = 4
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 5 \\
b = 4
\end{cases}
$$
