Find the number of ordered pairs $(a,b)$ of integers such that \frac{a + 2}{a + 1} = \frac{b}{8}.
\(\frac{a + 2}{a + 1} = \frac{b}{8} \)
Rewrite as
8(a + 2) / (a + 1) = b
As a approaches -infinity and + infinity, b approaches 8
a b
0 16
1 12
3 10
7 9
-2 0
-3 4
-5 6
-9 7
+1725 
