+1439 Find the number of ordered pairs $(a,b)$ of integers such that
\frac{a + 2}{a + 1} = \frac{b}{12}.
+1926 Cross multiplying, we get
\(12a+24=ab+b\)
Moving ab to the other side and factoring, we have
\(a\left(12-b\right)=b-24\)
Dividng both sides 12-b, we get
\(a=\frac{b-24}{12-b}\)
So, everything works, and we have \(a=\frac{b-24}{12-b};\quad \:b\ne \:12\)
So technically, there are infinite solutions.
Thanks! :)
