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Find the number of ordered pairs $(a,b)$ of integers such that
\frac{a + 2}{a + 1} = \frac{b}{3}.

 Jul 24, 2024
 #1
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We can do some rearrangements to do the following:

\(\frac{a + 2}{a + 1} = \frac{b}{3}\)

\(\frac{3a + 6}{a + 1} = b\)

\(\frac{3a + 3}{a + 1} + \frac{3}{a + 1} = b\)
\(3 + \frac{3}{a + 1} = b\).

From this, we have to find values of a for which \(\frac{3}{a + 1}\) is an integer. Since there's no limitation on positive or negative integers, we find that if
\(a = 2 \rightarrow b = 4\)
\(a = -4 \rightarrow b = 2\)

 Jul 24, 2024

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