swerve wrote:If a and b are positive integers, what is the value of a+b?
(1) a/b = 5/8.
(2) The greatest common divisor of a and b is 1.

Good (conceptual) problem!
\[a,b\,\, \geqslant 1\,\,\,{\text{ints}}\,\,\,\left( * \right)\]
\[? = a + b\]
\[\left( 1 \right)\frac{a}{b} = \frac{5}{8}\,\,\,\left\{ \begin{gathered}
\left( {a,b} \right) = \left( {5,8} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 13 \hfill \\
\left( {a,b} \right) = \left( {10,16} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 26 \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,a,b\,\,\,{\text{relatively}}\,\,{\text{prime}}\,\,\,\,\left\{ \begin{gathered}
\left( {a,b} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 2 \hfill \\
\left( {a,b} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 3 \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\]
\[\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,k\,\,{\text{technique}}\,\,\,\left\{ \begin{gathered}
a = 5k \hfill \\
b = 8k \hfill \\
\end{gathered} \right.\,\,\,\,\,\left( {k \ne 0\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,k \geqslant 1\,\,\operatorname{int} } \right)\]
\[\left[ {{\text{Reason}}\,\,{\text{for}}\,\,k\,\,\operatorname{int} \,\,:\,\,\,k = 2 \cdot \left( {8k} \right) - 3 \cdot \left( {5k} \right) = 2b - 3a = \operatorname{int} - \operatorname{int} = \operatorname{int} } \right]\]
\[\left( 1 \right) + \left( 2 \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,k = 1\,\,\,\, \Rightarrow \,\,\,\,? = a + b = 5 + 8 = 13\,\,\,\, \Rightarrow \,\,\,{\text{SUF}}.\,\,\,\,\,\,\,\]
\[\left( {**} \right)\,\,\,k \geqslant 2\,\,\, \Rightarrow \,\,\,GCD\left( {a,b} \right) = k \ne 1\,\,\, \Rightarrow \,\,\,\left( 2 \right)\,\,{\text{contradicted}}{\text{,}}\,\,\,{\text{impossible}}\,\,\,\,\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.