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If
\(\dfrac{a + b}{a} = \dfrac{a}{b} = x\), and \(x\) is positive, then find \(x\).

\(\begin{array}{|rcll|} \hline \dfrac{a + b}{a} &=& \dfrac{a}{b} \\\\ \dfrac{a}{a}+\dfrac{b}{a} &=& \dfrac{a}{b} \\\\ 1+\dfrac{b}{a} &=& \dfrac{a}{b} \quad | \quad \dfrac{a}{b} = x,\ \dfrac{b}{a} = \dfrac{1}{x} \\\\ 1+\dfrac{1}{x} &=& x \quad | \quad \times x \\\\ x+ 1 &=& x^2 \\ x^2-x-1 &=& 0 \\\\ x &=& \dfrac{1\pm \sqrt{1-4(-1)}} {2} \\ x &=& \dfrac{1\pm \sqrt{5}} {2} \quad | \quad x\ \text{is positive} \\ \mathbf{x} &=& \mathbf{\dfrac{1 + \sqrt{5}} {2}} \qquad (x=1.61803398875) \\ \hline \end{array}\)