The following relation $\frac{a}{b} =\frac{b}{a+b}$ is satisfied for $a = 1.6b$ approximately. How to reach this result?

Question

The following relation $\frac{a}{b}=\frac{b}{a+b}$ is satisfied for $a = 1.6 b$ approximately. My question is how to reach this result, I need a detailed explanation of each step of how $\frac{a}{b}=\frac{b}{a+b}$ became $a=1.6b$, because I have the answer that is $a=1.6 b$, but I don't know how the book who wrote this final answer got this result.

I need an explanation of this process. If there is a property (for example $\frac{e}{f}=\frac{d}{c}$ $\to$ equality property of rational numbers $\to$ $ec = df$), you can tell me because I might not know them.

I need an answer that makes me understand the concepts and why each step is that way, so I can learn and be able to solve similar problems.

Answer 1

Let $a = kb$. Then solve for $k$.

\begin{align} \dfrac ab &= \dfrac{b}{a+b} \\ \dfrac{kb}{b} &= \dfrac{b}{kb+b} \\ k &= \dfrac{1}{k+1} \\ k^2+k - 1 &= 0 \\ k &= \dfrac{-1 \pm \sqrt{1-4(1)(-1)}}{2} \\ k &= \dfrac{-1 \pm \sqrt{5}}{2} \end{align}

In response to your comment...

\begin{align} \dfrac ab &= \dfrac{b}{a+b} \\ \dfrac ab &= \dfrac{b \cdot \dfrac 1b}{(a+b)\dfrac 1b} \\ \dfrac ab &= \dfrac{1}{\dfrac ab + 1} \\ \left(\dfrac ab\right)^2+\left(\dfrac ab\right) - 1 &= 0 \\ etc \end{align}

Answer 2

Let's assume $b$ is non-zero and write $r=a/b$, a rational number. Since $r+1=\tfrac{a+b}{b}$, your equation turns into $$ r=\frac{1}{r+1}, $$ which becomes a quadratic equation: $$ r(r+1)=1, $$ and can be solved using square roots by completing the square.

However, from your question phrasing I think you actually meant to write down the equation for the golden ratio given at the wiki page, specifically this picture. Note that the golden ratio $\phi\approx 1.6$, and the derivation of its value is described in detail at the wiki page I linked.