The simplification is:
$$ \frac {r (1 + x)} {1 + 2x} $$
but I don't understand how one can arrive at the simplified formula from:
$$ \frac { r } {1 + (1/(1+(1/x)))} $$
I've tried multiplying inverses and multiplying the whole fraction by some other value but nothing I solves to the simplified answer. If someone could take me through the process or hint me in the right direction it would be much appreciated.
$\endgroup$4 Answers
$\begingroup$$$ \cfrac r {1 + \cfrac 1 {1 + \cfrac 1 x}} $$First concentrate on the part that appears in $\Big($parentheses$\Big)$ below:$$ \cfrac r {1 + \left( \cfrac 1 {1 + \cfrac1x}\right) } $$In the fraction $\cfrac 1 {1 + \cfrac1x},$ if you multiply the numerator by $x$ you get $x.$ The denominator is two terms:$$ 1 + \frac 1 x. $$Multiplying the first term by $x$ yields $x;$ multiplying the second term by $x$ yields $1$ since the $x$s cancel. Then you have$$ \cfrac r {1 + \left( \cfrac x {x+1} \right)}. $$Next we will multiply the numerator and denominator by $x+1.$ In the numerator, this yields $r(x+1).$ In the denominator, there are two terms:$$ 1 + \frac x {x+1}. $$Multiplying the first term by $x+1$ yields $x+1.$ Multiplying the second term by $x+1$ yields a cancellation so that you just get $x.$ Then the denominator is$$ (x+1) +x. $$Simplify this to $2x+1.$ Then you have$$ \frac{r(x+1)}{2x+1}. $$
$\endgroup$ 3 $\begingroup$$\dfrac r {1+\dfrac1{1+\frac 1x}}=\dfrac r{1+\dfrac x{x+1}}=\dfrac r {\left(\dfrac{2x+1}{x+1}\right)}.$
Can you take it from here?
$\endgroup$ 2 $\begingroup$Start by building the expression from the inside out. Let's successively form and simply the expression in the following sequence:
- First: Simplify $1+(1/x)$
- Second: $1/(1+(1/x))$ by simplifying $1/(\textrm{first result})$
- Third: $1+(1/(1+(1/x)))$ by simplifying $1+\textrm{ second result}$
- Fourth: $\dfrac{r}{1+(1/(1+(1/x)))}$ by simplifying $r/(\textrm{third result})$
Here we go:$$1 + (1/x) = 1 + \frac1x = \frac xx + \frac1x = \frac{x+1}x\tag{first}$$Note we had to get a common denominator to do the addition of fractions above.$$1/(1+(1/x)) = \frac{1}{1+(1/x)} = \frac{1}{\frac{x+1}x} = \frac 11\cdot \frac{x+1}x= \frac x{x+1}\tag{second}$$Note we divided fractions above by flipping the divisor and multiplying instead. We also created a fraction by supplying the implicit denominator $1$ if is isn't present.$$1+(1/(1+(1/x))) = 1 + \frac x{x+1} = \frac{x+1}{x+1} + \frac x{x+1} = \frac{2x+1}{x+1}\tag{third}$$Again, we had to get a common denominator above in order to add fractions.$$\dfrac{r}{1+(1/(1+(1/x)))}=\frac r{\frac{2x+1}{x+1}}= \frac r1\cdot\frac{x+1}{2x+1} = \frac{r(x+1)}{2x+1}\tag{fourth}$$Again, we perform division by flipping the divisor and multiplying instead; and we supplied the implicit denominator of $1$ where needed.
$\endgroup$ $\begingroup$$$\begin{align}\frac{r}{1+\frac{1}{1+\frac{1}{x}}}&=\frac{r}{1+\frac{1}{\frac{x+1}{x}}}\\&= \frac{r}{1+\frac{x}{x+1}}\\&=\frac{r}{\frac{x+1+x}{x+1}}\\&=\frac{r}{\frac{2x+1}{x+1}}\\&=\frac{r(x+1)}{2x+1} \end{align}$$
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