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I don't know what to do to derive the right side from the left side: $$\frac{B}{1+r} = B - \frac{r B}{1+r}.$$

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    Instead, derive the left side from the right side (with a least common denominator), then work backwards.2011-07-05

2 Answers 2

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Hint: Write the numerator of the left hand side as $B(1+r-r)$ and use the fact that $$\frac{a+c}{d}=\frac{a}{d}+\frac{c}{d}$$ for $d\neq 0.$


Added:
Observe that $\,\,$ $B=B(1+r-r)=B(1+r)-rB.$
So $$ \begin{align*}\frac{B}{1+r}&=\frac{B(1+r)-rB}{1+r}\\ &= \frac{B(1+r)}{1+r}-\frac{rB}{1+r}\\ &= B-\frac{rB}{1+r}. \end{align*}$$

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    Could you please show the whole transformation?2011-07-05
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    @Georg: I have added to my answer.2011-07-05
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    Great, thanks a lot!2011-07-05
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HINT $\ $ It's simply $\rm\displaystyle\ 1\ =\ \frac{1+r}{1+r}\ =\ \frac{1}{1+r}\ +\ \frac{r}{1+r}\ $ rearranged, then scaled by $\rm\:B\:.$