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Question Number 116507 by bemath last updated on 04/Oct/20

Answered by bobhans last updated on 04/Oct/20

$$\mathrm{let}\frac{1}{\frac{1}{\mathrm{x}}+\frac{1}{2}}=\mathrm{r}\Rightarrow \frac{1}{\frac{2+\mathrm{x}}{2\mathrm{x}}}=\mathrm{r}$$$$\Rightarrow \frac{2\mathrm{x}}{2+\mathrm{x}}=\mathrm{r}.\mathrm{the}\mathrm{equation}\mathrm{equivalent}$$$$\mathrm{to}\frac{1}{[\frac{1}{\mathrm{r}+\mathrm{r}}+\frac{1}{\mathrm{r}+\mathrm{r}}]}=\frac{\mathrm{x}}{36}\Rightarrow \frac{1}{(\frac{1}{2\mathrm{r}}+\frac{1}{2\mathrm{r}})}=\frac{\mathrm{x}}{36}$$$$\Rightarrow \frac{1}{\left(\frac{1}{\mathrm{r}}\right)}=\frac{\mathrm{x}}{36};\mathrm{r}=\frac{\mathrm{x}}{36}$$$$\Rightarrow \frac{2\mathrm{x}}{2+\mathrm{x}}=\frac{\mathrm{x}}{36};\mathrm{x}\ne 0$$$$\Rightarrow \frac{2}{2+\mathrm{x}}=\frac{1}{36}\Rightarrow \mathrm{x}=70$$

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