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Question Number 189792 by normans last updated on 21/Mar/23

Answered by HeferH last updated on 22/Mar/23

Commented by HeferH last updated on 22/Mar/23

$$\frac{\mathrm{a}}{5-\mathrm{a}}=\frac{9-\mathrm{a}}{\mathrm{a}};$$$${\mathrm{a}}^2=5(9-\mathrm{a})-\mathrm{a}(9-\mathrm{a})$$$${\mathrm{a}}^2=45-5\mathrm{a}-9\mathrm{a}+{\mathrm{a}}^2$$$$\mathrm{a}=\frac{45}{14}$$$$\ast \frac{9}{\mathrm{a}}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{n}};\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{q}}$$$$\ast \frac{\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{q}}{\mathrm{p}}$$$$9\cdot \frac{14}{45}=\frac{\mathrm{m}}{\mathrm{n}}+1$$$$\frac{14}{5}-1=\frac{\mathrm{m}}{\mathrm{n}}$$$$\ast \frac{\mathrm{n}}{\mathrm{m}}=\frac{5}{9}=\frac{\mathrm{p}}{\mathrm{q}}$$$$\mathrm{x}=(\frac{\mathrm{p}}{\mathrm{q}}+1)\frac{45}{14}$$$$\mathrm{x}=\frac{14}{9}\cdot \frac{45}{14}=5$$

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