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
$$\frac{\mathrm{a}}{\mathrm{5}−\mathrm{a}}\:=\:\frac{\mathrm{9}−\mathrm{a}}{\mathrm{a}};\:\: \\ $$$$\:\mathrm{a}^{\mathrm{2}} =\mathrm{5}\left(\mathrm{9}−\mathrm{a}\right)−\mathrm{a}\left(\mathrm{9}−\mathrm{a}\right) \\ $$$$\:\mathrm{a}^{\mathrm{2}} =\:\mathrm{45}−\mathrm{5a}−\mathrm{9a}+\mathrm{a}^{\mathrm{2}} \: \\ $$$$\:\mathrm{a}\:=\:\frac{\mathrm{45}}{\mathrm{14}} \\ $$$$\:\ast\:\frac{\mathrm{9}}{\mathrm{a}}\:=\:\frac{\mathrm{m}+\mathrm{n}}{\mathrm{n}};\:\:\:\frac{\mathrm{x}}{\mathrm{a}}\:=\:\frac{\mathrm{p}+\mathrm{q}}{\mathrm{q}} \\ $$$$\:\ast\:\frac{\mathrm{m}}{\mathrm{n}}\:=\:\frac{\mathrm{q}}{\mathrm{p}} \\ $$$$\:\mathrm{9}\:\centerdot\:\frac{\mathrm{14}}{\mathrm{45}}\:=\:\frac{\mathrm{m}}{\mathrm{n}}\:+\:\mathrm{1}\: \\ $$$$\:\frac{\mathrm{14}}{\mathrm{5}}−\mathrm{1}=\frac{\mathrm{m}}{\mathrm{n}} \\ $$$$\:\ast\:\frac{\mathrm{n}}{\mathrm{m}}\:=\:\frac{\mathrm{5}}{\mathrm{9}}\:=\:\frac{\mathrm{p}}{\mathrm{q}} \\ $$$$\:\mathrm{x}\:=\:\left(\frac{\mathrm{p}}{\mathrm{q}}\:+\mathrm{1}\right)\frac{\mathrm{45}}{\mathrm{14}} \\ $$$$\:\mathrm{x}\:=\:\frac{\mathrm{14}}{\mathrm{9}}\:\centerdot\frac{\mathrm{45}}{\mathrm{14}}\:=\:\mathrm{5}\: \\ $$