Question Number 166943 by cortano1 last updated on 03/Mar/22
Commented by blackmamba last updated on 03/Mar/22
$$\:{By}\:{Ladder}\:{theorem} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{4}+\mathrm{5}+\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{4}+\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{5}+\mathrm{10}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}+\mathrm{19}}=\frac{\mathrm{1}}{\mathrm{14}}+\frac{\mathrm{1}}{\mathrm{15}}−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}+\mathrm{19}}=\frac{\mathrm{4}}{\mathrm{105}}\:;\:{x}=\frac{\mathrm{29}}{\mathrm{4}}=\mathrm{7}.\mathrm{25}\: \\ $$
Commented by Tawa11 last updated on 03/Mar/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 03/Mar/22
$$\frac{{x}_{\mathrm{1}} }{\mathrm{4}}=\frac{{x}+\mathrm{5}}{\mathrm{4}+\mathrm{10}}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{4}\left({x}+\mathrm{5}\right)}{\mathrm{4}+\mathrm{10}} \\ $$$$\frac{{x}_{\mathrm{2}} }{\mathrm{5}}=\frac{{x}+\mathrm{4}}{\mathrm{5}+\mathrm{10}}\:\Rightarrow{x}_{\mathrm{2}} =\frac{\mathrm{5}\left({x}+\mathrm{4}\right)}{\mathrm{5}+\mathrm{10}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} ={x}=\frac{\mathrm{4}\left({x}+\mathrm{5}\right)}{\mathrm{4}+\mathrm{10}}+\frac{\mathrm{5}\left({x}+\mathrm{4}\right)}{\mathrm{5}+\mathrm{10}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{29}}{\mathrm{4}}=\mathrm{7}.\mathrm{25} \\ $$
Commented by Tawa11 last updated on 03/Mar/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Ari last updated on 03/Mar/22
Where are you based on creating reports sir