Question Number 206074 by MATHEMATICSAM last updated on 06/Apr/24
$$\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{2}} \left({x}_{\mathrm{4}} \:+\:{x}_{\mathrm{5}} \:−\:{x}_{\mathrm{2}} \right) \\ $$$$\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{3}} \:+\:{x}_{\mathrm{4}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{3}} \left({x}_{\mathrm{5}} \:+\:{x}_{\mathrm{1}} \:−\:{x}_{\mathrm{3}} \right) \\ $$$$\left({x}_{\mathrm{3}} \:−\:{x}_{\mathrm{4}} \:+\:{x}_{\mathrm{5}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{4}} \left({x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:−\:{x}_{\mathrm{4}} \right) \\ $$$$\left({x}_{\mathrm{4}} \:−\:{x}_{\mathrm{5}} \:+\:{x}_{\mathrm{1}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{5}} \left({x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \:−\:{x}_{\mathrm{5}} \right) \\ $$$$\left({x}_{\mathrm{5}} \:−\:{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{1}} \left({x}_{\mathrm{3}} \:+\:{x}_{\mathrm{4}} \:−\:{x}_{\mathrm{1}} \right) \\ $$$$\mathrm{Find}\:\frac{\mathrm{2}{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} }{\mathrm{3}{x}_{\mathrm{4}} \:−\:{x}_{\mathrm{5}} }\:. \\ $$
Commented by A5T last updated on 06/Apr/24
$$\mathrm{2}.\:{Equality}\:{when}\:{x}_{\mathrm{1}} ={x}_{\mathrm{2}} ={x}_{\mathrm{3}} ={x}_{\mathrm{4}} ={x}_{\mathrm{5}} \\ $$