Question Number 108769 by bemath last updated on 19/Aug/20
$$\:\:\:\frac{\iddots\mathcal{B}{e}\mathcal{M}{ath}\ddots}{\bigstar} \\ $$$$\:\mathrm{G}{iven}\:\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{6}}\:=\:\frac{\mathrm{5}{y}}{\mathrm{5}{y}+\mathrm{25}}\:=\:\frac{\mathrm{4}{z}}{\mathrm{4}{z}+\mathrm{16}} \\ $$$${and}\:{xy}\:+\:{yz}\:+\:{xz}\:=\:\mathrm{188}\:.\:\mathrm{F}{ind}\:{the} \\ $$$${solution} \\ $$
Answered by john santu last updated on 19/Aug/20
$$\:\:\:\frac{\multimap{JS}\multimap}{\heartsuit} \\ $$$$\Rightarrow\:\frac{{x}}{{x}+\mathrm{3}}\:=\:\frac{{y}}{{y}+\mathrm{5}}\:=\:\frac{{z}}{{z}+\mathrm{4}} \\ $$$$\Rightarrow\:\frac{{x}+\mathrm{3}}{{x}}\:=\:\frac{{y}+\mathrm{5}}{{y}}\:=\:\frac{{z}+\mathrm{4}}{{z}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}}{{x}}\:=\:\frac{\mathrm{5}}{{y}}\:=\:\frac{\mathrm{4}}{{z}}\:\rightarrow\begin{cases}{{y}\:=\:\mathrm{5}{b}}\\{{x}\:=\:\mathrm{3}{b}\:}\\{{z}\:=\:\mathrm{4}{b}}\end{cases} \\ $$$$\Rightarrow\mathrm{15}{b}^{\mathrm{2}} \:+\:\mathrm{12}{b}^{\mathrm{2}} \:+\:\mathrm{20}{b}^{\mathrm{2}} \:=\:\mathrm{188} \\ $$$$\Rightarrow\mathrm{47}{b}^{\mathrm{2}} \:=\:\mathrm{188}\:;\:\rightarrow\begin{cases}{{b}=\mathrm{2}}\\{{b}=−\mathrm{2}}\end{cases} \\ $$$${case}\left(\mathrm{1}\right)\:{b}\:=\:\mathrm{2}\:\rightarrow\:\begin{cases}{{x}=\mathrm{6}}\\{{y}=\mathrm{10}}\\{{z}=\mathrm{8}}\end{cases} \\ $$$${case}\left(\mathrm{2}\right)\:{b}=−\mathrm{2}\:\rightarrow\begin{cases}{{x}=−\mathrm{6}}\\{{y}=−\mathrm{10}}\\{{z}=−\mathrm{8}}\end{cases} \\ $$
Commented by bemath last updated on 19/Aug/20
$${jooss} \\ $$