Question Number 171443 by cortano1 last updated on 15/Jun/22
Answered by Rasheed.Sindhi last updated on 15/Jun/22
$$\begin{cases}{\frac{{y}+{z}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}}\\{\frac{{x}+{z}}{{y}}=\frac{\mathrm{1}}{\mathrm{3}{y}−\mathrm{1}}}\\{\frac{{x}+{z}}{{z}}=\frac{\mathrm{1}}{\mathrm{5}{z}−\mathrm{1}}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{\frac{{y}+{z}}{{x}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}+\mathrm{1}}\\{\frac{{x}+{z}}{{y}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}{y}−\mathrm{1}}+\mathrm{1}}\\{\frac{{x}+{z}}{{z}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{5}{z}−\mathrm{1}}+\mathrm{1}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{\frac{{x}+{y}+{z}}{{x}}=\frac{\mathrm{2}{x}}{\mathrm{2}{x}−\mathrm{1}}}\\{\frac{{x}+{y}+{z}}{{y}}=\frac{\mathrm{3}{y}}{\mathrm{3}{y}−\mathrm{1}}}\\{\frac{{x}+{y}+{z}}{{z}}=\frac{\mathrm{5}{z}}{\mathrm{5}{z}−\mathrm{1}}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{{x}+{y}+{z}=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}}\\{{x}+{y}+{z}=\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{3}{y}−\mathrm{1}}\:\:\:\:\:\:\:\:}\\{{x}+{y}+{z}=\frac{\mathrm{5}{z}^{\mathrm{2}} }{\mathrm{5}{z}−\mathrm{1}}}\end{cases}\:\:\:\: \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}=\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{3}{y}−\mathrm{1}}=\frac{\mathrm{5}{z}^{\mathrm{2}} }{\mathrm{5}{z}−\mathrm{1}} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }=\frac{\mathrm{3}{y}−\mathrm{1}}{\mathrm{3}{y}^{\mathrm{2}} }=\frac{\mathrm{5}{z}−\mathrm{1}}{\mathrm{5}{z}^{\mathrm{2}} }={k}\left({say}\right) \\ $$$$\begin{cases}{\mathrm{2}{kx}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0}}\\{\mathrm{3}{ky}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{1}=\mathrm{0}}\\{\mathrm{5}{kz}^{\mathrm{2}} −\mathrm{5}{z}+\mathrm{1}=\mathrm{0}}\end{cases}\:\:\: \\ $$$$\begin{cases}{{x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{8}{k}}}{\mathrm{4}{k}}}\\{{y}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{12}{k}}}{\mathrm{6}{k}}}\\{{z}=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{20}{k}}}{\mathrm{10}{k}}}\end{cases}\: \\ $$$$\begin{cases}{{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{2}{k}}}{\mathrm{2}{k}}\:\rightarrow\:\mathrm{1}−\mathrm{2}{k}>\mathrm{0}\Rightarrow{k}<\frac{\mathrm{1}}{\mathrm{2}}}\\{{y}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{12}{k}}}{\mathrm{6}{k}}\rightarrow\:\mathrm{9}−\mathrm{12}{k}>\mathrm{0}\Rightarrow{k}<\frac{\mathrm{3}}{\mathrm{4}}}\\{{z}=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{20}{k}}}{\mathrm{10}{k}}\rightarrow\mathrm{25}−\mathrm{20}{k}>\mathrm{0}\Rightarrow{k}<\frac{\mathrm{5}}{\mathrm{4}}}\end{cases}\: \\ $$$$\Rightarrow{k}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{search}\:\mathrm{rational}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{then} \\ $$$$\:\mathrm{k}=−\mathrm{60}\:\mathrm{provides}\:\mathrm{those}. \\ $$
Answered by MJS_new last updated on 15/Jun/22
$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\frac{\mathrm{1}}{{p}+{q}}=\mathrm{2}{x}−\mathrm{1}\:\Rightarrow\:{x}=\frac{{p}+{q}+\mathrm{1}}{\mathrm{2}\left({p}+{q}\right)} \\ $$$$\frac{{p}}{{q}+\mathrm{1}}=\mathrm{3}{px}−\mathrm{1}\:\Rightarrow\:{x}=\frac{{p}+{q}+\mathrm{1}}{\mathrm{3}{p}\left({q}+\mathrm{1}\right)} \\ $$$$\frac{{q}}{{p}+\mathrm{1}}=\mathrm{5}{qx}−\mathrm{1}\:\Rightarrow\:{x}=\frac{{p}+{q}+\mathrm{1}}{\mathrm{5}\left({p}+\mathrm{1}\right){q}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\left({p}+{q}\right)=\mathrm{3}{p}\left({q}+\mathrm{1}\right)=\mathrm{5}\left({p}+\mathrm{1}\right){q} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{11}}\wedge{q}=\frac{\mathrm{1}}{\mathrm{19}} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{239}}{\mathrm{60}}\wedge{y}=\frac{\mathrm{239}}{\mathrm{660}}\wedge{z}=\frac{\mathrm{239}}{\mathrm{1140}} \\ $$
Commented by infinityaction last updated on 15/Jun/22
$${nice}\:{sir}\: \\ $$
Commented by MJS_new last updated on 15/Jun/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jun/22
$$\mathcal{P}{erfect}\:\boldsymbol{{sir}}! \\ $$