Question Number 177763 by lapache last updated on 08/Oct/22
$${Solve}\:{in}\:\mathbb{R}\:{the}\:{system} \\ $$$$\begin{cases}{{x}+{y}={z}}\\{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{z}}}\end{cases} \\ $$
Commented by mr W last updated on 08/Oct/22
$${no}\:{solution}\:{for}\:{x},{y},{z}\in{R}. \\ $$$${x},\:{y}\:{are}\:{roots}\:{of}\:{t}^{\mathrm{2}} −{zt}+{z}^{\mathrm{2}} =\mathrm{0},\:{but}\:{this} \\ $$$${has}\:{no}\:{real}\:{roots},\:{because} \\ $$$${t}^{\mathrm{2}} −{zt}+{z}^{\mathrm{2}} =\left({t}−\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}{z}^{\mathrm{2}} }{\mathrm{4}}\neq\mathrm{0}\:{for}\:{z}\neq\mathrm{0}. \\ $$