Answer: x∈∅ (there is no decision).
Step-by-step explanation:
[tex]\left \{ {{x+y=5} \atop {\frac{1}{x} +\frac{1}{y}=\frac{9}{14} }} \right.\ \ \ \ \ \left \{ {{x+y=5} \atop {\frac{y+x}{x*y}=\frac{9}{14} \ |*14*x*y\ (x\neq 0;\ y\neq 0) }} \right. \ \ \ \ \left \{ {{x+y=5} \atop {14*(x+y)=9*x*y}} \right. \ \ \ \ \left \{ {{x+y=5} \atop {14*5=9*x*y}} \right.\ \ \ \ \left \{ {{x+y=5} \atop {9*x*y=70}} \right. \\[/tex]
[tex]\left \{ {{y=5-x} \atop {9*x*y=70\ |:9}} \right. \ \ \ \ \left \{ {{y=5-x} \atop {x*(5-x)=\frac{70}{9} }} \right.\ \ \ \ \left \{ {{y=5-x} \atop {5x-x^2=\frac{70}{9} }} \right. \ \ \ \ \left \{ {{y=5-x} \atop {x^2-5x+\frac{70}{9} =0\ |*9}} \right.\ \ \ \ \left \{ {{y=5-x} \atop {9x^2-45x+70=0}} \right.[/tex]
[tex]\left \{ {{y=5-x} \atop {D=-495}} \right. \ \ \ \ \Rightarrow\ \ \ \ x\in\varnothing.[/tex]
