Answer:
Step-by-step explanation:
[tex]\left \{ {{xy=5} \atop {4+y=-x}} \right.\\\\<=> \left \{ {{xy=5} \atop {x=-4-y}} \right.\\\\=> \left \{ {{(-4-y)y=5} \atop {x=-4-y}} \right. \\\\<=> \left \{ {{-y^{2}-4y =5} \atop {x=-4-y}} \right. \\\\<=> \left \{ {{y^{2}+4y+5 =0} \atop {x=-4-y}} \right.[/tex]
because y² + 4y + 5 = (y+2)² + 1 > 0, ∀ y ∈ R
=> y² + 4y + 5 = 0 is unreasonable
=> no solutions