Question

Solve each system of equations using substitution. x×y=5 4+y=-x​

Answer 1

Explanation:

`\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.`

because y² + 4y + 5 = (y+2)² + 1 > 0, ∀ y ∈ R

=> y² + 4y + 5 = 0 is unreasonable

=> no solutions