Answer: The numbers are 1 and 3.
Step-by-step explanation:
Let x = smaller number , y= larger number.
As per given,
[tex]x=\dfrac{y}{3}[/tex] ...(i)
[tex]x+y=y^2-5[/tex] ...(ii)
Put value of x from (i) in (ii)
[tex]\dfrac{y}{3}+y=y^2-5\\\\\Rightarrow\ \dfrac43y=y^2-5\\\\\Rightarrow\ 3y^2-4y-15=0\\\\\Rightarrow\ 3y^2-9y+5y-15=0\\\\\Rightarrow\ 3y(y-3)+5(y-3)=0\\\\\Rightarrow\ (y-3)(3y+5)=0\\\\\Rightarrow\ y=3 \ \ or\ y=\dfrac{-5}{3}[/tex]
Since numbers are positive , so y=3 is correct.
And x will be 1 [from (i)]
Hence, the numbers are 1 and 3.
