[tex]x;\ y-the\ numbers\\\left\{\begin{array}{ccc}x+y=3&|subtract\ y\ from\ both\ sides\\xy=-40\end{array}\right\\\left\{\begin{array}{ccc}x=3-y&(1)\\xy=-40&(2)\end{array}\right\\\\subtitute\ (1)\ to\ (2):\\(3-y)y=-40\\3y-y^2=-40\ \ \ |add\ 40\ to\ both\ sides\\-y^2+3y+40=0\\-y^2+8y-5y+40=0\\-y(y-8)-5(y-8)=0\\(y-8)(-y-5)=0\iff y-8=0\ or\ -y-5=0\\therefore\ \boxed{y=8\ or\ y=-5}[/tex]
[tex]subtitute\ the\ values\ of\ "y"\ to\ (1)\\\\x=3-8=-5\ or\ x=3-(-5)=3+5=8\\\\Answer:\\\boxed{\left\{\begin{array}{ccc}x=-5\\y=8\end{array}\right\ or\ \left\{\begin{array}{ccc}x=8\\y=-5\end{array}\right;\ conclusion:-5\ and\ 8}[/tex]
