[tex]x,y-the\ numbers\\\\\left\{\begin{array}{ccc}x+y=16&\to y=16-x\\x:y=-9\end{array}\right\\\\\text{substitute}\\\\x:(16-x)=-9\\\\\dfrac{x}{16-x}=-9\ \ \ \ |\cdot(16-x)\neq0\\\\x=-9(16-x)\ \ \ \ |\text{use distributive property}\\\\x=(-9)(16)+(-9)(-x)\\\\x=-144+9x\ \ \ \ |-9x\\\\-8x=-144\ \ \ \ |:(-8)\\\\\boxed{x=18}\\\\\text{substitute the value of x to the first equation}\\\\y=16-18\\\\\boxed{y=-2}\\\\Answer:\ 18\ and\ -2,[/tex]
