Answer:
[tex]x+ y = 10 ---(1)\\ x\cdot y = 9 ---(2)\\\\from(2) \ y = \frac{9}{x}\\\\substitute \ y \ in (1) \\x + \frac{9}{x} = 10\\\\x^2 + 9 - 10x = 0\\x^2 -10x + 9 = 0\\x^2 -9x -x +9 = 0\\x(x-9)-1(x-9)=0\\(x-1)(x-9)=0\\x = 1 \ or \ 9\\[/tex]
substitute x in y = 9/x
=> y = 9 or 1
So the number are 1 and 9 or 9 and 1
