Two numbers are : x and y
Sum is 16
⇒x+y = 16
⇒x = 16-y
Sum of reciprocals = 1/3
⇒[tex] \frac{1}{x} + \frac{1}{y} = \frac{1}{3} [/tex]
⇒[tex] \frac{1}{16-y} + \frac{1}{y} = \frac{1}{3} [/tex]
⇒[tex] \frac{y+(16-y)}{y(16-y)} = \frac{1}{3} [/tex]
⇒[tex] \frac{16}{y(16-y)} = \frac{1}{3} [/tex]
⇒16×3 = 16y - y²
⇒ y² - 16y + 48 = 0
⇒ y² - 12y - 4y + 48 = 0
⇒ (y-12)(y-4) = 0
⇒ y=4 or 12
⇒x=12 or 4
So numbers are 4,12
