Question

The sum of two numbers is 16 and the sum of their reciprocals is $\frac{1}{3}$.Find the numbers.

Answer 1

Let the two numbers be x and y
Therefore $x+y=16$....(1)
Again
$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$
$\frac{x+y}{xy}=\frac{1}{3}\Rightarrow xy=3\times 16$
⇒$xy=48$
⇒$x=\frac{48}{y}$
On putting value of x in eq(1) we get
$\frac{48}{y}+y=16$
⇒$48+y^2=16y$
⇒$y^2-16y+48=0$
⇒$y^2-12y-4y+48=0$
⇒ $(y-12)(y-4)=0$
⇒ $y=12or4$
Now for $y=12,x=16-12=4$
and for $y=4,x=16-4=12$
so 12 and 4 are the answer.