#### Answer

3,6

#### Work Step by Step

Let one number be $x$ and the other number be $y$.
If the sum of the two numbers is 9, this means that $x+y=9$.
Rearranging this equation to make $y$ the subject of the equation,
$x+y=9$ becomes $y=9-x$
This is the first of the two simultaneous equations.
Next, we know that the sum of the reciprocals of the two numbers is $\frac{1}{2}$. This means that:
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$
Now we simplify the above equation,
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$
$\frac{y+x}{xy}=\frac{1}{2}$
$xy(1)=2(y+x)$
$xy=2y+2x$
This is the second of the two simultaneous equations.
Therefore, we now have a pair of simultaneous equations to solve:
Substituting the first equation in the second equation, we obtain:
$x(9-x)=2(9-x)+2x$
$9x-x^{2}=18-2x+2x$
$9x-x^{2}=18$
$9x-x^{2}-18=0$
$x^{2}-9x+18=0$
We now solve this equation using the rules of factoring trinomials:
$x^{2}-3x-6x+18=0$
$x(x-3)-6(x-3)=0$
$(x-3)(x-6)=0$
$(x-3)=0$ or $(x-6)=0$
$x=3$ or $x=6$
Since the sum of two numbers is $9$, $x=3$ means that $y=6$ or $x=6$ means that $y=3$. Therefore, the two numbers are 3 and 6.