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# The sum of two numbers is 9. The sum of their reciprocals is $\frac{1}{2}$. Find the numbers.

Given:

The sum of two numbers is 9. The sum of their reciprocals is $\frac{1}{2}$.

To do:

We have to find the numbers.

Solution:

Let one of the numbers be $x$.

This implies,

The other number $=9-x$.

According to the question,

$\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}$

$\frac{1(9-x)+1(x)}{x(9-x)}=\frac{1}{2}$

$\frac{9-x+x}{9x-x^2}=\frac{1}{2}$

$\frac{9}{9x-x^2}=\frac{1}{2}$

$2(9)=1(9x-x^2)$ (On cross multiplication)

$18=9x-x^2$

$x^2-9x+18=0$

Solving for $x$ by factorization method, we get,

$x^2-6x-3x+18=0$

$x(x-6)-3(x-6)=0$

$(x-6)(x-3)=0$

$x-6=0$ or $x-3=0$

$x=6$ or $x=3$

If $x=6$, then $9-x=9-6=3$.

If $x=3$, then $9-x=9-3=6$

The required numbers are $3$ and $6$.

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