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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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