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The sum of two numbers is 18. The sum of their reciprocals is $\frac{1}{4}$. Find the numbers.
Given:
The sum of two numbers is 18. The sum of their reciprocals is $\frac{1}{4}$.
To do:
We have to find the numbers.
Solution:
Let the two numbers be $x$ and $18-x$.
According to the question,
$\frac{1}{x}+\frac{1}{18-x}=\frac{1}{4}$
$\frac{1(18-x)+1(x)}{x(18-x)}=\frac{1}{4}$
$\frac{18-x+x}{x(18-x)}=\frac{1}{4}$
$\frac{18}{18x-x^2}=\frac{1}{4}$
$4(18)=1(18x-x^2)$
$72=18x-x^2$
$x^2-18x+72=0$
Solving for $x$ by factorization method,
$x^2-12x-6x+72=0$
$x(x-12)-6(x-12)=0$
$(x-12)(x-6)=0$
$x-12=0$ or $x-6=0$
$x=12$ or $x=6$
The required numbers are $6$ and $12$.
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