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Sam

asked 2016-06-02 09:05:39

Sum of two numbers is $18$. Sum of their reciprocals is $\dfrac{1}{4}$. Find the numbers.

jiju

answered 2016-06-04 19:01:35

$12$ and $6$

Let numbers be $x$ and $18-x$

$\dfrac{1}{x}+\dfrac{1}{18-x}=\dfrac{1}{4}\\

4(18-x)+4x=x(18-x)\\

72-4x+4x=18x-x^2\\

x^2-18x+72=0\\

(x-12)(x-6)=0\\

x=12\text{ or }6$

Numbers are $12$ and $18-12=6$

Let numbers be $x$ and $18-x$

$\dfrac{1}{x}+\dfrac{1}{18-x}=\dfrac{1}{4}\\

4(18-x)+4x=x(18-x)\\

72-4x+4x=18x-x^2\\

x^2-18x+72=0\\

(x-12)(x-6)=0\\

x=12\text{ or }6$

Numbers are $12$ and $18-12=6$

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