Question:
The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Solution:
Let first numbers be $x$ and other $(8-x)$
Then according to question
$\left\{\frac{1}{x}+\frac{1}{(8-x)}\right\} \times 15=8$
$\left\{\frac{8-\not x+\not x}{x(8-x)}\right\}=\frac{8}{15}$
$x(8-x)=15$
$x^{2}-8 x+15=0$
$x^{2}-5 x-3 x+15=0$
$x(x-5)-3(x-5)=0$
$(x-5)(x-3)=0$
$x=5$
Or
$(x-3)=0$
$x=3$
Thus, two consecutive number be 3,5
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