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The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Given:
The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8.
To do:
We have to find the numbers.
Solution:
Let one of the numbers be $x$.
This implies,
The other number is $8-x$.
The reciprocals of $x$ and $8-x$ are $\frac{1}{x}$ and $\frac{1}{8-x}$ respectively.
According to the question,
$15(\frac{1}{x}+\frac{1}{8-x})=8$
$15(\frac{8-x+x}{x(8-x)})=8$
$15(\frac{8}{8x-x^2})=8$
$15(8)=8(8x-x^2)$
$120=64x-8x^2$
$8x^2-64x+120=0$
$8(x^2-8x+15)=0$
$x^2-8x+15=0$
Solving for $x$ by factorization method, we get,
$x^2-8x+15=0$
$x^2-5x-3x+15=0$
$x(x-5)-3(x-5)=0$
$(x-5)(x-3)=0$
$x-5=0$ or $x-3=0$
$x=5$ or $x=3$
The required numbers are $3$ and $5$.