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 $5$.

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Updated on: 10-Oct-2022

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