Question

Find two numbers whose sum is 25 and the sum of whose reciprocals is 1/6

Answer 1

Answer:

10 and 15

Step-by-step explanation:

Let 'x' and 'y' are the numbers we need to find.

x + y = 25 (two numbers whose sum is 25)

(1/x) + (1/y) = 1/6 (the sum of whose reciprocals is 1/6)

The solutions of the this system of equations are the numbers we need to find.

x = 25 - y

1/(25 - y) + 1/y = 1/6 multiply both sides by 6(25-y)y

6y + 6(25-y) = (25-y)y

6y + 150 - 6y = 25y - (y^2)

y^2 - 25y + 150 = 0 quadratic equation has 2 solutions

y1 = 15

y2 = 10

Thus we have :

First solution: for y = 15, x = 25 - 15 = 10

Second solution: for y = 10, x = 25 - 10 = 15

The first and the second solution are in fact the same one solution we are looking for: the two numbers are 10 and 15 (since the combination 10 and 15 is the same as 15 and 10).

Answer 2

The required first number is 15 and second number be 10.

Given that,

Two number whose sum is 25,

and sum of whose reciprocal is 1\6.

We have to find,

What are the two numbers.

According to the question,

Let, The first number be x,

And the second number be y.

Then,

The sum of two number is 25,

$x + y = 25$

And the sum of its reciprocal is 1\6,

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6}$

On solving both equation,

$x + y = 25\\\\\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6}$

From equation 1,

$x= 25-y$

Substitute the value of x in the equation 2,

$\dfrac{1}{25-y} + \dfrac{1}{y} = \dfrac{1}{6}\\\\\dfrac{y+25-y}{y.(25-y)} = \dfrac{1}{6}\\\\6 \times 25 = 1 \times y.(25-y)\\\\150 = 25y - y^2\\\\y^2 - 25y +150= 0 \\\\y^2 - 15y -10y + 150 = 0\\\\y (y-15) -10(y-15)=0 \\\\(y-15) (y-10)= 0\\\\When \ y -15 = 0 , \ y = 15\\\\when \ y -10 = 0 ,\ y = 10$

Therefore,

$When \ y = 10 \ then \ x = 25 - 10 , x = 15\\\\And \ y = 15\ then\ x = 25 - 15, \ x = 10$

Hence, The required first number is 15 and second number be 10.

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