Question

the difference of two natural number is 4 and the difference of their reciprocal is 1 upon 8 find the number

Answer 1

$\sf Solution:$

As the diffence of two natural number is 4, let the number be x and x + 4, where x ∈ N.

We note that x + 4 > x ⇒ $\sf\frac{1}{x + 4}$ < $\sf\frac{1}{x}$ ⇒

According to given, $\sf\frac{1}{x}$ - $\sf\frac{1}{x+4}$ = $\sf\frac{1}{8}$

⇒  $\sf \frac{(x+4 )-x}{x(x+4)}$ = $\sf \frac{1}{8}$   ⇒ $\sf\frac{4}{x(x+4)}$ = $\sf\frac{1}{8}$

⇒ x (x + 4) = 32 ⇒ x² + 4x - 32 = 0

⇒ (x - 4) (x + 8) = 0  ⇒ x - 4 = 0 or x + 8 = 0

⇒x = 4 or x = -8 but x ∈ N

⇒ x = 4

When x = 4, x + 4 = 4 + 4= 8

Hence, the required numbers are 8,4.

Hope it helps! :)

Answer 2

Step-by-step explanation:

As the diffence of two natural number is 4, let the number be x and x + 4, where x ∈ N.

We note that x + 4 > x ⇒ \sf\frac{1}{x + 4}

x+4

1

< \sf\frac{1}{x}

x

1

⇒

According to given, \sf\frac{1}{x}

x

1

- \sf\frac{1}{x+4}

x+4

1

= \sf\frac{1}{8}

8

1

⇒ \sf \frac{(x+4 )-x}{x(x+4)}

x(x+4)

(x+4)−x

= \sf \frac{1}{8}

8

1

⇒ \sf\frac{4}{x(x+4)}

x(x+4)

4

= \sf\frac{1}{8}

8

1

⇒ x (x + 4) = 32 ⇒ x² + 4x - 32 = 0

⇒ (x - 4) (x + 8) = 0 ⇒ x - 4 = 0 or x + 8 = 0

⇒x = 4 or x = -8 but x ∈ N

⇒ x = 4

When x = 4, x + 4 = 4 + 4= 8

Hence, the required numbers are 8,4.