Question

Sum of two natural numbers is 8 and the difference of their reciprocals is $\bold{\frac{2}{15} }$ . Find the numbers??

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Answer 1

Answer:-

$\red{\bigstar}$ The numbers are $\large\leadsto\boxed{\rm\purple{3 \: and \: 5}}$

• Given:-

Sum of two natural numbers is 8.

Difference of their reciprocals is 2/15

• To Find:-

The numbers

• Solution:-

Let the numbers be 'x' and 'y'.

• According to the question:-

✴ $\sf x + y = 8 \dashrightarrow\bf\red{[eqn.i]}$

✴ $\sf \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \dashrightarrow\bf\red{[eqn.ii]}$

• Taking eqn[i]:-

$\longrightarrow \sf x + y = 8$

$\longrightarrow \sf x = 8 - y$

• Substituting this value of x in eqn[ii]:-

$\\ \longrightarrow \sf \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15}$

$\\ \longrightarrow \sf \dfrac{1}{(8-y)} - \dfrac{1}{y} = \dfrac{2}{15}$

• Taking LCM:-

$\\ \longrightarrow \sf \dfrac{y - (8-y)}{y(8-y)} = \dfrac{2}{15}$

$\\ \longrightarrow \sf \dfrac{y - 8 + y}{8y-y^2} = \dfrac{2}{15}$

$\\ \longrightarrow \sf \dfrac{2y - 8}{8y-y^2} = \dfrac{2}{15}$

• Cross multiplying:-

$\\ \longrightarrow \sf 15 \times(2y-8) = 2 \times (8y-y^2)$

$\\ \longrightarrow \sf 30y - 120 = 16y - 2y^2$

$\\ \longrightarrow \sf 30y - 16y - 120 = - 2y^2$

$\\ \longrightarrow \sf 2y^2 - 14y - 120$

$\\ \longrightarrow \sf y^2 - 7y - 60$

• Splitting the middle term:-

$\\ \longrightarrow \sf y^2 - 5y + 12y - 60$

$\\ \longrightarrow \sf y(y - 5) + 12(y - 5)$

$\\ \longrightarrow \sf (y-5)(y+12)$

Now,

$\\ \bigstar \: \sf y - 5 = 0$

$\\ \longrightarrow \sf y = 5$

$\\ \bigstar \: \sf y + 12 = 0$

$\\ \longrightarrow \sf y = -12$

Here, y = -12 is not possible.

Hence,

$\pink{\bigstar}$ $\bf\green{y = 5}$

• Substituting the value of y in eqn[i]:-

$\\ \longrightarrow \sf x + y = 8$

$\\ \longrightarrow \sf x + 5 = 8$

$\\ \longrightarrow \sf x = 8 - 5$

$\pink{\bigstar}$ $\bf\green{x = 3}$

Therefore, the numbers are 3 and 5.

Answer 2

Answer:

Given :

• Sum of two natural numbers is 8 .

• the difference of their reciprocals is 2/15 .

To Find :

• Find the numbers??

Solution :

Let the numbers be a and b

According to the Question :

a + b = 8

• b = 8 - a .......( i )

1/b - 1/a = 2/15

• a - b = (2/15) × ab .........( ii )

Adding (I) and (ii), we get :

2a = 8 + 2ab/15

30a = 120 + 2ab

Substitute the value of b ;

Therefore,

30a = 120 + 2a(8 - a)

30a = 120 + 16a - 2a²

30a - 16a + 2a² = 120

2a² + 14a - 120 = 0

a² + 7a - 60 = 0

• [ dividing by 2 throughout ]

a² + 12a - 5a = 0

(a + 12)(a - 5) = 0

a = -12 or a = 5

As "a" is a natural number, a = -12 is inadmissible then a = 5

Using (i), we get b = 8 - a

Substitute all Values :

b = 8 - 5

b = 3

The required numbers are 5 and 3

Verification:

Sum of given numbers = 5 + 3 = 8

Difference of reciprocals = 1/3 - 1/5 = 2/15