Question

The Sum of two natural numbers is 9 and the sum of their reciprocals is 9/20 find the numbers​

Answer 1

Given

Sum of two natural numbers is 9

sum of their reciprocals is 9/20

To find

We have to find the numbers

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Let the first number be 'x'

Second number be 'y'

According to the question:

➙x+y= 9

x= 9-y―――❶

Sum of their reciprocals is 9/20

Reciprocals of numbers be 1/x & 1/y respectively.

➙1/x+1/y= 9/20

from equation 1

➙1/9-y+1/y= 9/20

➙y+9-y/y(9-y)= 9/20

➙9/9y-y²=9/20

since 9 lies on both sides on the numerator so,it gets cancelled

➙20×1= 9y-y²

➙20-9y+y²

➙y²-9y+20

➙y²-5y-4y+20

➙y(y-5)-4(y-5)

➙(y-4)(y-5)=0

➙y-4=0 & y-5=0

➙y= 4 & y = 5

Put y = 4 ,5 in Equation 1

At y = 4

➙x= 9-y= 9-4=5

x= 5

At y = 5

➙x= 9-5= 4

x= 4

Hence,the numbers are either 5 & 4 or 4 & 5.

Answer 2

Answer:

⇒ Let one number be x, then other number will be 9−x

According to the question,

⇒

x

1

+

9−x

1

=

2

1

⇒

x(9−x)

9−x+x

=

2

1

⇒

9x−x

2

9

=

2

1

⇒ 2(9)=9x−x

2

⇒ 18−=9x−x

2

⇒ x

2

−9x+18=0

⇒ x

2

−6x−3x+18=0

⇒ x(x−6)−3(x−6)=0$

⇒ (x−6)(x−3)=0

⇒ x−6=0 and x−3=0

∴ x=6 and x=3

⇒ One number is 6 and other number is 3

⇒ The sum of squares of the numbers =(6)

2

+(3)

2

=36+9=45