Question

The sum of of two numbers is 18 . And the sum of their reciprocals is 1/4. Find the numbers.

Answer 1

Let the first number be a

Second number = 18 - a

According to question,

1/a + 1/(18-a) = 1/4

=> ( 18 - a + a) / (a) (18 - a) = 1/4

=> 18 / ( 18a - a^2) = 1 /4

=> 72 = 18a - a^2

=> a^2 - 18a + 72 = 0

=> a^2 - 6a - 12a + 72 = 0

=> a( a - 6) - 12(a - 6) = 0

=> (a-6) (a-12) = 0

First number = 6

Second number = 12

Answer 2

Let the first number be a

Second number = 18 - a

According to question,

1/a + 1/(18-a) = 1/4

=> ( 18 - a + a) / (a) (18 - a) = 1/4

=> 18 / ( 18a - a^2) = 1 /4

=> 72 = 18a - a^2

=> a^2 - 18a + 72 = 0

=> a^2 - 6a - 12a + 72 = 0

=> a( a - 6) - 12(a - 6) = 0

=> (a-6) (a-12) = 0

First number = 6

Second number = 12

hope it helped ❤️