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

Let the required numbers be x and y It is given that the sum of two numbers is 18 ⇒ x + y = 18 or y = 18 - x It is given that the sum of their reciprocals is 1 4 Then, 1 x + 1 18 - x = 1 4 ⇒ 18 - x + x x 18 - x = 1 4 ⇒ 18 18 x - x 2 = 1 4 ⇒ 18 x - x 2 = 72 So, the quadratic equation formed is, ⇒ x 2 - 18 x + 72 = 0 On splitting the middle term of x 2 - 18 x + 72 = 0 , we get, ⇒ x 2 - 12 x - 6 x + 72 = 0 ⇒ x x - 12 - 6 x - 12 = 0 ⇒ x - 12 x - 6 = 0 ⇒ x - 12 = 0 or x - 6 = 0 ⇒ x = 12 or x = 6 When x = 12 , then the required numbers are 12 and 6 When x = 6 , then the required numbers are 6 and 12 Hence, the required numbers are 6 and 12

The sum of two numbers a and b is 15 and the sum of their reciprocals 1 a and 1 b is 3 10 . Find the numbers a and b .

Given: Two required numbers are a and b So, a + b = 15 ⇒ a = 15 - b .... 1 Then, 1 a + 1 b = 3 10 ⇒ b + a a b = 3 10 ⇒ 10 a + 10 b = 3 a b .... 2 Substituting the value of a from eq 1 in eq 2 , we get 10 ( 15 - b ) + 10 b = 3 ( 15 - b ) b ⇒ 150 - 10 b + 10 b = 45 b - 3 b 2 ⇒ 3 b 2 - 45 b + 150 = 0 So, the quadratic equation formed is, ⇒ b 2 - 15 b + 50 = 0 On splitting the middle terms of b 2 - 15 b + 50 = 0 , we get, ⇒ b 2 - 5 b - 10 b + 50 = 0 ⇒ b b - 5 - 10 b - 5 = 0 ⇒ b - 5 b - 10 = 0 ⇒ b - 5 = 0 or b - 10 = 0 ⇒ b = 5 or b = 10 When b = 5 , then a = 15 - 5 = 10 When b = 10 , then a = 15 - 10 = 5 Hence, the required numbers are a = 5 , b = 10 or a = 10 , b = 5