Question

find the pairs of consecutive even positive integers both of which are smaller than 10 and their sum is more than 11​

Answer 1

Step-by-step explanation:

x < 10, x + 2 < 10 and x + (x + 2) > 11 ⇒ x < 10, x < 8 and 2x > 9 x < 10, (∵ x < 8 automatically smallest of the lesser than) ... (1) and x > (9/2) ... (2) From (1) and (2), we get 9 (9/2) < x < 8 Also, x is an odd positive integer. x can take values 5 and 7. So, the required possible pairs will be (x, x + 2) = (5, 7), (7, 9)

hope it will be helpful to you