 Hi and welcome to the session. The question says, find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11. Now let us discuss this question stepwise. Now here let a pair of consecutive odd integers p x plus 2, right? Also we are given that the sum of the odd positive integers is more than 11. So therefore we get the inequality. Now according to the condition given to us in the question it says that their sum, sum of the pair of positive consecutive odd integers is more than 11 since it's not given to us that it can be equal to 11 also. So we will only have this greater than sign. So x plus x plus 2 is greater than 11. Let us solve it. We have 2 x plus 2 is greater than 11. Now on subtracting 2 from both the sides we have 2 x plus 2 minus 2 is greater than 11 minus 2. After simplification 2 x is greater than 9. On dividing both sides by positive 2 we have 2 x divided by 2 is greater than 9 by 2 which gives us that x is greater than 4.5. Hence each positive odd integer is greater than 4.5 and given is that each integer of the pair of consecutive odd number is less than 10. So the required pair of consecutive odd integers 5, 7 and 7 and 9. They both are greater than 4.5 and less than 10. So this is our required answer. I hope you enjoyed it well. Bye for now.