 How the question says, find all pairs of consecutive even positive integers both of which are larger than 5 such that their sum is less than 23. Now let us discuss this question. It says, find all the pair of consecutive even positive integers. So let a pair of consecutive even positive integers be x and x plus 2, right? The condition is both of which are larger than 5 that means x has to be greater than 5, right? And such that their sum is less than 23. And their sum that is the sum of x and x plus 2 has to be less than 23. Now we will simplify the second inequality. We have 2x plus 2 is less than 23. Further on subtracting 2 from both the side we have 2x plus 2 minus 2 is less than 23 minus 2. They both will get cancelled out with each other. 2x is less than 21 dividing both the sides by positive 2 we have 2x divided by 2 is less than 21 divided by 2 which gives us the value of x is less than 10.5, right? And also the second condition has to be considered. Now the positive even integer that is less than 10.5 but greater than 5 becomes that x has to be greater than 5 but less than 10.5. So the even positive integers are 6, 8 and 10, right? So the required pair of consecutive even integers first of all 6 and 6 plus 2 becomes 8, 8 plus 2 becomes 10, next is 10 and 10 plus 2 becomes 12. So these are the three required pair. So this completes the question that was given to us. I hope you enjoyed. Take care.