 Hello and welcome to the session. In this session we discuss the following question which says simplify 3 1 upon 7 minus square bracket open 5 2 upon 3 minus curly bracket open 4 upon 5 minus round bracket open 1 upon 3 plus 3 upon 7 minus 4 upon 15 round bracket closed curly bracket closed and square bracket closed. Let's move on to the solution. We need to simplify 3 1 upon 7 minus square bracket open 5 2 upon 3 minus curly bracket open 4 upon 5 minus round bracket open 1 upon 3 plus 3 upon 7 minus 4 upon 15 round bracket closed curly bracket closed and square bracket closed. This is further equal to, now we can write 3 1 upon 7 as 20 2 upon 7 minus square bracket open now 5 2 upon 3 can be written as 17 upon 3 minus curly bracket open 4 upon 5 minus round bracket open. Now we will simplify the terms in the round bracket first since this is the innermost bracket. So, we take the LCM of the denominator of the fractions inside the round bracket or the parenthesis. We get the LCM here as 105. Now 3 35 times as 105 and 35 multiplied by 1 is 35 plus 7 15 times as 105 15 multiplied by 3 is 45 minus 15 7 times as 105 and 7 multiplied by 4 is 28 round bracket closed curly bracket closed and square bracket closed. This further is equal to 22 upon 7 minus square bracket open 17 upon 3 minus curly bracket open 4 upon 5 minus on simplifying this in the parenthesis we get 52 upon 105 curly bracket closed round bracket closed. Now we will simplify the curly bracket. This is further equal to 22 upon 7 minus square bracket open 17 upon 3 minus curly bracket open. Now we take the LCM of the denominator of the fractions inside the curly bracket and the LCM comes out to be equal to 105. Now 5 21 times as 105 and 21 multiplied by 4 is 84 minus 105 1 times as 105 and 1 multiplied by 52 is 52 curly bracket closed square bracket closed and so we get this is further equal to 22 upon 7 minus square bracket open 17 upon 3 minus. Now simplifying this expression in the curly bracket we get 32 upon 105 square bracket closed and this is further equal to 22 upon 7 minus square bracket open. Now we will simplify these terms in the square bracket. We take the LCM of the denominator of the fractions in the square bracket which is 105. Now 3 35 times as 105 and 35 multiplied by 17 is 595 minus 105 1 times as 105 and 1 multiplied by 32 is 32 and so this is further equal to 22 upon 7 minus 563 upon 105. Further we take the LCM of the denominator of the two fractions and so this is equal to 105 now 7 35 times as 105 and 35 multiplied by 22 is 330 minus 105 1 times as 105 and 1 multiplied by 563 is 563 and so this is further equal to minus 233 upon 105. So on simplifying the given expression we get the final answer as minus 233 upon 105. This completes the session hope you have understood the solution of this question.