 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that given a is equal to 3 into x plus 1 is less than 4x minus 1 where x belongs to the set of integers and b equal to 5x minus 3 less than equal to 9 minus x where x belongs to the set of integers then find the elements of a intersection b. Here a is equal to 3 into x plus 1 less than 4x minus 1 where x belongs to the set of integers so consider 3 into x plus 1 less than 4x minus 1. This implies 3x plus 3 is less than 4x minus 1. As adding or subtracting any positive number on both sides of the equation will not change the equation so subtract 3x and add 1 on both sides of the equation we obtain 3x plus 3 minus 3x plus 1 is less than 4x minus 1 minus 3x plus 1. This implies 4 is less than x as 3x minus 3x is 0 and 3 plus 1 is 4 and 4x minus 3x is x minus 1 plus 1 is 0. So a is equal to all those x such that 4 is less than x where x belongs to the set of integers. Therefore the solution set of a is equal to 5, 6, 7, 8 and so on. First we shall graph the solution set of a on the number line. First mark all the points 0, 1, 2, 3, 4 and so on and minus 1, minus 2, minus 3, minus 4 and so on on the number line at equal distances. Now as 4 is strictly less than x so mark 4 by an open dot. Now shade the number line on the right of 4. So here the open dot on 4 shows that 4 is not included in the solution set and the shaded portion shows the solution set of a. Now b is equal to 5x minus 3 less than equal to 9 minus x where x belongs to the set of integers. Consider the in equation 5x minus 3 less than equal to 9 minus x. Now add x and 3 on both sides of the in equation as adding or subtracting any positive number to the in equation will not change the in equation. So we get 5x minus 3 plus x plus 3 is less than equal to 9 minus x plus x plus 3. So now 5x plus x is 6x and minus 3 plus 3 is 0. So 6x is less than equal to 9 plus 3 is 12 and minus x plus x is 0. So we get 6x is less than equal to 12. This implies x is less than equal to 12 upon 6 which is equal to 2 on dividing both sides by 6. So the solution set of b is equal to x less than equal to 2 where x belongs to the set of integers. Now let us graph the solution set of b on the number line. So draw a number line and mark all points 0, 1, 2, 3, 4 and so on minus 1, minus 2, minus 3, minus 4 and so on on the number line at equal distances. As x is less than equal to 2 therefore mark 2 with a closed dot which shows that 2 is included in the solution set. Now shade the number line on the left of 2. This shows the graph of b on the number line. Now looking at the graph of a again it shows all points on the right of 4 that is greater than 4. And the graph of b shows all points on the left of 2 that is less than equal to 2. So we can see that there are no elements common in the solution set of a and b. This implies that a intersection b is empty. So there are no elements in the solution set of a intersection b. A intersection b is empty which is our answer. This completes our session. Hope you enjoyed the session.