 Hello and welcome to the session. In this session we will discuss a question which says that solve the following inequalities and drop the solution set on the number line. First part is absolute value of 2x plus 1 is greater than 5 and second part is absolute value of 2x minus 3 is less than minus 1. Now before starting the solution of this question we should know some results and that is basic absolute inequalities where k is greater than 0 and in this table you can see the absolute value in equality, its equivalent inequality, its solution set and representation of its solution set on the number line. For the inequality absolute value of x is greater than k, its equivalent inequality is given as x is greater than k or x is less than minus k, its solution set is given as open interval minus infinity to minus k union, open interval k to infinity and this is the representation of its solution set on the number line. Then for the inequality absolute value of x is greater than equal to k, its equivalent inequality is given as x is greater than equal to k or x is less than equal to minus k, its solution set is given as sunny open interval minus infinity to k union, sunny closed interval k to infinity and this is the representation of its solution set on the number line. Then for the inequality absolute value of x is less than k, its equivalent inequality is given as minus k is less than x is less than k, its solution set is given as the open interval minus k to k and this is the representation of solution set. Now for the inequality absolute value of x is less than equal to k, its equivalent inequality is given as minus k is less than equal to x is less than equal to k, its solution set is given by the closed interval minus k to k and this is the representation of its solution set on the number line. Now these results will work out as a key idea for solving all the given question. Now let us start with the solution of the given question. In the first point we are given the inequality absolute value of 2x plus 1 is greater than 5. Now this inequality is of the time absolute value of x is greater than k, so its equivalent inequality is given as x is greater than k or x is less than minus k. So its equivalent inequality is given as 2x plus 1 is greater than 5 or 2x plus 1 is less than minus 5. Now here subtracting 1 from both sides of the inequality we have 2x plus 1 minus 1 is greater than 5 minus 1 or and for this inequality again we subtract 1 from both sides and here we have 2x plus 1 minus 1 is less than minus 5 minus 1. This implies 2x is greater than now 5 minus 1 is 4 or 2x is less than now minus 5 minus 1 is minus 6. Now this implies now here dividing both sides of this inequality by 2 we have 2x upon 2 is greater than 4 upon 2 or here also dividing both sides of this inequality by 2 we have 2x upon 2 is less than minus 6 upon 2 which implies x is greater than 2 or x is less than minus 3. So the solution set of the given inequality is the union of the solution sets of these two simple inequalities. Now the solution set of the inequality x is greater than 2 is the open interval 2 to infinity and the solution set of inequality x is less than minus 3 is given by the open interval minus infinity to minus 3. So the solution of the given inequality is union of these two solution sets that is open interval minus infinity to minus 3 union open interval 2 to infinity. Now let us plot this solution set on the number line. First of all we will draw a number line. Now for plotting this solution we will shape that portion of number line which is to the left of minus 3 that is all this portion which is to the left of minus 3 and that portion which is to the right of 2 so we will shape all this portion and we will put 2 volu circles at the point minus 3 and at the point 2 which indicate that the points minus 3 and 2 are not included in the solution set. So the shaded portion of the number line represents the solution set that is the open interval minus infinity to minus 3 union. The open interval 2 to infinity that is the solution set contains all real numbers that are less than minus 3 and greater than 2. Now let us start with the second part. In the second part we are given the inequality that is absolute value of 2x minus 3 is less than minus 1. Now we can write an equivalent inequality only when k is greater than 0 that is when k is positive. In this inequality we have minus 1 and right inside that is we have k is equal to minus 1 which is less than 0. So we do not write its equivalent inequality since absolute value of any quantity is greater than or equal to 0. So no value of x can make this absolute value less than minus 1. So solution set of the given inequality is the empty set and this is the solution of the given question. That is all for this session. Hope you all have enjoyed the session.