 Hello and welcome to the session. In this session we will discuss a question which says that find the range of real values of x for which x-3 over x-4 is less than x-1 over x-2. Now before starting the solution of the question, we shall know a method. And that is the method of intervals. In this method the first step is to factorize the quadratic expression whose coefficient of x square is positive, express left hand side of the inequality in the form x-alpha the whole into x-beta the whole where y is less than beta. Secondly, plot the points alpha and beta on the number line and thus dividing the number line into three parts. Now after plotting the points alpha and beta on the number line from the very right point the sign says plus, minus and plus such that the expression x-alpha the whole into x-beta the whole is non-negative in the region to the right of beta. Now when x-alpha the whole into x-beta the whole is less than 0 then the required range is alpha is less than x is less than beta. As x-alpha the whole into x-beta the whole is negative so the value of x will lie in between this range as this region is negative or this region is less than 0. Now this method will work out as a key idea for solving out this question and now we will start with the solution. Now it is given that x-3 over x-4 is less than x-1 over x-2. This implies x-3 over x-4 minus x-1 over x-2 is less than 0. This implies on taking the LCM the LCM here is x-4 the whole into x-2 the whole and in the numerator it will be x-3 the whole into x-2 the whole minus x-1 the whole into x-4 the whole which is less than 0. Further this implies on multiplying these two expressions this gives x square minus 5x plus 6 minus on multiplying these two expressions this gives x square minus 5x plus 4 whole upon x-4 the whole into x-2 the whole and this is less than 0. Further this implies x square minus 5x plus 6 minus x square plus 5x minus 4 whole upon x-4 the whole into x-2 the whole is less than 0. Now this implies x square is cancelled with x square and 5x is cancelled with 5x so it will be 6 minus 4 which is 2 over x-4 the whole into x-2 the whole and this is less than 0. Now multiplying both sides by x-4 whole square into x-2 whole square this implies 2 into x-4 whole square into x-2 whole square whole upon x-4 the whole into x-2 the whole is less than 0. Further on solving this implies 2 into x-4 the whole into x-2 the whole is less than 0. Now dividing both sides by 2 this implies x-4 the whole into x-2 the whole is less than 0. Now putting each factor equal to 0 we get x is equal to 4 and x is equal to 2. Now by the method of intervals we will plot these points on the number line. So now we have plotted the points to 4 on the number line. Now starting from the right put the signs of plus minus and plus as x-2 the whole into x-4 the whole is less than 0. So according to the formula given in the key idea if x-alpha the whole into x-beta the whole is less than 0 then the required range is alpha is less than x is less than beta. So here the required range is 2 is less than x is less than 4 as alpha is here 2 and beta is 4. This means that the expression x-2 the whole into x-4 the whole is negative when x is lying in this range. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.