 Hello and welcome to the session. In this session we will learn how to solve simple rational equations. We know that a rational expression is of the form f of x upon g of x where g of x is not equal to 0 and f of x and g of x are polynomials in variable x. Now we are going to learn about rational equations. Rational equations are the equations that contain rational expressions. 2 upon x plus 1 is equal to 3 by x is an example of a rational equation. Now we are going to discuss steps for solving a rational equation. Now first of all we will make all the denominators 1 in the given equation by multiplying both sides of the equation by the least common denominator of all rational expressions. Then we solve the simplified equation from the previous step. Then we check our solution to make sure that the proposed solution does not cause any of the denominators in original equation equal to 0. That is we eliminate extraneous solution that is a false solution that forces division by 0 in the original equation. Now let us discuss an example. Solve p plus 1 whole upon 3 minus p by 5 is equal to 3. To solve this rational equation first of all we will find the least common denominator of the rational expressions in the given equation. Now the least common denominator of the given rational expressions in this equation will be given by 3 into 5 that is 15. So we multiply each side of the given equation by the least common denominator. We have 15 into p plus 1 whole upon 3 minus p upon 5 the whole is equal to 15 into 3. This implies that 15 into p plus 1 whole upon 3 the whole minus 15 into p by 5 is equal to 45. This implies that 5 into p plus 1 the whole minus 3p is equal to 45 which implies that 5 into p that is 5p plus 5 into 1 that is 5 minus 3p is equal to 45. This further implies that now 5p minus 3p is 2p plus 5 is equal to 45. This implies that 2p is equal to 45 minus 5 which further implies that 2p is equal to 40. Now dividing both sides by 2 we get 2p upon 2 is equal to 40 upon 2 which implies that p is equal to 20. Now we have got the value of p as 20. Now let us check our answer. For this we put p is equal to 20 in the original equation that is p plus 1 whole upon 3 minus p upon 5 is equal to 3. So we have 20 plus 1 whole upon 3 minus 20 upon 5 is equal to 3 which implies that 21 upon 3 minus 4 is equal to 3 which further implies that 7 minus 4 is equal to 3 that is 3 is equal to 3 which is true. So p is equal to 20 is the solution of the given equation. Let us consider one more example. Solve 2p upon x minus 4 is equal to 8 upon x minus 4 plus 1. Here we see that the least common denominator will be equal to x minus 4. So we multiply both sides of the equation by x minus 4 and we get x minus 4 the whole into 2x upon x minus 4 the whole is equal to x minus 4 the whole into 8 upon x minus 4 plus 1 the whole. This implies that now here x minus 4 cancels with x minus 4 so we are left with 2x on the left hand side of the equation and this is equal to now here we write x minus 4 the whole into 8 upon x minus 4 the whole plus x minus 4 the whole into 1 that is equal to x minus 4. This further implies that 2x is equal to now x minus 4 cancels with x minus 4 and here we get 8 plus x minus 4. This further implies that 2x is equal to now 8 minus 4 is 4 plus x that is 2x minus x is equal to 4 which further implies that x is equal to 4. So we have got the value of x as 4 so now we will check our answer for this we put x is equal to 4 in the original equation. That is 2x upon x minus 4 is equal to 8 upon x minus 4 plus 1. Here we get 2 into 4 upon 4 minus 4 is equal to 8 upon 4 minus 4 plus 1. This implies that 8 upon 0 is equal to 8 upon 0 plus 1. Here you can see that x is equal to 4 results in division by 0 in the original equation that is x is equal to 4 causes the denominators of the rational equation equal to 0. So x is equal to 4 is an extraneous solution that means x is equal to 4 is not the solution of the given equation hence the given equation has no solution. We must note that we can use cross products to solve rational equations with a single fraction on each side of the equal sign. For example if we have this rational equation that is 2 upon x is equal to 3 upon x plus 1 on cross multiplication we get 2 into x plus 1 the whole is equal to 3 into x. This further implies that 2 into x is 2x plus 2 into 1 is 2 is equal to 3 into x that is 3x. This implies that 2 is equal to 3x minus 2x. This further implies that 2 is equal to x or we can write it as x is equal to 2 thus x is equal to 2 is the solution of this rational equation. Thus in this session we have learnt how to solve simple rational equations. This completes our session. Hope you enjoyed this session.