 Hello and welcome to the session. In this session we discussed the following question which says solve 1 upon P plus Q plus X is equal to 1 upon P plus 1 upon Q plus 1 upon X Where we have X not equal to 0 and X not equal to minus of P plus Q So we need to solve the given equation that is we need to find the value of X So let's see the solution. We have 1 upon P plus Q plus X is equal to 1 upon P plus 1 upon Q plus 1 upon X So first of all We transfer this term 1 upon X from the right-hand side to the left-hand side. So we have 1 upon P plus Q plus X Minus 1 upon X is equal to 1 upon P plus 1 upon Q Further we take LCM on the left-hand side and on the right-hand side So we get on the left-hand side. We get the denominator as P plus Q plus X whole into X in the numerator. We have X minus P plus Q plus X. This is equal to taking LCM on the right-hand side We get PQ in the denominator and Q plus P in the numerator This gives us X minus P plus Q minus X upon P plus Q plus X whole into X is equal to P plus Q upon PQ Now X and minus X cancels so we have minus of P plus Q upon P plus Q plus X whole into X is equal to P plus Q upon PQ Now dividing both sides by P plus Q We get minus 1 upon P plus Q plus X whole into X is equal to 1 upon PQ Now We cross multiply and this gives us Minus PQ is equal to X into P plus Q plus X Or you can say this gives us X square plus PX plus QX plus PQ is equal to 0 Now making pairs From the first pair we take out X common inside the bracket We have X plus P from the second pair we take out Q common inside the bracket. We have X plus P so This gives us X into X plus P plus Q into X plus P is equal to 0 or you can say X plus P into X plus Q is equal to 0 which means Either X plus P is equal to 0 or X plus Q is equal to 0 This gives us X equal to minus P or X equal to minus Q So we have got X equal to minus P or X equal to minus Q which means minus P and minus Q are the roots of the given equation So we have solved the given equation this complete C session. Hope you have understood the solution of this question