 Hello friends welcome to the session I am Malka. We are going to discuss how to form the pair of linear equations in the following problems and find the solutions if they exist by any algebraic method. A problem is Yash scored 40 marks in a test getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks being deducted for each incorrect answer then Yash would have scored 50 marks. How many questions were there in the list? So let's start with the solution. The number of correct answer and let the number of wrong answer be Y. Now according to question 3 marks for correct answer and minus 1 for wrong answer then Yash obtained 40 marks. Therefore it can be written as 3x minus y equal to 40. This is our first equation. Now again from the question we see that 4 marks for correct and minus 2 for wrong answers Yash scored 50 marks. This can be written as 4x minus 2y equal to 50 or it can be written as 2x minus y equal to 25. This is our second equation. Thus the equations are minus y equal to 40 which is our first equation and 2x minus y equal to 25. This is our second equation. Now on subtracting equation second from first we get minus minus y and minus of minus y equal to 40 minus 25. This implies x minus y plus y equal to 15. This implies y by cancel out x equal to 15. Now on substituting x equal to 15 in equation first we get our equation first is 3x minus y equal to 40. Now we substitute x equal to 15. This will give us 3 into 15 minus y equal to 40. This implies minus y equal to 40 minus 45. This implies minus y equal to minus 5 minus minus cancel out. This implies y equal to 5. The two equations are 3x minus y equal to 40 and 2x minus y equal to 25 where x and y are the number of right answer and wrong answer respectively and total number of question equal to x plus y equal to 15 plus y equal to 20. Hope you understood the solution and enjoyed the session. Goodbye and take care.