 Hello friends, so welcome to another session on problem solving on linear equations in two variables This is a word problem or the application of linear equation into problems here. We are dealing with fractions So let us first see what a fraction is. So a fraction will be of the form of n by d where n is numerator So this is a numerator and D will be denominator Isn't it denominator and where n and d both are Integers Integers, this is very important Okay. Now Let us solve this problem. What does the problem say the sum of the numerator? Let us now translate as we read also some the moment you see the word some so you have to add Something right. So this is plus some of the numerator and denominator of a fraction. So n Plus D. This is some is not the moment. You say is that means it is equal to Three less than so three is less than something. So that means minus three Twice the denominator. So that means 2d So three less than twice the denominator is some of the numerator and the denominator So, let us simplify this first equation. So you'll get D Minus n 2d minus D and n goes on the right-hand side and then I bring it all on the right-hand side You'll go left-hand side. You'll get D minus n is equal to 3 so this d goes here becomes sub you know reduces 2d 2d and And also goes here on the right-hand side and this 3 comes here. So Yeah, so this will be D minus n equals to 3 equation number one Right, then let's go to the second statement. What is the second statement? It says if the numerator and denominator are reduced by 1 The numerator becomes half or becomes means again becomes equal to half the denominator. That means if numerator and Denominator are reduced by 1 The numerator becomes half the denominator. So n minus 1 becomes half D minus 1 Okay, this is the so numerator once attracted from numerator and one subtracted from denominator So it says once you do this the numerator will become half the denominator. So let us simplify this You'll get 2n minus 2 is equal to D minus 1 That means 2n minus D is equal to 1 This is equation number 2 Right, let us highlight our equations. So this is equation number 1 and this is equation number 2 Now you know how to solve linear equation in two variables and these are very simple looking equations So what you can do is add 1 plus 2. Why am I adding 1 plus 2 because in Equation 1 there is D and in equation 2 there is minus D. So they will cancel out the moment you add them So you'll get D minus n plus 2n minus D is equal to 3 plus 1 So you'll get n equals 4, right? If n equals 4 then from 1 From 1 D minus n equals 4 Sorry D minus D minus n was equal to 3 3 so D minus 4 was equal to 3 because n we just found out is 4 So D is equal to 7 right, so fraction Required fraction you should write required fraction is 3 upon I'm sorry 4 upon 7 4 upon 7 4 upon 7 Let us check whether it is true or not so first condition was that Some of the numerator and denominator fraction is 3 less than twice the denominator So some this is checking. Let us check Some of this will be 4 plus 7 equals 11. Now this must be 2 times 7 minus 3 twice the denominator minus 3 which is 11 So both are matching. So hence it is correct. Let us also check the second one. So if Denominator is reduced by 1 so it becomes 3 and denominator is reduced by 1 it also it becomes 6 So 3 by 6 is half Which is true. That is what was mentioned in the question. So numerator becomes half the denominator So this is our fraction related problems in linear equation into variables will be solved. Thank you