 Hello friends. So welcome to another session of problem solving on consistency of system of linear equation into variables. So the question is find the value of a and b for which the given system of equations has an infinite number of solutions and the equations are 2x plus 3y equals 7. So this is equation number 1 and a plus b plus 1x plus a plus 2b plus 2y equals 4 a plus b plus 1. This is equation number 2 and for consistency is consistency means whether the given system of linear equations has a solution or it has infinite solution or it has no solution. So we had three conditions, right? Now in this case it is asking whether they have infinite number of solutions and for infinite number of solutions for an equation which is given by a1x plus b1y plus c1 equals 0 and a2x plus b2y plus c2 equals 0. Let us say these were the two pair of linear equations given to us. Then for infinite solutions, infinite solutions that also means that the lines are coincident. So hence in that the condition is a1 by a2 must be equal to b1 by b2 and this also should be equal to c1 by c2. Where a1 and a2 are coefficients of x, b1 and b2 are coefficients of y and c1 and c2 are the constant terms. Mind you the two equations have been expressed in such a way that all the terms are on in the LHS left hand side and only 0 is on the right hand side, right? So make sure that this standard form is maintained only then you can write this condition. This condition is also valid if the constant terms, both the constant terms of the two equations are on the right hand side. So you have to make sure that either all of them are, all the terms are on the LHS like this or you could have also got the same condition if this was the case. That is, let us say if you have expressed it like that also, then in this case also this condition would be valid. Okay, now let us now take these two given equations and apply these this condition of infinite solutions and then solve for a and b. So what is a1? If you see a1 is 2 and a2 is a plus b plus 1, isn't it? And this must be equal to 3 which is b1 divided by a plus 2b plus 2 and this must be equal to. So hence here if you see 7 is on the right hand side. So to get it in a standard form, you should have written it as 2x plus 3y minus 7 equals 0. Okay, and same with here also. But then since both of them are on the right hand side, I can use the same condition, right? So I can write very much as 7 upon 4 times a plus b plus 1. Because both 7 and this term is on the right hand side. If both of them would have been on the left hand side also, then also the same condition would have been valid. Now we have got these this equation. So there are three equations basically in this. So this item is equal to this one. You can get one equation from here and then you can equate these two. You can get another equation from here or you can equate these two together. Yeah, so whichever way you want you can equate. Let us take first two terms. So hence I can equate first two. So hence a plus b plus 1 equals 3 upon a plus 2b plus 2. So cross multiplying you will get 2 times a plus 2b plus 2 is equal to 3 times a plus b plus 1. Isn't it? So hence if you solve it, it is 2a plus 4b plus 4 is equal to 3a plus 3b plus 3. Simplifying you will get a and then minus b and then so basically what I am trying to do is you can shift all of them on the right hand side and you can get 3a minus 2a that is a. 3b minus 4b is b and 3 minus 4 is minus 1 equals 0. Let us say this is equation number 3. Okay. And now let us also equate these two, the last two. So hence what will I get? I will get 3 upon a plus 2b plus 2 is equal to 7 upon 4a plus 4b plus 1. Okay. Now let us cross multiply once again. So this will imply 3 times 4a plus 4b plus 1 equals 7 times a plus 2b plus 2b. So this will be nothing but 12a plus 12b plus 3 this will be equal to 7a plus 14b plus 14. So let us simplify once again. So this implies 5a 12 minus 7a and then 12b minus 14b is minus 2b and plus 3 minus 14 is minus 11 equals 0. Let it be equation number 4. Right? So let us rewrite the two equations. I have a minus b minus 1 equals 0. This is equation number 3 and I have 5a minus 2b minus 11 equals to 0. This is equation number 4. Let us multiply the first equation by 2. This will eliminate. Now let us, this particular example can also give us opportunity to solve a linear equation in two variables. Here a and b are the variables. So if you multiply the third equation with 2 you will get 2a minus 2b minus 2 equals 0. Now you subtract third and let us say this is 5. Yeah. So if you subtract this will become minus this will become plus. This will become plus and hence if you add you will get 3a. This will become 0 and this will become minus 11 plus 2 is minus 9 and this is equal to 0. So hence 3a equals 9. So a we get as 3. Okay. Now the moment we get a is equal to 3. We can deploy this in the equation number 3. So from 3 I can write from 3 what can I write? I can write 2 into 3. No. 2 is not required actually. So the equation was simply what was the equation? Equation was a minus b minus 1 equals 0. So hence a is already we found out 3 minus b minus 1 equals 0. So b is equal to 2. Is it? So let us check whether this is correct or not. So let us find out all the values here in this case here. So this will become 2 upon a plus b. So a 3 plus 2 plus 1 should be equal to 3 upon 3 plus 2b is 4 plus 2. Okay. And this should be equal to 7 upon 4 times a plus b is 3 plus 2 plus 1. So if you see this is equal to I am just doing it here. So 3 upon 6 3 upon no sorry this is 2 upon 6. This is 2 upon 6 which is equal to 3 upon 9 which is equal to 7 upon 21 which is all or equal to 1 by 3. So hence that means our value of a and b are correct. So a is equal to 3 and b is equal to 2 is the solution. So for a equals to 3 and b equals to b equals 2 this system of linear equation will have infinitely many solution infinitely many solution. Okay. For what value of a? a is 3 and b is 2. Hope you understood the process. Thank you.