 Hello friends welcome to this session of problem solving on linear equations in two variables in this session We are going to take case where equations are not linear, but they can be reduced to a linear linear form and then hence solving the system of linear equations So if you can see here in the question, it's given that solve for x and y 3a upon x minus 2b upon y plus 5 equals 0 and a upon x plus 3b upon y minus 2 equals 0 and it is given that x is not equal to 0 and y is not equal to 0 This condition is important. Otherwise, we will not be able to solve these equations because if x and y become 0 then the values are undefined Because the division by 0 is not allowed now so first question is are these Equations linear so are these equations equations Linear and clearly they are not why because if you see the first equation is of the form 3a x to the power negative 1 minus 2b y to the power negative 1 Plus 5 equals 0 So if you see the power is negative 1 on the variable But for linear equation we have studied that power the power must be 1 power is equal to 1 for linear equation linear Equation if it if it is anything but 1 then it is not a linear equation So how do we reduce it to linear form then? So this is how it is done. So You assume 1 upon x to be equal to you Okay, and 1 upon y to be equal to v now rewrite the equations after substituting for x and y. So you'll get 3a u minus 2b v Equals, I'm sorry not equals plus 5 equals 0 so Plus 5 this equals 0. Let us say this is equation number 1 and the second one will be a u Plus 3b times v because 1 upon y is v Minus 2 equals 0. Let us say this is the equation number 2 Okay, now, how do we solve this is now a regular system of linear equations in two variables You can use any of the methods Substitution elimination cross multiplication, but let us go for Elimination and let's first eliminate you so to eliminate you you have to multiply this second equation by 3 Right that will yield 3 a u Plus 9 b v Minus 6 equals 0. This is the equation number 3 and let us rewrite the first equation 3 a u minus 2 b v Plus 5 equals 0. Let us this was equation number 1. So operating work 3 Minus 1 will give you 11 b v Minus 11 Equals 0 So v is equal to 11 upon 11 b right, which is nothing but 1 upon b. This is V right so if we is known we can find out you So you substitute This value of v in 1 so from 1 We know a times u plus 3 times b times v and v is 1 upon b minus 2 equals 0. So this will give you a u Plus 3 minus 2 equals 0. This implies a u is equal to minus 1 Right, so you will be equal to minus 1 upon a So if that is the case, then what is x and y now? I know 1 upon x was u This implies x will be equal to 1 upon u reciprocal. So this is equal to nothing but negative a because a was Minus 1 upon a Similarly 1 by y was v. So this implies y y is equal to 1 upon v and V was what 1 upon b. So hence it is simply b. So what do we get the solution as so x is negative a and Y as b is the solution correct. How do we check? Let's put in any of these equation Let us take the first equation. So the first equation was first equation was 3 a u and plus 9 b minus 6 plus 9 b v minus 6 equals 0. Let us put x or Yeah, so rather it was 3 a by x So it was rather 3 upon x and 9 b upon y minus 6 equals to 0, isn't it? So this is nothing but 3 a is 3 into a by x is minus a plus 9 into b is b and y is b minus 6 equals minus 3 plus 9 minus 6 which is equal to 0. So hence our solution x equals to minus a and y equals to b looks like they are correct Okay, so this is how you solve or you reduce a system of linear equation which doesn't appear to be linear But you can convert them into a linear equation and then solve using any of the standard methods. Thank you