Hello everyone so here in this problem we have to solve the equations by Cramer’s rule and find out the value of X and Y now before starting with the problem will observe the given equations then they are not in linear form and whenever we want to apply Cramers rule that time we want the equations in linear form so first of all I’ll convert the given equations in linear form so for that we’ll consider 1 upon X minus 2 as A and 1 upon y minus 3 as B so because of this the equation number 1 will become 3 next 3 into 1 upon X minus 2 is a so that will become 3 a plus 2 into 1 upon y minus 3 so 1 upon y minus 3 is B so that will become 2 B is equal to 5 so let’s say this is the equation number 1 next the second given equation will become 4 1 upon X minus 2 is a so 4 a minus 1 upon Y minus 3 is B so minus B is equal to 3 so this is the equation number 2 now if you’ll observe then we got the equation in linear form so now let’s start with the crema sole so to apply the Kramer’s rule we have to find out the value of capital D then da and DB by using ceruse rule so first of all I’ll apply ceruse rule to find out D da and DB so therefore D is nothing but determinant of coefficient of vary so you have the variables are a and B so D will be the determinant of coefficient of a and B so here the coefficient of a is three and four so 3/4 and the coefficient of B is two and minus one so the determinant of these four elements will be or 2×2 order mattress will be 3 into minus 3 that is minus 3 middle sign is minus and 4 into 2 that is 8 so we’ll get minus of 11 next let’s find out the value of da so da is nothing but the determinant of constants and coefficient of B so here the constants are 5 & 3 so I will write it down 5/3 and the coefficient of B are 2 and minus 1 now friends here you can remember one more tree that whenever you want to find out D a so that time you have to hide the coefficients of a so in the determinant D or in the matrix D we were having the four elements three four and two minus one where three and four were the coefficients of a and – and – and coefficients of B so when V 1 da will hide the coefficient of a so here the coefficient of a are 3 & 4 so we’ll hide this 3 & 4 and in place of this will replace the elements by the constant so here the constants are 5 n 3 so I have replaced it by 5 n 3 so by that you can get the determinant D a so by multiplying here we’ll get 5 into minus 1 that is minus 5 minus 3 to the 6 so again we got – of 11 next let’s find out D B so for D B I’ll again use the same trick I hide the elements of B that is 2 and minus 1 so my first column will be 3 and 4 and instead of this – n replace it by the constant so that is nothing but 5 n 3 so I’ll get 3 4 & 5 3 so therefore DB is equal to 3 4 & 5 3 next so by solving this here we’ll get 3 3 the 9-5 forza 20 so 9 minus 20 is minus 11 so we got the value of BB as minus 11 so we got the values of d da and DB now let’s apply the Kramer’s rule to find out the values of a and B so therefore here I’ll say by Cramer’s rule we’ll get a as the formula of climbers rule is da upon D and we’ll get B by the formula D B upon B so here the value of DA is minus 11 and value of D is minus 11 so here we’ll get minus 11 upon minus 11 and which is nothing but 1 similarly the value of DB is minus 11 and the value of D is also minus 11 so again we’ll get one it means we got two values of a and B as one but if you’ll observe the question that in the question we have the variables x and y it means by Kramer’s rule we have to find out the value of x and y so here we got the value of a and B and we have considered that 1 upon X minus 2 is a and 1 upon Y minus 3 is B so now I’ll replace the a and B by the values of a and B so here we’ll get so here I’ll say but a is 1 upon X minus 2 and B is 1 upon y minus 3 so therefore here we’ll get a is 1 is equal one upon X minus two and B is equal to or here I’ll say one so one is equal to one upon Y minus three so this will become by cross multiplication X minus 2 equal to 1 and y minus 3 is equal to 1 so here we’ll get X as 1 plus 2 that is 3 and Y as 1 plus 3 that is 4 so here we got the value of x and y is 3 & 4 so we’ll write down the solution set therefore the solution set is always written inside the curly brackets and the values of x and y are 3 comma 4 so we have solved the given equations by Kramer’s rule and we got the value of x and y as 3 & 4 thank you

Thnkuu sir

How to solve 3equation with 2 unknown variable using Cramer's rule

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