 Hello and welcome to this session. In this session we discussed the following question which says solve 2 upon A is equal to minus 1 minus 3 upon B and 9 upon B is equal to 4 upon A minus 7. Let's proceed with the solution now. We are given two simultaneous linear equations. 2 upon A is equal to minus 1 minus 3 upon B. Let this be equation 1 and the other equation is 9 upon B is equal to 4 upon A minus 7. Let this be equation 2. Now we will rewrite the equations 1 and 2 in such a way that we have the terms containing the variables on the left hand side and the constant terms on the right hand side. So now rewriting equation 1 we get 2 upon A plus 3 upon B is equal to minus 1. Let this be equation 3. In the same way we rewrite the equation 2. So now rewriting equation 2 we get 4 upon A minus 9 upon B is equal to 7. Let this be equation 4. Now we will solve the equations 3 and 4 for the values of A and B. For this we take let 1 upon A be equal to x and 1 upon B be equal to y. Then we have equation 3 can be written as 2x plus 3y equal to minus 1 and equation 4 can be written as 4x minus 9y equal to 7. So we get this equation from the equation 3 and this equation from the equation 4. Let this be equation 5 and this be equation 6. Now we will solve the equations 5 and 6 to get the values for x and y. For this we will multiply the equation 5 by 2 so that the coefficient of x in the equations 5 and 6 are equal. So now multiplying equation 5 by 2 we get 2 multiplied by 2x plus 3y is equal to 2 multiplied by minus 1 that is minus 2. So this gives us 4x plus 6y is equal to minus 2. So let this be equation 7 and we have the equation 6 as 4x minus 9y equal to 7. Now we would subtract equation 6 from equation 7. So subtracting equation 6 from equation 7 we get 4x plus 6y minus 4x minus 9y is equal to minus 2 minus 7. Further we get 4x plus 6y minus 4x plus 9y is equal to minus 9. Now this 4x and minus 4x cancels and 6y plus 9y is 15y is equal to minus 9. To get the value for y we divide both sides by 15. So this 15 cancels with 15 and 3 3 times is 9 and 3 5 times is 15. So we get the value for y as minus 3 upon 5. Now that we have got the value for y we can easily find out the value for x by substituting this value for y in either of the equations 6 or 7. So now substituting y equal to minus 3 upon 5 in equation 6 we get 4x minus 9 into the value for y which is minus 3 upon 5 is equal to 7. So this further gives us 4x plus 27 upon 5 is equal to 7. Now taking the constant term 27 upon 5 to the right hand side we get 4x is equal to 7 minus 27 upon 5. Now further we take the LCM on the right hand side we get the LCM as 5. Now 5 multiplied by 7 is 35 minus 27 upon 5. So now we have 4x is equal to 8 upon 5. Now to get the value for x we divide both sides by 4 and so now we have this 4 cancels with 4 and 4 2 times is 8. So we have x is equal to 2 upon 5 that we get the value for x as 2 upon 5 and the value of y that we obtained was minus 3 upon 5. As you know we had assumed 1 upon a to be x and 1 upon b to be y. So we have 1 upon a is equal to x which is 2 upon 5 and 1 upon b is equal to y which is minus 3 upon 5. This gives us a is equal to 5 upon 2 and this gives us b is equal to minus 5 upon 3. So we now have the values for a and b. So this is our final answer. a is 5 upon 2 and b is minus 5 upon 3. Now we can also check the solution that we have obtained in each equation. So a solution is correct only when it satisfies each equation in the given system. So we will now see the checking of the solution. The equations given to us were 2 upon a is equal to minus 1 minus 3 upon b. This was equation 1 and the other equation was 9 upon b is equal to 4 upon a minus 7. This was equation 2. Now we will check whether a equal to 5 upon 2 and b equal to minus 5 upon 3 satisfies both these equations or not. Now putting a equal to 5 upon 2 and b equal to minus 5 upon 3 in equation 1 we get 2 upon a which is 5 upon 2 is equal to minus 1 minus 3 upon b which is minus 5 upon 3 or further we have 4 upon 5 is equal to minus 1 plus 9 upon 5 further. We take LCM on the right side we get the LCM as 5. 5 multiplied by minus 1 is minus 5 plus 9 and this whole upon 5. So this means we have 4 upon 5 is equal to 4 upon 5 which is true. So this means a equal to 5 upon 2 and b equal to minus 5 upon 3 satisfies the equation 1 but for the solution to be correct it should satisfy the second equation also. So next putting a equal to 5 upon 2 and b equal to minus 5 upon 3 in equation 2 we get 9 upon b which is minus 5 upon 3 is equal to 4 upon a which is 5 upon 2 minus 7. This gives us minus 27 upon 5 is equal to 8 upon 5 minus 7. Further we take the LCM on the right hand side we get 5 as the LCM and now 5 one time is 5 and 1 multiplied by 8 is 8 minus 7 multiplied by 5 which is 35. Now we have minus 27 upon 5 is equal to minus 27 upon 5 which is true. Thus a equal to 5 upon 2 and b equal to minus 5 upon 3 satisfies both the equations 1 and 2 hence the solution a equal to 5 upon 2 and b equal to minus 5 upon 3 is correct. So thus we find that the solution that we have found out is correct. So this completes the session hope you have understood the solution of this question.