 Hello and welcome to the session. In this session, we discussed the following question which says, solve dy by dx is equal to 3x minus 3y plus 1 the whole upon minus 6x plus 6y plus 5. Now, if we have the equation of we form dy by dx equal to ax plus dy plus c upon capital Ax plus capital By plus capital C and if we have the case that A upon capital A is equal to B upon capital B. And each be equal to say r then we will take capital Ax plus capital By as z. This is the key idea that we use for this question. Let's now move on to the solution. The given equation that we have to solve is dy by dx is equal to 3x minus 3y plus 1 the whole upon minus 6x plus 6y plus 5. Let this be equation number one. Now this equation is of the form dy by dx is equal to ax plus by plus c upon capital Ax plus capital By plus capital C. Now let's see if A upon capital A is equal to B upon capital B. Now A upon capital A is equal to 3 upon minus 6 and this is equal to minus 1 upon 2. And now B upon capital B is equal to minus 3 upon 6 and this is equal to minus 1 upon 2. So we have A upon capital A is equal to B upon capital B and each is equal to minus 1 upon 2. That is we have r is equal to minus 1 upon 2. Now as A upon capital A is minus 1 upon 2 so we can write A as minus capital A upon 2 and B as minus capital B upon 2. So from equation one we have dy by dx is equal to minus 1 upon 2 into capital A which is minus 6 into x. That is in place of small a we write minus capital A upon 2. Now in place of small b we put minus capital B upon 2. So here we have minus 1 upon 2 into capital B which is 6 into y plus 1 and this whole upon the denominator which is minus 6x plus 6y plus 5. That is the denominator remains as it is. Now further we have dy by dx is equal to taking minus 1 upon 2 common from the numerator. We have minus 1 upon 2 into minus 6x plus 6y minus 2 the whole and this whole upon minus 6x plus 6y plus 5. Now putting minus 6x plus 6y as z. And differentiating both sides with respect to x we have minus 6 plus 6 into dy by dx is equal to dz by dx and from here we get the value of dy by dx as 1 upon 6 into dz upon dx plus 6 the whole. Let this equation be equation one. Now substituting the value of dy by dx in equation one we get 1 upon 6 into dz by dx plus 6 the whole is equal to minus 1 upon 2 into z minus 2 upon z plus 5 the whole and from here we have dz upon dx plus 6 is equal to minus 6 upon 2 into z minus 2 upon z plus 5 the whole. Now here 2 3 times is 6 so we now have dz upon dx is equal to minus 3 into z minus 2 upon z plus 5 the whole minus 6. Next we have dz upon dx is equal to minus 3z plus 6 minus 6z minus 30 and this whole upon z plus 5 that is we have taken the Lcm here. So further we get dz upon dx is equal to minus 9z minus 24 upon z plus 5. Now we can separate the variables so we have z plus 5 upon minus 9z minus 24 the whole dz is equal to dx. Or we can also say this as z plus 5 upon 9z plus 24 the whole dz is equal to minus dx we can write further z plus 5 upon 9 into z plus 24 upon 9 the whole dz is equal to minus dx. Now 3 3 times is 9 and 3 8 times is 24 so we further have 1 upon 9 into z plus 5 the whole upon z plus 8 upon 3 the whole dz is equal to minus dx. Now when we divide z plus 5 by z plus 8 upon 3 we get 1 plus 7 upon 3 into 1 upon z plus 8 upon 3. So this expression can now be written as 1 plus 7 upon 3 into 1 upon z plus 8 upon 3 the whole dz is equal to minus 9 dx. Next we integrate both the sides so integrating both sides we get integral 1 plus 7 upon 3 into 1 upon z plus 8 upon 3 the whole dz is equal to minus 9 integral dx. Further we get z plus 7 upon 3 into log modulus z plus 8 upon 3 is equal to minus 9x plus the constant of integration c. We have taken the value of z as minus 6x plus 6y so now substituting the value of z we get minus 6x plus 6y plus 7 upon 3 into log modulus minus 6x plus 6y plus 8 upon 3 is equal to minus 9x plus the constant of integration c. Further we get 9x minus 6x is 3x plus 6y plus 7 upon 3 into log modulus minus 6x plus 6y plus 8 upon 3 is equal to the constant of integration c. So this is the required solution. So this is our final answer. This completes the session. Hope you have understood the solution of this question.