I absolutely love your calm while doing algebra :) Such a smooth reasoning - nice! It is cool to watch two great maths teachers peacefully agree on each other's method. The second method reminded me of a virtuoso artist expert in abstract art painting. Your method was for me more like a pilot flying confidently at night with instruments only. I like both methods. Thank you both for the interesting video!
@@ThePhysicsMathsWizard Well, actually, the function on the left hand side is not well-defined at x=-1 because the limit doesn’t exist. If you approach -1 from the right, the limit is -inf but if you approach -1 from the left, the limit is +inf. Similarly, the function on the right hand side is not well-defined at x=-3. So, x=-3 and x=-1 cannot be part of the solution set.
@@unconcernedbeast9190 You can do it with a long division, or note (for the first) that x=x+1-1 so x/(x+1) = (x+1-1)/(x+1) which you can split into the two fractions (x+1)/(x+1) and -1/(x+1). Then you can use a similar method for the second fraction, with x-2=x+3-5
