Investigate the possibility of solving x2-4x+2y2-yz = 1 for each of its variables in terms of the other two near the point (2,-1,3). Attempt: Ok so using the IFT I was able to determine that I can only slze for y,z. But in the question they ask me to solve for y and z. z was not problem, but y, they got a solution like this: y = (z - [z2+8(1-x2+4x)]1/2) / 4 as well they said x = 2 (+ or -) [5-2y2+yz]1/2 and the square root vanishes at (y,z) = (-1,3) so there are two values of “ for some nearby values of and and none for others. Question: How did they solve for y? (What bloody algebra trick did they use that I'm not seeing? and how did those conditions on x appear?