Once upon a time, when expressing (x/2)-(y/3)+5=0 in the form ax+by+c=0, I gave the value of b as 1/3. Not -1/3.
Another time, to solve a³+2a=3, I thought I could do a³=3-2a --> (a³/a)=3-(2a/a) --> a²=3-2 --> a=±1. Yes, while dividing both sides by a, I managed to overlook how I didn't divide 3...