Let’s say y=x
Now multiply both sides by x:
Now subtract a y[sup]2[/sup] from both sides:
xy- y[sup]2[/sup]= x[sup]2[/sup]- y[sup]2[/sup]
Factoring the left side and apply the difference of squares to the right side, we get:
dividing the common terms
I had one of my professors stumped on this one for awhile. It’s a fun one to torment math geeks a little.
Yes, there is an explanation for this, but at least spoiler box your answers for awhile so everyone can enjoy.
Dividing the common term (x-y) must assume x != y (because if they are, the term is = 0, and it doesn’t matter what y or (x+y) are.) Since in your case you’ve posited that x does = y, you shouldn’t be dividing by (x-y)
Well, I’ve seen puzzles similar to this before, and about 90% of the time it involves the same kind of error at some stage in the cvalculation. (The next most common problem is taking square roots, where you assume a positive square root, but the problem really requires the other, negative, root.
I think that you segmented the phrase at the wrong place. I think that Elendil’s Heirthe was alluding to the incompatibility between “Fun” and “Math Equation”. But I disagree. Math can be fun, but the solution to this thread’s particular math puzzle is so well known that it was indeed no fun.