Substituting an equation into itself, why such erratic behavior? Until now, I thought that substituting an equation into itself would $always$ yield $0=0$. What I mean by this is for example if I have $3x+4y=5$, If I substitute $y=\dfrac {5-3x}{4}$, I will eventually end up with $0=0$. However, consider the equation $\large{\sqrt {x+1}+\sqrt{x+2}=1}$ . If we multiply by the conjugate, we get $\dfrac {-1}{\sqrt{x+1}-\sqrt{x+2}}=1$, or $\large{\sqrt{x+2}-\sqrt{x+1}=1}$. Now we can set this equation equal to the original, so $\sqrt{x+2}-\sqrt{x+1}=\sqrt {x+1}+\sqrt{x+2}$ , and you get $0=2 \sqrt{x+1}$ which simplifies to $x=-1$ , which is actually a valid solution to the original! So how come I am not getting $0=0$ , but I am actaully getting useful information out of this? Is there something inherently wrong with this? Thanks.

You didn't actually substitute anything (namely a solution for $x$) into the original equation; if you would do that the $x$ would disappear. Instead you combined the equation with a modified form of itself to obtain a new equation that is implied by the original one; the new equation may or may not have retained all information from the original one. As it happens the new equation has a unique solution and it also solves the original equation; this shows the new equation implies the original one, and you did not in fact loose any information.
If you consider the operation of just adding a multiple of an equation to itself, you can see what can happen: in most cases you get something equivalent to the original equation, but if the multiple happened to be by a factor$~-1$ then you are left with $0=0$ and you have lost (in this case all) information contained in the equation.
