TITLE: Substituting an equation into itself, why such erratic behavior?
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: An equation will simplify to the tautology $0=0$ only if the equation is true for all values of its variables. This happens in your first case, because you're basically going:

This equation will be true if and only if $y$ is equal to this, and no matter what $x$ is. So now, if $y$ is equal to that, will the equation be true?

And the answer is of course: yes, no matter what $x$ is.
The question is therefore: why does your second manipulation not result in something that is true for every value of $x$?
You start with
$$\sqrt {x+1}+\sqrt{x+2}=1$$
Step 1: You multiply by the conjugate to obtain
$$(\sqrt {x+1}-\sqrt{x+2})(\sqrt {x+1}+\sqrt{x+2})=\sqrt {x+1}-\sqrt{x+2}$$
$$(x+1)-(x+2)=\sqrt {x+1}-\sqrt{x+2}$$
$$-1=\sqrt {x+1}-\sqrt{x+2}$$
$$1=\sqrt{x+2}-\sqrt {x+1}$$
Observing that you can set this equal to the original expression, you do so (Step 2), and are surprised to obtain something that is not a tautology, ie, not true for all $x$.
But watch more closely what happens in step 1. Why did we end up with that $1$ on the left hand side? Merely because the $2$ and the $1$ in the original expression happened to produce a 1. This was a coincidence. Had the original equation been:
$$\sqrt {x+5}+\sqrt{x+2}=1$$
That wouldn't have happened, and you wouldn't have been able to set the result equal to the original expression. In other words, step 2 was not something that would have worked for all $x$. It only worked for this particular case, for $x$ solving this particular equation. Therefore everything deduced from step 2 are equations equivalent to the original, non-tautologous, equation. In particular, you even happened to get an equivalent equation that made one of the solutions obvious.
Another way of seeing this is that your solution involved finding that any radical expression satisfying your equation must be equal to its conjugate. You then exploited that fact, which is not at all true in general, to get the solution.

REPLY [2 votes]: Notice the first equation has two variables, whereas your second equation has one variable. The performed substitution was not an authentic swap of variables; the equation obtained is  essentially $$\dfrac {\sqrt{x+1}-\sqrt{x+2}}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac {{(\sqrt{x+1}-\sqrt{x+2})}^2}{1}$$ 
for which we get
$\dfrac {1}{1}=1$, thus obtain 0=0.
As a bonus, if in the process you divided something by an expression that evaluates to zero, a no solution situation may result. Allegedly Einstein has done that :)