Plugging equation into itself; works only for finite-solution equations?

Question

In this old post, substituting a modified - but completely equivalent - form of the equation ${\sqrt {x+1}+\sqrt{x+2}=1}$ back into itself yields its solution, as opposed to doing it for a linear equation in two variables, where the result is always a tautology. The question asks how this difference arises.

The only reason I can think of for why this happens is this; equations with finite solutions can sometimes be manipulated in specific ways to produce roots, which is not the case with infinite-solution equations. For example, the quadratic formula to solve the general quadratic is just a manipulation we did by completing the square, but it does the trick and gives us the answer. The same goes for cubic equations (Cardano's formula). But for cases like $3x+2y=1$, you can't solve it in that sense because it doesn't have a practically obtainable set of solutions. The conjugate multiplication the OP did in the linked question is just another manipulation as in the first case, which yielded the solutions, and predictably, nothing happens with the two-variable equation.

I'm asking separately if this view is correct because none of the answers there seem to use this, and also because;

1. It doesn't seem to work for all equations with finite solutions, like $x^2+y^2=0$. So what exactly is the flaw with this line of reasoning? And,
2. Only some manipulations seem to yield solutions (if I had substituted $\sqrt {x+1}=1-\sqrt{x+2}$, nothing would have happened). If there is a reason for it, what is it, and what exactly makes particular manipulations fruitful?

Edit: I wanted to ask about solving single equations, sorry about the confusion.

Answer 1

(1) No, it works also for infinite-solution equations in $\mathbb{R}$, e.g., $$ \lvert\sqrt{x+1}+\sqrt{x}-1\rvert+\lvert y+z\rvert=0. $$

(2) The real reason is that the definition of $\sqrt{\dots}$ (say in the nonnegative reals $\mathbb{R}^+$ so it is unambiguous) means that $\sqrt{x+2}+\sqrt{x+1}=1$ is not just one equation but three: \begin{align} a+b&=1\tag{1}\\ a^2&=x+2\tag{2}\\ b^2&=x+1\tag{3} \end{align} with the constraints $a,b\in\mathbb{R}^+$ and we are really manipulating with these subequations instead, namely: $((2)-(3))\div (1)$ to get $a-b=1$. So this is really not "manipulating one equation and substitute back into itself" when you really think about it. (Alternatively, what you have done is applying an automorphism to an equation, but that is definitely too advanced for a question tagged algebra-precalculus)

Answer 2

$\require{cancel}$

1. All $\space x^2+y^2=0\space$ solution are complex.
2. Substitution for $\space ax+by=c\space$ ends up as $\space x=x\space$ but it does work for simultaneous equations.
3. Double squaring works for $this$ problem by eliminating square roots.

\begin{align*} \sqrt {x+1}+\sqrt{x+2}&=1\\ \\ \big(\sqrt {x+1} +\sqrt{x+2}\space \big)^2 &=2 x + 2 \sqrt{x + 1} \sqrt{x + 2} + 3=1^2\\ \big( 2 \sqrt{x + 1} \sqrt{x + 2}\space \big)^2 &=(1-3-2x)^2\\ \bcancel{4 x^2} + 12 x + 8 &= \bcancel{4 x^2} + 8 x + 4\\ \\4x=-4\implies x&=-1 \end{align*}