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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.

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    $\begingroup$ I agree with Marc van Leeuwen whose answer is this. $\endgroup$
    – mathlove
    Sep 13 at 19:07
  • $\begingroup$ @mathlove: is it that isolating a variable and substituting it back always yields a tautology (as with the linear equation) , and it's because it wasn't a variable substitution, but just some manipulation that the OP did on the other equation that a solution was obtained ? $\endgroup$
    – harry
    Sep 13 at 19:30
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    $\begingroup$ (1) Solving $\sqrt{x+1}+\sqrt{x+2}=1$ gives $x=-1$, and substituting $x=-1$ into the equation gives $0=0$. (2) From $\sqrt{x+1}+\sqrt{x+2}=1$, one obtains $\sqrt{x+2}-\sqrt{x+1}=1$. Subtracting the latter from the former, one gets $\sqrt{x+1}+\sqrt{x+2}=\sqrt{x+2}-\sqrt{x+1}$, i.e. $2\sqrt{x+1}=0$. I would not call this "substituting" since "you combined the equation with a modified form of itself to obtain a new equation that is implied by the original one" as Marc van Leeuwen said. $\endgroup$
    – mathlove
    Sep 14 at 7:08
  • $\begingroup$ @mathlove: okay, it wasn't substition. Would my earlier comment be correct in that case? Because even if it's a modified form that we're combining the original with, it doesn't seem rigorous enough that 'modification' should determine whether we obtain a tautology or solutions. $\endgroup$
    – harry
    Sep 14 at 7:28
  • $\begingroup$ If you have an equation $f(x)=g(x)$, and get a modified form $h(x)=g(x)$ where $f(x)\not=h(x)$, and combine the first equation with the second one to have $f(x)=h(x)$, then $f(x)=h(x)$ is an equation, and so you'll finally obtain solutions. $\endgroup$
    – mathlove
    Sep 14 at 16:10
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  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*}

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  • $\begingroup$ I'm sorry, I had meant single equations, not systems. I'll fix it, could you change your answer accordingly? Also, isn't (0, 0) a solution, in (1)? $\endgroup$
    – harry
    Sep 13 at 3:45
  • $\begingroup$ @harry I meant for example that $x=x=0=0$. For substitution in single equations, the result is always $0=0$. I just mentioned systems of equations for substitution because that's the only place it works, to my knowledge. $\endgroup$
    – poetasis
    Sep 13 at 3:49
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(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)

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  • $\begingroup$ I'm sorry, I had meant single equations, not systems. I'll fix it, could you change your answer accordingly? $\endgroup$
    – harry
    Sep 13 at 3:41
  • $\begingroup$ @harry Over $\mathbb{R}$ we have tools like $\sum a_i^2=0$ iff $a_i=0$ for all $i$, or using absolute value for the same thing, so a system of equations is no different from a single equation. $\endgroup$ Sep 13 at 6:22
  • $\begingroup$ Got it. But how exactly does substituting back work in the example used in your (1)? Also, in (2), aren't the extraneous roots $a=-\sqrt{x+2} $ and $b=-\sqrt{x+1}$ created? (I understand you've set the square root operation as being a function from $R^+\to R^+$, but this problem still persists.) $\endgroup$
    – harry
    Sep 13 at 17:49

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